$$\delta^{\frac{r}{\gamma}-1} \int\_{\frac{2\pi}{\sqrt{\delta}}}^{\infty} \frac{1}{x} \left(1 + \sqrt[r]{\delta} x\right)^{\gamma - r} dx = \delta^{\frac{r}{\gamma}-1} \int\_{1+2\pi}^{\infty} \frac{y^{\gamma - r}}{y - 1} dy = $$

$$= \delta^{\frac{r}{\gamma}-1} \int\_{1+2\pi}^{\infty} y^{\gamma - r - 1} \left(1 + \frac{1}{y - 1}\right) dy \le \left(1 + \frac{1}{2\pi}\right) \delta^{\frac{r}{\gamma} - 1} \int\_{1+2\pi}^{\infty} y^{\gamma - r - 1} dy = \mathsf{K} \delta^{\frac{r}{\gamma} - 1} \int\_{1+2\pi}^{\infty} y^{\gamma - r - 1} dy$$

from (13) and (12) we can write that

$$\int\_0^{\sqrt[\pi]{}} \left| \int\_0^{\infty} \rho(u) \cos \left( ut + \frac{\beta \pi}{2} \right) du \right| dt = O \left( \delta^{\frac{L}{\gamma} - 1} \right), \delta \to \infty. \tag{14}$$

One can analogously show that

$$\int\_{-\sqrt[\ell]{\delta}}^{0} \left| \int\_{0}^{\infty} \rho(u) \cos \left(ut + \frac{\beta \pi}{2}\right) du \right| dt = O\left(\delta^{\frac{L}{\gamma} - 1}\right), \delta \to \infty. \tag{15}$$

From the formulas (11), (14) and (15) we obtain

$$A(\varphi) = O\left(\delta^{\frac{r}{\gamma}-1}\right), \; \delta \to \infty.$$

Now we show the convergence of the integral

$$A(\mu) = \frac{1}{\pi} \int\_{-\infty}^{\infty} |\hat{\mu}\_{\beta}(t)| dt.$$

Integrating twice by parts and taking into account that *µ*(0) = *µ* 0 (0) = 0, lim*u*→∞ *µ*(*u*) = lim*u*→∞ *µ* 0 (*u*) = 0, we have

$$\int\_0^\infty \mu(u)\cos\left(ut + \frac{\beta\pi}{2}\right) du = -\frac{1}{t^2} \int\_0^\infty \mu''(u)\cos\left(ut + \frac{\beta\pi}{2}\right) du$$

and hence

$$\left| \int\_{0}^{\infty} \mu(u) \cos \left( ut + \frac{\beta \pi}{2} \right) du \right| \leq \frac{1}{t^2} \int\_{0}^{\infty} |\mu''(u)| du = \frac{1}{t^2} \left( \left( \int\_{0}^{\frac{1}{\sqrt{\delta}}} + \int\_{1}^{1} + \int\_{1}^{\infty} \right) |\mu''(u)| du \right). \tag{16}$$

Further we use the notations

$$W(\boldsymbol{u}) = \left(1 - e^{-\boldsymbol{u}^{\gamma}} - \boldsymbol{u}^{\gamma}\right) \boldsymbol{u}^{2 - \gamma}, \; \mathcal{W}(\boldsymbol{u}) = \left(1 - e^{-\boldsymbol{u}^{\gamma}} - \boldsymbol{u}^{\gamma}\right) \boldsymbol{u}^{1 - \gamma}.\tag{17}$$

Let us differentiate twice the functions *V*(*u*) and *W*(*u*):

$$V'(u) = \gamma u (e^{-u^{\gamma}} - 1) + (2 - \gamma) u^{1 - \gamma} \left(1 - e^{-\gamma} - u^{\gamma}\right),$$

$$\mathcal{W}'(u) = \gamma (e^{-u^{\gamma}} - 1) + (1 - \gamma) u^{-\gamma} \left(1 - e^{-u^{\gamma}} - u^{\gamma}\right),$$

$$V''(u) = \gamma \Big(e^{-u^{\gamma}} (1 - \gamma u^{\gamma}) - 1\Big) + (2 - \gamma) \Big((1 - \gamma) u^{-\gamma} \Big(1 - e^{-u^{\gamma}} - u^{\gamma}\Big) + \gamma (e^{-u^{\gamma}} - 1)\Big),$$

$$\begin{aligned} \mathcal{W}^{\prime\prime}(u) &= -\gamma^2 u^{\gamma - 1} e^{-u^{\gamma}} - \gamma(\gamma - 1) u^{-\gamma - 1} \left( 1 - e^{-u^{\gamma}} - u^{\gamma} \right) + \gamma(\gamma - 1) u^{-1} \left( e^{-u^{\gamma}} - 1 \right) . \\\\ \text{By virtue of the fact. that for } u &\in \left[ 0, \frac{1}{\sqrt{\gamma}} \right] \end{aligned}$$

By virtue of the fact, that for *u* ∈ 0, 1 √*γ δ*

$$
\mu''(u) = (\gamma - r - 1)\delta^{\frac{r+2}{\gamma}-1}V''(u) + (2 + r - \gamma)\delta^{\frac{r+1}{\gamma}-1}W''(u),\tag{18}
$$

we obtain

$$\int\_0^{\frac{1}{\sqrt{\rho}}} |\mu^{\prime\prime}(u)| du \le (r+1-\gamma)\delta^{\frac{r+2}{\gamma}-1} \int\_0^{\frac{1}{\sqrt{\rho}}} |V^{\prime\prime}(u)| du + (2+r-\gamma)\delta^{\frac{r+1}{\gamma}-1} \int\_0^{\frac{1}{\sqrt{\rho}}} |\mathcal{W}^{\prime\prime}(u)| du.$$

Taking into account, that for *u* ∈ h 0, 1 √*γ δ* i *V* 00(*u*) ≤ 0, *W*00(*u*) ≤ 0, and also the inequalities

$$1 - e^{-\mathfrak{u}^{\gamma}} \le \mathfrak{u}^{\gamma}, \ e^{-\mathfrak{u}^{\gamma}} + \mathfrak{u}^{\gamma} - 1 \le \frac{\mathfrak{u}^{2\gamma}}{2},\tag{19}$$

we have

$$\begin{split} \int\_{\sqrt{\delta}}^{\frac{1}{\sqrt{\delta}}} & \int\_{0}^{\frac{r+1}{\sqrt{\delta}}-1} |u''(u)| \, du \le (r+1-\gamma)\delta^{\frac{r+1}{\sqrt{\delta}}-1} \left( V'(0) - V'\left(\frac{1}{\sqrt[\gamma]{\delta}}\right) \right) + \\ & + (2+r-\gamma)\delta^{\frac{r+1}{\sqrt{\delta}}-1} \left( W'(0) - W'\left(\frac{1}{\sqrt[\gamma]{\delta}}\right) \right) = \\ & = (r+1-\gamma)\delta^{\frac{r+2}{\sqrt{\delta}}-1} \left( \frac{\gamma}{\sqrt[\gamma]{\delta}} \left( 1 - e^{-\frac{1}{\delta}} \right) + \frac{2-\gamma}{\left( \sqrt[\gamma]{\delta} \right)^{1-\gamma}} \left( e^{-\frac{1}{\delta}} + \frac{1}{\delta} - 1 \right) \right) + \\ & + (2+r-\gamma)\delta^{\frac{r+1}{\sqrt{\delta}}-1} \left( \gamma \{ 1 - e^{-\frac{1}{\delta}} \right) + \frac{1-\gamma}{\left( \sqrt[\gamma]{\delta} \right)^{-\gamma}} \left( e^{-\frac{1}{\delta}} + \frac{1}{\delta} - 1 \right) \right) \leq K\_6 \delta^{\frac{r+1}{\sqrt{\delta}}-2}. \tag{20} \\ & \text{which is shown } \gamma. \end{split}$$

Noting, that for *u* ≥ 1 √*γ δ*

$$\mu^{\prime\prime}(u) = r(r+1)\left(1 - e^{-u^{\gamma}} - u^{\gamma}\right)u^{-r-2} - 2\gamma u^{\gamma-r-2}\left(e^{-u^{\gamma}} - 1\right) + \cdots$$

$$+ \gamma\left((\gamma - 1)u^{\gamma-2}(e^{-u^{\gamma}} - 1) - \gamma u^{2\gamma-2}e^{-u^{\gamma}}\right)u^{-r} \dots$$

we can write

$$\begin{aligned} \label{1.5} \int & \left| \mu''(u) \right| du \leq r(r+1) \int\_{\frac{1}{\sqrt[r]{\delta}}}^{1} (e^{-u^{\gamma}} + u^{\gamma} - 1) u^{-r-2} du + \\ \varinjlim\_{\frac{1}{\sqrt[r]{\delta}}}^{1} \left( 1 - e^{-u^{\gamma}} \right) u^{\gamma - r-2} du + \varinjlim\_{\frac{1}{\sqrt[r]{\delta}}}^{1} \left| (\gamma - 1) u^{\gamma - 2} (e^{-u^{\gamma}} - 1) - \gamma u^{2\gamma - 2} e^{-u^{\gamma}} \right| u^{-r} du. \end{aligned}$$

The inequality (19) in combination with

$$\left| (\gamma - 1)u^{\gamma - 2}(e^{-u^{\gamma}} - 1) - \gamma u^{2\gamma - 2}e^{-u^{\gamma}} \right| \le (2\gamma - 1)u^{2\gamma - 2}, \ u \in [0, \infty), \tag{21}$$

yields

$$\begin{aligned} \label{eq:1} \int\_{\frac{1}{\sqrt{7}}}^{1} |\mu''(u)| du &\leq \left(\frac{r(r+1)}{2} + 2\gamma r + \gamma(2\gamma - 1)\right) \int\_{\frac{1}{\sqrt{7}}}^{1} u^{2\gamma - r - 2} du \leq\\ \begin{cases} \frac{r(r+1)}{2} + 2\gamma r + \gamma(2\gamma - 1) \end{cases} &\stackrel{\text{~}}{\sqrt[r]{\delta}} \int\_{\frac{1}{\sqrt{7}}}^{1} u^{2\gamma - r - 1} du = \begin{cases} \begin{array}{c} K\_{7} \delta^{\frac{1}{7}} + K\_{8} \delta^{\frac{r+1}{7} - 2}, & r \neq 2\gamma, \\\ K\_{8} \delta^{\frac{1}{7}} \ln \delta, & r = 2\gamma, \end{cases} \end{aligned} \end{aligned} \tag{22}$$

In the case *u* ∈ [1, ∞) we obtain

$$\int\_{1}^{\infty} |\mu''(u)| du \le r(r+1) \int\_{1}^{\infty} (e^{-u^{\gamma}} + u^{\gamma} - 1) u^{-r-2} du + \int\_{1}^{\infty} (1 - e^{-u^{\gamma}}) u^{\gamma - r - 2} du + \int\_{1}^{\infty} \left| (\gamma - 1) u^{\gamma - 2} (e^{-u^{\gamma}} - 1) - \gamma u^{2\gamma - 2} e^{-u^{\gamma}} \right| u^{-r} du.$$

$$\text{If } u \ge 0 \text{ and then } \gamma \text{ is a bi-invariant. Then } u(0) = 1.624, \quad u \ge 1.624.$$

Let 0 < *γ* < 1, then using the inequalities (19) and (21), we obtain

$$\int\_{1}^{\infty} |\mu''(u)| du \le \left(\frac{r(r+1)}{2} + 2\gamma r + \gamma(2\gamma - 1)\right) \int\_{1}^{\infty} u^{2\gamma - r - 2} du = K\_{10}.\tag{23}$$

Let further 1 ≤ *γ* ≤ 2. By virtue of the inequalities

$$(e^{-\imath\ell^{\gamma}} + \imath^{\gamma} - 1)\imath^{-2} \le 1,\\ \left(1 - e^{-\imath\ell^{\gamma}}\right)\imath^{\gamma-2} \le 1,$$

$$|(\gamma - 1)\imath^{\gamma-2}(e^{-\imath\ell^{\gamma}} - 1) - \gamma\imath^{2\gamma-2}e^{-\imath\ell^{\gamma}}| \le 2\gamma - 1,$$

we have

$$\int\_{1}^{\infty} |\mu''(u)| du \le \left(\frac{r(r+1)}{2} + 2\gamma r + \gamma(2\gamma - 1)\right) \int\_{1}^{\infty} u^{-r} du = \mathcal{K}\_{11}.\tag{24}$$

Therefore, combining the relations (23), (24), we obtain

$$\int\_{1}^{\infty} |\mu^{\prime\prime}(u)| du = O(1), \; \delta \to \infty. \tag{25}$$

In view of (16), taking into account (20), (22) and (25), we obtain

$$\int\_{|t| \ge \sqrt[r]{\delta}} \left| \int\_0^\infty \mu(u) \cos \left( ut + \frac{\beta \pi}{2} \right) du \right| dt = \begin{cases} O(1), & \gamma < r < 2\gamma, \\ O(\ln \delta), & r = 2\gamma, \\ O(\delta^{\frac{r}{\gamma} - 2}), & r > 2\gamma. \end{cases} \tag{26}$$

Let us further consider

$$\int\_0^{\sqrt{\delta}} \left| \int\_0^{\infty} \mu(u) \cos \left( ut + \frac{\beta \pi}{2} \right) du \right| dt \le \int\_0^{\sqrt{\delta}} \left| \int\_0^{\frac{1}{\sqrt{\delta}}} \mu(u) \cos \left( ut + \frac{\beta \pi}{2} \right) du \right| dt + \int\_{\frac{1}{\sqrt{\delta}}}^{\sqrt{\delta}} \left| \int\_0^{\infty} \mu(u) \cos \left( ut + \frac{\beta \pi}{2} \right) du \right| dt. \tag{27}$$

$$+ \int\_0^{\frac{\sqrt{\delta}}{\sqrt{\delta}}} \left| \int\_{\frac{1}{\sqrt{\delta}}}^{\infty} \mu(u) \cos \left( ut + \frac{\beta \pi}{2} \right) du \right| dt. \tag{27}$$

By the inequality (19), one can easily verify that the following relations hold

$$\int\_0^{\sqrt{\delta}} \left| \int\_0^{\frac{1}{\sqrt{\delta}}} \mu(u) \cos \left(ut + \frac{\beta \pi}{2}\right) du \right| dt \le \int\_0^{\sqrt{\delta}} \int\_0^{\frac{1}{\sqrt{\delta}}} |\mu(u)| du dt = K\_{12} \delta^{\frac{L}{\gamma} - 2}. \tag{28}$$

The function |*µ*(*u*)| is monotonically decreasing on the interval [*u*0, ∞], *u*0 ≥ 1, nonnegative and tends to zero as *u* → ∞. Then, by the equality (4.16) from ([1], p. 69), we obtain

$$\int\_{0}^{\sqrt{\delta}} \left| \int\_{0}^{\infty} \mu(u) \cos \left(ut + \frac{\beta \pi}{2}\right) du \right| dt = \int\_{0}^{\sqrt{\delta}} \left| \int\_{\frac{1}{\sqrt{\delta}}}^{\infty} |\mu(u)| \cos \left(ut + \frac{\beta \pi}{2}\right) du \right| dt \le \frac{1}{\sqrt{\delta}} \int\_{\frac{1}{\sqrt{\delta}}}^{\infty} |\mu(u)| \cos \left(ut + \frac{\beta \pi}{2}\right) du \Big| dt \le \frac{1}{\sqrt{\delta}} \int\_{\frac{1}{\sqrt{\delta}}}^{\infty} |\mu(u)| \frac{\sqrt{\delta}}{2} du \Big| \le \frac{1}{\sqrt{\delta}} \int\_{\frac{1}{\sqrt{\delta}}}^{\infty} |\mu(u)| \frac{\sqrt{\delta}}{2} du dt \le \frac{1}{\sqrt{\delta}} \int\_{\frac{1}{\sqrt{\delta}}}^{\infty} |\mu(u)| \frac{\sqrt{\delta}}{2} du dt \le \frac{1}{\sqrt{\delta}} \mathcal{I} \tag{29}$$

$$\le \int\_{0}^{\sqrt{\delta}} \int\_{\frac{1}{\sqrt{\delta}}}^{\infty} |\mu(u)| du dt + \int\_{0}^{\sqrt{\delta}} \int\_{u\_{0}}^{\infty} |\mu(u)| du dt \le \frac{1}{\sqrt{\delta}} \int\_{\frac{1}{\sqrt{\delta}}}^{\infty} |\mu(u)| du dt. \tag{20}$$

Let *n* ∈ N is such that 1 √*γ δ* + 2*π*(*n*−1) *t* ≤ *u*0 ≤ 1 √*γ δ* + 2*πn t* , then

$$\int\_{0}^{\sqrt[\Im]{}u\_{0}+\frac{2\pi}{t}} \int\_{0}^{\sqrt[\Im]{}} |\mu(u)| du dt \le \int\_{0}^{\sqrt{\Im}+\frac{2\pi(n+1)}{t}} \int\_{\frac{1}{\sqrt[\Im]{}}}^{\frac{1}{\sqrt[\Im]{}}+\frac{2\pi(n+1)}{t}} |\mu(u)| du dt. \tag{30}$$

We transform the latter integral using a change of variable and integration by parts (assume that *δ* > (2*π*(*n* + 1) + 1) *γ* )

$$\begin{split} \frac{\sqrt{\delta}}{\delta} \frac{\frac{1}{\sqrt{\delta}} + \frac{2\pi(n+1)}{t}}{\int\_{0}^{\infty} \frac{1}{\sqrt{\delta}} |\mu(u)| du dt} &= 2\pi(n+1) \int\_{\frac{2\pi(n+1)}{\sqrt{\delta}}}^{\infty} \int\_{\frac{1}{\sqrt{\delta}}}^{\infty} |\mu(u)| du \frac{dx}{\sqrt{x}} = \\ = 2\pi(n+1) \left( -\left(\frac{1}{\mathfrak{x}} \int\_{\frac{1}{\sqrt{\delta}}}^{\frac{1}{\sqrt{\delta}} + x} |\mu(u)| du \right) \Big|\_{\frac{2\pi(n+1)}{\sqrt{\delta}}}^{\infty} + \int\_{\frac{2\pi(n+1)}{\sqrt{\delta}}}^{\infty} \frac{1}{\mathfrak{x}} \Big| \mu\left(\frac{1}{\sqrt[\mathfrak{x}]{\delta}} + x\right) \Big| du \right) = \\ = 2\pi(n+1) \left( -\lim\_{x \to \infty} \frac{1}{x} \int\_{0}^{\frac{1}{\sqrt{\delta}} + x} \left( e^{-u^{\gamma}} + u^{\gamma} - 1 \right) u^{-r} du + \\ + \frac{\sqrt[\mathfrak{x}]{\delta}}{2\pi(n+1)} \int\_{\frac{1}{\sqrt{\delta}}}^{\frac{1+2\pi(n+1)}{\sqrt{\delta}}} \left( e^{-u^{\gamma}} + u^{\gamma} - 1 \right) u^{-r} du + \\ \overset{\frac{1}{\sqrt[\mathfrak{x}]{\delta}}}{\frac{1}{\sqrt[\mathfrak{x}]{\delta}}} \int\_{\frac{1}{\sqrt[\mathfrak{x}]{\delta}}}^{\frac{1}{\sqrt[\mathfrak{x}]{\delta}}} \left( e^{-u^{\gamma}} + u^{\gamma} - 1 \right) u^{-r} du + \end{split}$$

$$+\int\_{-\frac{2\pi(n+1)}{\sqrt{\delta}}}^{1-\frac{1}{\sqrt{\delta}}} \frac{1}{\mathfrak{x}} \left(e^{-\left(\frac{1}{\sqrt{\delta}}+\mathbf{x}\right)^{\gamma}} + \left(\frac{1}{\sqrt[\gamma]{\delta}}+\mathbf{x}\right)^{\gamma} - 1\right) \left(\frac{1}{\sqrt[\gamma]{\delta}}+\mathbf{x}\right)^{-r} d\mathbf{x} +$$

$$+\int\_{-\frac{1}{\sqrt{\delta}}}^{\infty} \frac{1}{\mathfrak{x}} \left(e^{-\left(\frac{1}{\sqrt{\delta}}+\mathbf{x}\right)^{\gamma}} + \left(\frac{1}{\sqrt[\gamma]{\delta}}+\mathbf{x}\right)^{\gamma} - 1\right) \left(\frac{1}{\sqrt[\gamma]{\delta}}+\mathbf{x}\right)^{-r} d\mathbf{x}\right).\tag{31}$$

Obviously,

$$\lim\_{\chi \to \infty} \frac{1}{\chi} \int\_{\frac{1}{\sqrt{\delta}}}^{\frac{1}{\sqrt{\delta}} + \chi} \left( e^{-\mu^\gamma} + \mu^\gamma - 1 \right) \psi(\sqrt[\gamma]{\delta} u) du = 0. \tag{32}$$

Since the second inequality from (19) holds, then

$$\frac{\sqrt[n]{\delta}}{2\pi(n+1)} \int\_{\frac{1}{\sqrt[n]{\delta}}}^{\frac{1+2\pi(n+1)}{\sqrt[n]{\delta}}} \left(e^{-u^{\gamma}} + u^{\gamma} - 1\right) u^{-r} du \le 0$$

$$\leq \frac{\sqrt[n]{\delta}}{4\pi(n+1)} \int\_{\frac{1}{\sqrt[n]{\delta}}}^{\frac{1+2\pi(n+1)}{\sqrt[n]{\delta}}} u^{2\gamma-r} du \leq \frac{\delta^{\frac{r+1}{\gamma}}}{4\pi(n+1)} \int\_{\frac{1}{\sqrt[n]{\delta}}}^{\frac{1+2\pi(n+1)}{\sqrt[n]{\delta}}} u^{2\gamma} du \leq \mathbb{K}\_{13} \delta^{\frac{r}{\gamma}-2}.\tag{33}$$

Using the second inequality from (19), we have

1− 1 *γ*√ Z *δ* 2*π*(*n*+1) *γ*√ *δ* 1 *x e* − 1 *γ*√ *δ* +*x γ* + 1 √*γ δ* + *x γ* − 1 1 √*γ δ* + *x* −*r dx* ≤ ≤ 1− 1 *γ*√ Z *δ* 2*π*(*n*+1) *γ*√ *δ* 1 *x* 1 √*γ δ* + *x* 2*γ*−*r dx* = *δ r γ* −2 1− 1 *γ*√ Z *δ* 2*π*(*n*+1) *γ*√ *δ* 1 *x* 1 + √*γ δx* 2*γ*−*r dx* = = *δ r γ* −2 √*γ* Z *δ* 1+2*π*(*n*+1) *y* 2*γ*−*r y* − 1 *dy* = *δ r γ* −2 √*γ* Z *δ* 1+2*π*(*n*+1) *y* 2*γ*−1*ψ*(*y*) 1 + 1 *y* − 1 *dy* ≤ ≤ 1 + 1 2*π*(*n* + 1) *δ r γ* −2 √*γ* Z *δ* 1+2*π*(*n*+1) *y* 2*γ*−1−*r dy* = ( *K*14 + *K*15*δ r γ* −2 , *r* 6= 2*γ*, *K*16 ln *δ*, *r* = 2*γ*. (34)

Considering the inequality

$$e^{-\mathfrak{u}^{\gamma}} + \mathfrak{u}^{\gamma} - 1 \le \mathfrak{u}^{\gamma}\_{\cdot \cdot}$$

we have

$$\int\_{1-\frac{1}{\sqrt[4]{\delta}}}^{\infty} \frac{1}{\pi} \left( e^{-\left(\frac{1}{\sqrt[4]{\delta}} + \mathbf{x}\right)^{\gamma}} + \left(\frac{1}{\sqrt[4]{\delta}} + \mathbf{x}\right)^{\gamma} - 1\right) \left(\frac{1}{\sqrt[4]{\delta}} + \mathbf{x}\right)^{-r} d\mathbf{x} \le \frac{1}{\delta}$$

$$\leq \int\_{1-\frac{1}{\sqrt[\delta]{\delta}}}^{\infty} \frac{1}{x} \left(\frac{1}{\sqrt[\delta]{\delta}} + x\right)^{\gamma - r} dx = \delta^{\frac{r}{\gamma} - 1} \int\_{1}^{\infty} \frac{1}{x} \left(1 + \sqrt[\delta]{\delta} x\right)^{\gamma - r} dx = \delta^{\frac{r}{\gamma} - 1} \int\_{1 - \frac{1}{\sqrt[\delta]{\delta}}}^{\infty} \frac{1}{y} \left(1 + \sqrt[\delta]{\delta} x\right)^{\gamma - r} dx = \delta^{\frac{r}{\gamma} - 1} \int\_{\sqrt[\delta]{\delta}}^{\infty} y^{\gamma - 1 - r} \left(1 + \frac{1}{y - 1}\right) dy \leq \delta^{\frac{r}{\gamma} - 1} \int\_{\sqrt[\delta]{\delta}}^{\infty} y^{\gamma - 1 - r} dy = K\_{18}. \tag{35}$$

$$\leq \left(1 + \frac{1}{\sqrt[\delta]{\delta} - 1}\right) \delta^{\frac{r}{\gamma} - 1} \int\_{\sqrt[\delta]{\delta}}^{\infty} y^{\gamma - 1 - r} dy \leq K\_{17} \delta^{\frac{r}{\gamma} - 1} \int\_{\sqrt[\delta]{\delta}}^{\infty} y^{\gamma - 1 - r} dy = K\_{18}. \tag{36}$$

From (27), taking into account (28) and (29)–(35), we can write the estimation

$$\int\_0^{\sqrt[\beta]{\delta}} \left| \int\_0^{\infty} \mu(u) \cos \left(ut + \frac{\beta \pi}{2}\right) du \right| dt = \begin{cases} O(1), & \gamma < r < 2\gamma, \\ O\left(\ln \delta\right), & r = 2\gamma, \\ O\left(\delta^{\frac{r}{\gamma} - 2}\right), & r > 2\gamma. \end{cases} \tag{36}$$

Similarly, we can show that

$$\int\_{-\sqrt[\gamma]{\delta}}^{0} \left| \int\_{0}^{\infty} \mu(u) \cos \left(ut + \frac{\beta \pi}{2}\right) du \right| dt = \begin{cases} O(1), & \gamma < r < 2\gamma, \\ O\left(\ln \delta\right), & r = 2\gamma, \\ O\left(\delta^{\frac{r}{\gamma} - 2}\right), & r > 2\gamma. \end{cases} \tag{37}$$

Combining formulas (26), (36) and (37), we obtain

$$A(\mu) = \begin{cases} \mathcal{O}(1), & \gamma < r < 2\gamma, \\ \mathcal{O}(\ln \delta), & r = 2\gamma, \\ \mathcal{O}(\delta^{\frac{L}{\gamma}-2}), & r > 2\gamma. \end{cases} \tag{38}$$

Similarly to [27] we can show that the following equality holds

$$f(\mathbf{x}) - P\_{\mathcal{V}}(\delta, f, \mathbf{x}) = \frac{1}{(\sqrt[r]{\delta})^r} \int\_{-\infty}^{\infty} f\_{\beta}^{\psi} \left( \mathbf{x} + \frac{t}{\sqrt[r]{\delta}} \right) \widehat{\mathbf{r}}\_{\beta}(t) dt \,\omega$$

where

$$\mathfrak{t}\_{\beta}(t) = \mathfrak{t}(t,\beta) = \frac{1}{\pi} \int\_{0}^{\infty} \mathfrak{r}(u) \cos \left(ut + \frac{\beta \pi}{2} \right) du.$$

Thence

$$\begin{split} \mathcal{E}\left(W^{r}\_{\beta,\infty};P\_{\boldsymbol{\gamma}}(\boldsymbol{\delta})\right)\_{\mathbb{C}} &= \sup\_{\boldsymbol{f}\in\mathcal{W}^{r}\_{\beta,\infty}} \left\| \frac{1}{(\sqrt[r]{\delta})^{r}} \int\_{-\infty}^{\infty} \boldsymbol{f}\_{\beta}^{r} \left(\boldsymbol{x} + \frac{\boldsymbol{t}}{\sqrt[r]{\delta}}\right) \hat{\boldsymbol{\gamma}}\_{\beta}(\boldsymbol{t}) d\boldsymbol{t} \right\|\_{\mathbb{C}} = \\ &= \sup\_{\boldsymbol{f}\in\mathcal{W}^{r}\_{\beta,\infty}} \left\| \frac{1}{(\sqrt[r]{\delta})^{r}} \int\_{-\infty}^{\infty} \boldsymbol{f}\_{\beta}^{r} \left(\boldsymbol{x} + \frac{\boldsymbol{t}}{\sqrt[r]{\delta}}\right) (\hat{\boldsymbol{\rho}}\_{\beta}(\boldsymbol{t}) + \hat{\boldsymbol{\mu}}\_{\beta}(\boldsymbol{t})) d\boldsymbol{t} \right\|\_{\mathbb{C}} \leq \\ &\leq \sup\_{\boldsymbol{f}\in\mathcal{W}^{r}\_{\beta,\infty}} \left\| \frac{1}{(\sqrt[r]{\delta})^{r}} \int\_{-\infty}^{\infty} \boldsymbol{f}\_{\beta}^{r} \left(\boldsymbol{x} + \frac{\boldsymbol{t}}{\sqrt[r]{\delta}}\right) \boldsymbol{\theta}\_{\beta}(\boldsymbol{t}) d\boldsymbol{t} \right\|\_{\mathbb{C}} + \frac{1}{(\sqrt[r]{\delta})^{r}} \int\_{-\infty}^{\infty} |\boldsymbol{\mu}\_{\beta}(\boldsymbol{t})| d\boldsymbol{t}. \end{split}$$

Therefore,

$$\mathcal{E}\left(\mathcal{W}^{r}\_{\hat{\rho},\infty^{r}};\mathcal{P}\_{\gamma}(\delta)\right)\_{\mathbb{C}} = \sup\_{f\in\mathcal{W}^{r}\_{\hat{\rho},\infty}} \left\| \frac{1}{(\sqrt[r]{\delta})^{r}} \int\_{-\infty}^{\infty} f^{r}\_{\hat{\rho}}\left(\mathbf{x} + \frac{t}{\sqrt[r]{\delta}}\right) \phi\_{\hat{\rho}}(t)dt \right\|\_{\mathbb{C}} + O\left(\frac{1}{(\sqrt[r]{\delta})^{r}} A(\mu)\right). \tag{39}$$

Similarly to the work [28] , we can show that the Fourier series of the function *fϕ*(*x*) = R∞ −∞ *f r β x* + *t* √*γ δ ϕ*ˆ *β*(*t*)*dt* has the form:

$$S\left[f\_{\varphi}\right] = \sum\_{k=1}^{\infty} \frac{k^{\gamma}}{(\sqrt[r]{\delta})^{\gamma-r}} (a\_k \cos kx + b\_k \sin kx)\_k$$

where *ak* , *bk* are the Fourier coefficients of the function *f* . Therefore

$$\int\_{-\infty}^{\infty} f\_{\beta}^{\gamma} \left( x + \frac{t}{\sqrt[\gamma]{\delta}} \right) \phi\_{\beta}(t) dt = \frac{1}{(\sqrt[\gamma]{\delta})^{\gamma - r}} f\_{0}^{\gamma}(x), \tag{40}$$

where *f γ* 0 (*x*) is (*r*, *β*)—derivative in the Weyl-Nagy sense for *r* = *γ*, *β* = 0. Substituting (40) into (39), we obtain

$$\mathcal{E}\left(\mathcal{W}\_{\mathfrak{f},\infty}^{r};P\_{\gamma}(\delta)\right)\_{\mathbb{C}}=\frac{1}{\delta}\sup\_{f\in\mathcal{W}\_{\mathfrak{f},\infty}^{r}}\left\Vert f\_{0}^{\gamma}(\cdot)\right\Vert\_{\mathbb{C}}+O\left(\frac{1}{(\sqrt[r]{\delta})^{r}}A(\mu)\right),\ \delta\to\infty.\tag{41}$$

Substituting (38) into (41), we obtain the equation (5). The theorem is proved.

#### **3. Conclusions**

One of the extremal problems of approximation theory, namely the problem of studying the asymptotic properties of linear summation methods of Fourier series, has been considered in the paper. Among the linear summation methods, on the one hand, there are methods that are defined by infinite numerical matrices, and on the other hand, methods that are defined by the set of functions of the natural argument that depend on the real parameter *δ*. This work is devoted to the study of the approximation properties of the methods of the last type, namely, generalized Poisson integrals. The Kolmogorov–Nikol'skii problem takes a special place among the extremal problems of the approximation theory. We have considered the problem of asymptotic equalities finding for the value of the exact upper limits of deviations of generalized Abel-Poisson integrals from functions of the Weyl-Nagy classes in the uniform metric. In particular, the asymptotic equality (5) for arbitrary *r* > *γ*, 0 < *γ* ≤ 2, has been written in the paper, providing the solution of the corresponding Kolmogorov–Nikol'skii problem. The importance of this type of problems in the theory of decision making, in signal transmission, in the study of mathematical models and in the coding and reproduction of images has been noted. Regarding further research in this direction, we note that similar problems can be considered in the broader classes of functions, such as Stepanets classes and classes of non-periodic locally summable functions.

**Author Contributions:** Conceptualization, I.K. and Y.K.; methodology, I.K. and Y.K.; formal analysis, I.K. and Y.K.; writing—original draft preparation, I.K. and Y.K.; writing—review and editing, I.K. and Y.K. All authors have read and agreed to the published version of the manuscript.

**Funding:** This work was supported by Grant of the Ministry of Education and Science of Ukraine, 0120U102630.

**Institutional Review Board Statement:** Not applicable.

**Informed Consent Statement:** Not applicable.

**Data Availability Statement:** Not applicable.

**Conflicts of Interest:** The authors declare no conflict of interest.

#### **References**

## *Article* **Approximate Optimal Control for a Parabolic System with Perturbations in the Coefficients on the Half-Axis**

**Olena A. Kapustian 1,2 , Oleksiy V. Kapustyan 3 , Anton Ryzhov 3,\* and Valentyn Sobchuk 3**

**Abstract:** In this paper, we use the averaging method to find an approximate solution in the optimal control problem of a parabolic system with non-linearity of the form *f*(*t*/*ε*, *y*) on an infinite time interval.

**Keywords:** parabolic system; optimal control; averaging method; approximate solution

#### **1. Introduction**

Many results in the theory of asymptotic approximations have been obtained from 1930 onwards. Indeed, there were a lot of results on integral manifolds, equations with retarded argument, quasi- or almost-periodic equations etc. Earlier work on this theory has been presented in the famous book [1].

Averaging is a valuable method to understand the long-term evolution of dynamical systems characterized by slow dynamics and fast periodic or quasi-periodic dynamics. In [2], a transparent proof of the validity of averaging in the periodic case is presented. Different proofs for both the periodic and the general case are provided by [3,4]. In the last paper, moreover, the relation between averaging and the multiple time-scales method is established.

The averaging method for constructing approximate solutions in the theory of ODEs is presented in [5,6]. In [7], the asymptotic analysis of nonlinear dynamical systems is developed.

The work [8] is devoted to using an asymptotic method for studying the Cauchy problem for a 1D Euler–Poisson system, which represents a physically relevant hydrodynamic model but also a challenging case for a bipolar semiconductor device by considering two different pressure functions. In [9], the averaging results for ordinary differential equations perturbed by a small parameter are proved. Here, authors assume only that the right-hand sides of the equations are bounded by some locally Lebesgue integrable functions with the property that their indefinite integrals satisfy a Lipschitz-type condition.

In [10], the authors prove that averaging can be applied to the extremal flow of optimal control problems with two fast variables, that is considerably more complex because of resonances.

The averaging method is one of the most effective tools for constructing approximate solutions, including optimal control problems for ODEs [11] and PDEs [12], where the autors consider the optimal control problem in coefficients in the so-called class of Hadmissible solutions.

The Krasnoselski–Krein theorem and its various modifications [13–15] play an essential role in all such considerations, since it guarantees the limit transition in perturbed problem with fast-oscillating coefficients of the form *a t ε* as *ε* → 0.

**Citation:** Kapustian, O.A.; Kapustyan, O.V.; Ryzhov, A.; Sobchuk, V. Approximate Optimal Control for a Parabolic System with Perturbations in the Coefficients on the Half-Axis. *Axioms* **2022**, *11*, 175. https://doi.org/10.3390/ axioms11040175

Academic Editor: Clemente Cesarano

Received: 17 February 2022 Accepted: 12 April 2022 Published: 14 April 2022

**Publisher's Note:** MDPI stays neutral with regard to jurisdictional claims in published maps and institutional affiliations.

**Copyright:** © 2022 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (https:// creativecommons.org/licenses/by/ 4.0/).

The typical averaging problem may be defined as follows: one considers an unperturbed problem in which the slow variables remain fixed. Upon perturbation, a slow drift appears in these variables which one would like to approximate independently of the fast variables.

In the present paper we use this approach to nonlinear parabolic system with fastoscillating (w.r.t. time variable) coefficients *f t ε* , *y* on an infinite time interval. We prove that the optimal control of the problem with averaging coefficients can be considered to be "approximately" optimal for the initial perturbed system.

#### **2. Statement of the Problem**

Let Ω ⊂ R*d* be a bounded domain. In cylinders *Q* = (0, +∞) × Ω we consider an initial boundary-value problem for a parabolic system [16,17]

$$\begin{cases} \frac{\partial y}{\partial t} = A \Delta y + f\left(\frac{t}{\varepsilon'}y\right) + g(y) \cdot u(t, x), (t, x) \in Q\_{\prime} \\ y|\_{\partial \Omega} = 0, \\ y|\_{t=0} = y\_0(x). \end{cases} \tag{1}$$

Here *ε* > 0 is a small parameter, *A* is a real *N* × *N* matrix, *f* is a given vectorvalued mapping, *g* is a given matrix-valued mapping, *y* = (*y*1, . . . , *yN*) is an unknown state function, *u* = (*u*1, . . . , *uM*) is an unknown control function, which are determined by requirements

$$
\mu \in \mathcal{U} \subseteq (L^2(\mathbb{Q}))^M. \tag{2}
$$

$$J(y, u) = \int\_{\mathcal{Q}} e^{-\gamma \cdot t} \cdot q(\mathbf{x}, y(t, \mathbf{x})) dt d\mathbf{x} + \int\_{\mathcal{Q}} \sum\_{i=1}^{M} \mathbf{a}\_i \cdot \mathbf{u}\_i^2(t, \mathbf{x}) dt d\mathbf{x} \to \inf,\tag{3}$$

where *γ*, *α*1, . . . , *αM* are positive constants.

Under the natural assumptions on *A*, *f* , *g*, *U*, *q* we prove, that the optimal control problem (1)–(3) has a solution {*y*¯ *ε* , *u*¯ *ε*}, i.e., for every *u* ∈ *U* and for any solution *y ε* of (1) with control *u* we have

$$J(\mathfrak{y}^{\varepsilon}, \mathfrak{u}^{\varepsilon}) \le J(\mathfrak{y}^{\varepsilon}, \mathfrak{u})$$

In what follows we consider the problem of finding an approximate solution of (1)–(3) by transition to averaged coefficients. For this purpose we assume that uniformly w.r.t. *y* ∈ R*N* there exists

$$f(y) := \lim\_{T \to \infty} \frac{1}{T} \int\_0^T f(s, y) ds. \tag{4}$$

We consider the following optimal control problem

$$\begin{cases} \frac{\partial y}{\partial t} = A \Delta y + \tilde{f}(y) + g(y) \cdot u(t, x), (t, x) \in Q, \\ y|\_{\partial \Omega} = 0, \\ y|\_{t=0} = y\_0(x), \end{cases} \tag{5}$$

$$
\mu \in \mathcal{U} \subseteq (L^2(\mathbb{Q}))^M,\tag{6}
$$

$$J(y, u) = \int\_{Q} e^{-\gamma \cdot t} \cdot q(\mathbf{x}, y(t, \mathbf{x})) dt d\mathbf{x} + \int\_{Q} \sum\_{i=1}^{M} \mathbf{a}\_{i} \cdot \mathbf{u}\_{i}^{2}(t, \mathbf{x}) dt d\mathbf{x} \to \inf. \tag{7}$$

It should be noted that the transition to the averaging parameters can essentially simplify the problem. In particular, if ¯ *f* does not depend on *y* then in some cases exact solution of (1)–(3) can be found [18,19]. Another approaches for finding exact solutions of optimal control problems and approximate procedures can be found in [20,21].

Assume that {*y*¯, *u*¯} is a solution of (5)–(7). The main goal of the paper is to prove the limit equality

$$J(\vec{y}^{\varepsilon}, \vec{u}^{\varepsilon}) - J(\vec{y}, \vec{u}) \to 0, \varepsilon \to 0. \tag{8}$$

As a consequence of (8) we will prove that the control *u*¯ is approximately optimal for the problem (1)–(3) in the following sense:

$$J(\mathfrak{y}^{\varepsilon}, \mathfrak{a}^{\varepsilon}) - J(\mathfrak{y}^{\varepsilon}, \mathfrak{a}) \to 0, \varepsilon \to 0,$$

where *y ε* is a solution of (1) with control *u*¯.

#### **3. Assumptions, Notations and Basic Results**

We assume the following conditions hold.

**Assumption 1.** 1 2 (*A* + *A* ∗ ) ≥ *v* · *I, where v* > 0 *and I is a unit matrix;*

**Assumption 2.** *f* : R+ × R*N* 7→ R*N is continuous and bounded:*

∃*C*1 > 0 ∀*t* ≥ 0 ∀*y* ∈ R *N* k *f*(*t*, *y*)kR*N* ≤ *C*1;

**Assumption 3.** *g* : R*N* 7→ R*N*×*M is continuous and bounded:*

$$\exists \mathsf{C}\_2 > 0 \quad \forall y \in \mathbb{R}^N \quad \|g(y)\|\_{\mathbb{R}^{N \times M}} \le \mathsf{C}\_2;$$

**Assumption 4.** *U* ⊆ *L* 2 (*Q*) *M is closed and convex,* 0 ∈ *U;*

**Assumption 5.** *q* : Ω × R*N* 7→ R *is a Carathéodory function,* ∃*K*1, *K*2 ∈ *L* 1 (Ω)*,* ∃*C*3 > 0 *such that* ∀*x* ∈ Ω*,* ∀*ξ* ∈ R*N*

$$|q(\mathfrak{x}, \mathfrak{f})| \le \mathsf{C}\_{\mathfrak{Z}} ||\mathfrak{f}||\_{\mathbb{R}^N}^2 + \mathsf{K}\_2(\mathfrak{x}), \quad q(\mathfrak{x}, \mathfrak{f}) \ge \mathsf{K}\_1(\mathfrak{x}).$$

*Here,* k*ξ*kR*N denotes the Euclidean norm of ξ* ∈ R*N.*

For *u* ∈ *U* and *y*0 ∈ *L* 2 (Ω) *N* we understand solution of (1) in weak (or generalized) sense on every finite time interval, i.e., *y* is a solution of (1) if

$$y \in L\_{loc}^2\left(0, +\infty, (H\_0^1(\Omega))^N\right) \bigcap L\_{loc}^\infty\left(0, +\infty, (L^2(\Omega))^N\right),$$

such that ∀*T* > 0, ∀*ϕ* ∈ (*H*1 0 (Ω))*N*, ∀*η* ∈ *C* ∞ 0 (0, *T*) the following equality holds:

$$-\int\_0^T (y,\boldsymbol{\varrho})\_H \cdot \boldsymbol{\eta}^t dt + \int\_0^T (A \nabla y, \nabla \boldsymbol{\varrho})\_H \boldsymbol{\eta} dt = \int\_0^T \left( f\left(\frac{t}{\varepsilon}, y\right), \boldsymbol{\varrho} \right)\_H \boldsymbol{\eta} dt + \int\_0^T (\boldsymbol{g}(\boldsymbol{y}) \cdot \boldsymbol{u}, \boldsymbol{\varrho})\_H \boldsymbol{\eta} dt.\tag{9}$$

Here and after we denote by k · k*H* and (·, ·)*H* the classical norm and scalar product in *H* := (*L* 2 (Ω))*N*, by k · k*V* and (·, ·)*V* the classical norm and scalar product in *V* := (*H*1 0 (Ω))*N*, by k · k*U* the norm in *L* 2 (*Q*))*M*, and by *V* ∗ the dual space to *V*.

Due to the Assumptions 1–3, every solution of (1) satisfies

$$\frac{\partial y}{\partial t} \in L\_{loc}^2(0, +\infty, V^\*).$$

It means that ∀*T* > 0 every solution of (1) is an absolutely continuous function from [0, *T*] to *H*, and equality (9) is equivalent to the following one [16]:

$$\frac{d}{dt}(y,\boldsymbol{\varrho})\_H + (A\nabla y, \nabla \boldsymbol{\varrho})\_H = \left(f\left(\frac{t}{\varepsilon'}y\right), \boldsymbol{\varrho}\right)\_H + (\boldsymbol{g}(y)\cdot\boldsymbol{u}, \boldsymbol{\varrho})\_H \tag{10}$$

for almost all (a.a.) *t* > 0.

It is known [16,17] that, under Assumptions 1–3, for every *y*0 ∈ *H*, *u* ∈ *U* there exists at least one solution of (1), and for a.a. *t* > 0

$$\frac{1}{2}\frac{d}{dt}\|y(t)\|\_{H}^{2} + v \cdot \|y(t)\|\_{V}^{2} \le \mathbb{C}\_{1} \|y(t)\|\_{H} + \mathbb{C}\_{2} \|y(t)\|\_{H} \cdot \|u(t)\|\_{\left(L^{2}(\Omega)\right)^{M}}.\tag{11}$$

**Remark 1.** *Uniqueness of solution of (1) is not guaranteed. This can be done under some additional assumptions, e.g., [16]* ∀*s* ≥ 0*,* ∀*y* ∈ R*N,* ∀*ω* ∈ R*N*

$$\left(f\_y'(s,y)\omega,\omega\right)\_{\mathbb{R}^N} \ge -\mathsf{C}\_4 \cdot \|\omega\|\_{\mathbb{R}^N}$$

In the sequel, we denote by F*ε* (or F¯) a set of all pairs {*y*, *u*}, where *y* is a solution of (1) (or (5)) with control *u*.

The following Lemma gives us a result about the solvability of the optimal control problem (1)–(3) and it also provides some useful inequalities.

**Lemma 1.** *Under the Assumptions 1–5 for every ε* > 0 *the problem (1)–(3) has a solution* {*y*¯ *ε* , *u*¯ *ε*}*, that is,*

$$J(\mathfrak{y}^{\varepsilon}, \mathfrak{u}^{\varepsilon}) \le J(y, u) \quad \forall \{y, u\} \in \mathcal{F}^{\varepsilon}.$$

**Proof of Lemma 1.** Fix *ε* > 0 and suppress index *ε* throughout the proof. The idea of the proof is to derive a priori estimates for the minimizing sequence. Obtained estimates allow us to pass to the limit in problem (1)–(3).

From (11), Poincare inequality k*y*k 2 *V* ≥ *λ*k*y*k 2 *H* , *y* ∈ *V*, and Young inequality we derive that for some *δ* > 0, *C*5 > 0 (not depending on *ε*) for every {*y*, *u*} ∈ F*ε* for a.a. *t* > 0

$$\frac{d}{dt} \|y(t)\|\_{H}^{2} + \delta \|y(t)\|\_{H}^{2} \le \mathbb{C}\_{5} \left(1 + \|u(t)\|\_{\left(L^{2}(\Omega)\right)^{M}}^{2}\right).$$

Therefore using Gronwall inequality we get for all *t* > 0

$$\|\|y(t)\|\|\_{H}^{2} \le e^{-\delta \cdot t} \left\{ \|y\_{0}(t)\|\|\_{H}^{2} + \mathbb{C}\_{5} \int\_{0}^{t} \left(1 + \|u(s)\|\|\_{\left(L^{2}(\Omega)\right)^{M}}^{2}\right) e^{\delta \cdot s} ds \right\},\tag{12}$$

$$\|y(t)\|\_{H}^{2} \le e^{-\delta \cdot t} \|y\_{0}\|\_{H}^{2} + \frac{\mathsf{C}\_{5}}{\delta} + \mathsf{C}\_{5} \cdot \|u\|\_{H}^{2}.\tag{13}$$

From the inequality (13) and the first inequality from the Assumption 5 we have that for some *C*6 > 0

$$J(y, u) \le \mathcal{C}\_6 \left( 1 + \|y\_0\|\_H^2 + \|u\|\_{\mathcal{U}}^2 \right). \tag{14}$$

Now let {*yn*, *un*} be a minimizing sequence, that is,

$$\lim\_{n \to \infty} J(y\_{n\prime} u\_n) = \inf\_{\{y, u\} \in \mathcal{F}^{\varepsilon}} J(y, u) =: \overline{f}^{\varepsilon}. \tag{15}$$

Note that due to the Assumption 5 ∀{*y*, *u*} ∈ F*ε*

$$J(y, u) \ge \frac{\|K\_1\|\_{L^1}}{\gamma} \Rightarrow \mathcal{J}^\varepsilon \ge \frac{\|K\_1\|\_{L^1}}{\gamma} > -\infty.$$

From (15) for sufficiently large *n*

$$J(y\_n, u\_n) \le \overline{f}^\varepsilon + 1.\tag{16}$$

On the other hand, for *α* := min 1≤*i*≤*M αi* > 0

$$J(y\_n, u\_n) \ge \frac{\|K\_1\|\_{L^1}}{\gamma} + \mathfrak{a} \cdot \|u\_n\|\_{\mathcal{U}}^2. \tag{17}$$

Inequalities (16) and (17) imply that {*un*} is bounded in (*L* 2 (*Q*))*M*, so for subsequence

$$
\mu\_n \to \mu \text{ weakly in } (L^2(\mathbb{Q}))^M. \tag{18}
$$

Due to convexity of *U* we have inclusion *u* ∈ *U*. From (11) over (0, *T*) and using (13) we we obtain from (5) that {*yn*} is bounded in

$$L^2(0,T;V)\bigcap L^\infty(0,T;H)\_V$$

n *∂yn ∂t* o is bounded in *L* 2 (0, *T*; *V* ∗ ). Using Compactness Lemma [22] we conclude that up to subsequence ∀*T* > 0

$$y\_n \to y \text{ weakly in } L^2(0, T; V)\_\prime$$

$$y\_n \to y \text{ in } L^2(0, T; H)\_\prime \tag{19}$$

$$\forall t \ge 0 \; y\_n(t) \to y(t) \text{ weakly in } H\_\prime$$

$$y\_n(t, x) \to y(t, x) \text{ a.a. in } Q.$$

From (19) and Lebesgue's Dominated Convergence Theorem we can pass to the limit in the equality (9) applied to {*yn*, *un*}, and obtain that {*y*, *u*} ∈ F*ε* . Due to pointwise convergence

$$e^{-\gamma \cdot t} \cdot q(\mathfrak{x}, y\_n(t, \mathfrak{x})) \to e^{-\gamma \cdot t} q(\mathfrak{x}, y(t, \mathfrak{x})) \text{ a.a. in } Q\_{\lambda}$$

Fatou's lemma and weak convergence (18) we have

$$\overline{f}^{\varepsilon} = \lim\_{n \to \infty} f(y\_n, u\_n) \ge \varliminf\_{Q} \int\_Q e^{-\gamma \cdot t} q(\mathbf{x}, y\_n(t, \mathbf{x})) dt d\mathbf{x} + \varliminf\_{Q} \int\_0^M \mathbf{a}\_i(\mu\_i^{\mathbf{n}}(t, \mathbf{x}))^2 dt d\mathbf{x} \ge f(y, \mathbf{u}).$$

Therefore {*y*, *u*} is a solution of (1)–(3).

#### **4. Main Results**

We assume that ∀*η* > 0 ∃*δ* > 0 ∀*t* ≥ 0, ∀*y*, *z* ∈ R*N*

$$\|\|y - z\|\|\_{\mathbb{R}^N} < \delta \Rightarrow \|\|f(t, y) - f(t, z)\|\|\_{\mathbb{R}^N} < \eta. \tag{20}$$

Assumption (20) implies that the averaged function ¯ *f* : R*N* 7→ R*N* from (4) is a continuous function and the Assumption 2 holds. It means that under conditions (4), (20) the optimal control problem (5)–(7) has a solution {*y*¯, *u*¯}.

The main result of the paper is the following

**Theorem 1.** *Suppose that the Assumptions 1–5 and (4), (20) hold and, moreover, the problem (5) has a unique solution for every u* ∈ *U. Let* {*y*¯ *ε* , *u*¯ *ε*} *be a solution of (1)–(3). Then*

$$J(\mathfrak{g}^{\varepsilon}, \mathfrak{a}^{\varepsilon}) \to J(\mathfrak{g}, \mathfrak{a}), \varepsilon \to 0,\tag{21}$$

*and up to subsequence*

$$
\overline{y}^{\varepsilon} \to \overline{y} \text{ in } L^2\_{loc}(0, +\infty; H),
$$

$$
\overline{u}^{\varepsilon} \to \overline{u} \text{ in } (L^2(\mathbb{Q}))^M. \tag{22}
$$

*where* {*y*¯, *u*¯} *is a solution of (5)–(7).*

**Proof.** Let *εn* → 0, {*y*¯ *n* , *u*¯ *n*} be a solution of (1)–(3) for *ε* = *εn*. Due to the optimality of {*y*¯ *n* , *u*¯ *n*} we have

$$J(\mathfrak{J}^n, \mathfrak{u}^n) \le J(y\_n, 0),$$

where *yn* is a solution of (1) with *ε* = *εn* and *u* ≡ 0. Then from (14)

$$\frac{1}{\gamma} \|K\_1\|\_{L^1} + \mathfrak{a} \|\mathfrak{u}^n\|\_{\mathcal{U}}^2 \le \mathsf{C}\_6 \cdot (1 + \|y\_0\|\_{H}^2).$$

Repeating arguments used in the prof of Lemma 1 conclude that on subsequence for some *y*ˆ, *u*ˆ:

$$
\mathfrak{u}^n \to \mathfrak{A} \text{ weakly in } (\operatorname{L}^2(\mathbb{Q}))^M, n \to \infty,
$$

$$
\mathfrak{y}^n \to \mathfrak{y} \text{ in the sense of (19), } n \to \infty,\tag{23}
$$

Let us prove that {*y*ˆ, *u*ˆ} ∈ F¯, i.e., *y*ˆ is a solution of the averaged problem (5) with control *u*ˆ. For this purpose it is sufficient to make a limit transition in the equality

$$(\bar{y}^n, \boldsymbol{\phi})\_H - (y\_0, \boldsymbol{\phi})\_H + \int\_0^T (A \nabla \bar{y}^n, \nabla \boldsymbol{\phi})\_H = \int\_0^T \left( f\left(\frac{t}{\varepsilon\_n}, \bar{y}^n\right), \boldsymbol{\phi} \right)\_H + \int\_0^T (g(\bar{y}^n)\bar{u}^n, \boldsymbol{\phi})\_{H^s} \tag{24}$$

for arbitrary *ϕ* ∈ *V* and *T* > 0.

Limit transition in the left part of (24) is a direct consequence of (23). From the Dominated Convergence Theorem we see that

$$g(\mathfrak{y}^n) \to g(\mathfrak{y}) \text{ in } L^2(0, T; H), n \to \infty.$$

Then (23) implies convergence in the last term of (24). Let us prove that ∀*T* > 0, ∀*ϕ* ∈ *V*

$$\int\_{Q\_T} \sum\_{i=1}^N f\_i \left( \frac{t}{\varepsilon\_n}, \mathfrak{J}^n(t, \mathbf{x}) \right) \varphi\_i(\mathbf{x}) dt d\mathbf{x} \to \int\_{Q\_T} \sum\_{i=1}^N f\_i(\mathfrak{J}(t, \mathbf{x})) \varphi\_i(\mathbf{x}) dt d\mathbf{x}, n \to \infty,\tag{25}$$

where *QT* = (0, *T*) × Ω. Due to the Dominated Convergence Theorem ∀0 < *a* < *b*, ∀*ψ* ∈ *H*

$$\int\_{a}^{b} \int\_{\Omega} \sum\_{i=1}^{N} \left( f\_{i} \left( \frac{t}{\varepsilon\_{n}}, \psi(\mathbf{x}) \right) - \bar{f}\_{i}(\psi(\mathbf{x})) \right) \varphi\_{i}(\mathbf{x}) d\mathbf{x} dt \to 0, n \to \infty. \tag{26}$$

Due to Egorov's theorem ∀*δ* > 0 ∃*Qδ* 1 ⊂ *QT* such that *µ*(*Qδ* 1 ) < *δ* and

$$
\bar{y}^n \to \hat{y} \text{ uniformly on } Q\_T \backslash Q\_1^\delta \text{ as } n \to \infty. \tag{27}
$$

Here *µ* is Lebesgue's measure on R2 . On the other hand there exists a sequence of step functions

$$y^m(t, \mathbf{x}) = \sum\_{k=1}^m y\_k^m(\mathbf{x}) \cdot \chi\_{A\_k^m}(t) \prime \{ y\_k^m \} \subset H\_\prime$$

{*A m k* = (*a m k* , *b m k* )} is a covering of (0, *T*) such that

$$y^m \to \mathfrak{Y} \text{ in } L^2(0, T; H) \text{ and a.e. in } Q\_T.$$

$$\text{Moreover } \forall \delta > 0 \, \exists Q\_2^\delta \subset Q\_T \text{ such that } \mu(Q\_2^\delta) < \delta \text{ and }$$

$$\dots \qquad \dots \qquad \dots \qquad \dots \qquad \dots \qquad \dots$$

$$y^m \to \mathcal{Y} \text{ uniformly on } Q\_T \backslash Q\_2^\delta \text{ as } m \to \infty.$$

Let us denote

$$I\_1^{(n)} := \int\_{\mathcal{Q}\_T} \sum\_{i=1}^N \left( f\_i\left(\frac{t}{\varepsilon\_n}, \mathfrak{z}^n(t, \mathbf{x})\right) - f\_i\left(\frac{t}{\varepsilon\_n}, \mathfrak{z}(t, \mathbf{x})\right) \right) \varrho\_i(\mathbf{x}) dt d\mathbf{x} \,\mu$$

$$I\_2^{(n)} := \int\_{\mathcal{Q}\_T} \sum\_{i=1}^N \left( f\_i\left(\frac{t}{\varepsilon\_n}, \mathfrak{z}(t, \mathbf{x})\right) - f(\mathfrak{z}(t, \mathbf{x})) \right) \varrho\_i(\mathbf{x}) dt d\mathbf{x} \,\mu$$

Then due to (27)

$$I\_1^{(n)} \le \int\_{\mathbb{Q}\Gamma \backslash \mathbb{Q}\_1^{\mathbb{Q}}} \left\| f\left(\frac{t}{\varepsilon\_n}, \mathfrak{z}^n(t, \mathbf{x})\right) - f\left(\frac{t}{\varepsilon\_n}, \mathfrak{z}(t, \mathbf{x})\right) \right\|\_{\mathbb{R}^{\mathbb{R}^N}} \cdot \|\varrho(\mathbf{x})\|\_{\mathbb{R}^N} dt d\mathbf{x} + 2\mathbb{C}\_1 \cdot \|\varrho\|\_{H}^{\frac{1}{2}} \cdot \delta^{\frac{1}{2}}.\tag{28}$$

Due to (20) for a given *δ* > 0 ∃*λ* ∀*n* ≥ 1, ∀*t* ≥ 0

$$\|\|y - z\|\|\_{\mathbb{R}^N} < \lambda \Rightarrow \left\| \left| f\left(\frac{t}{\varepsilon\_n}, y\right) - f\left(\frac{t}{\varepsilon\_n}, z\right) \right| \right\| \leq \delta^{\frac{1}{2}}.$$

Therefore, choosing *n*1 such that ∀*n* ≥ *n*1

$$\sup\_{(t,\varkappa)\in Q\_T\backslash Q\_1^{\delta}} \|\bar{y}^n(t,\varkappa) - \hat{y}(t,\varkappa)\|\_{\mathbb{R}^N} < \lambda$$

we get from (28) that ∀*n* ≥ *n*1

$$\|I\_1^{(n)} \le \delta^{\frac{1}{2}} \cdot \mu^{\frac{1}{2}}(Q\_T) \cdot \|\varphi\|\_{H}^{\frac{1}{2}} + 2\mathcal{C}\_1 \cdot \|\|\varphi\|\_{H}^{\frac{1}{2}} \cdot \delta^{\frac{1}{2}} \le \mathcal{C}\_7(T)\delta^{\frac{1}{2}}.\tag{29}$$

On the other hand, for every step function *y m*(*t*, *x*) we have due to (26): ∀*m* ≥ 1

$$\begin{split} &\int\_{Q\_T} \sum\_{i=1}^N \left( f\_i \left( \frac{t}{\varepsilon\_n}, y^m(t, \mathbf{x}) \right) - \bar{f}\_i(y^m(t, \mathbf{x})) \right) \varphi\_i(\mathbf{x}) dt d\mathbf{x} \\ &= \sum\_{k=1}^m \int\_{A\_k^m} \int \sum\_{i=1}^N \left( f\_i \left( \frac{t}{\varepsilon\_n}, y\_k^m(\mathbf{x}) \right) - \bar{f}\_i(y\_k^m(\mathbf{x})) \right) \varphi\_i(\mathbf{x}) dt d\mathbf{x} \to 0, n \to \infty. \end{split} \tag{30}$$

So ∀*m* ≥ 1, ∃*n*2 = *n*2(*m*), ∀*n* ≥ *n*2

$$\left| \int\_{\mathcal{Q}\_T} \sum\_{i=1}^N \left( f\_i \left( \frac{t}{\varepsilon\_n}, y^m(t, \mathbf{x}) \right) - \bar{f}\_i(y^m(t, \mathbf{x})) \right) \rho\_i(\mathbf{x}) dt d\mathbf{x} \right| < \delta. \tag{31}$$

Furthermore, ∃*m*0, ∀*m* ≥ *m*0, ∀*n* ≥ 1

Z *QT*\*Qδ* 2 *f t εn* , *y*ˆ(*t*, *x*) − *f t εn* , *y m*(*t*, *x*) R*N* · k*ϕ*(*x*)kR*N dtdx* ≤ *δ* 1 2 · *µ* 1 2 (*QT*)· k*ϕ*k 1 2 *H* , (32) Z 1

$$\int\_{\mathbb{Q}\Gamma\backslash\mathbb{Q}\_{2}^{\mathbb{S}}} \lVert \bar{f}(\hat{y}(t, \mathbf{x})) - \bar{f}(y^{m}(t, \mathbf{x})) \rVert\_{\mathbb{R}^{N}} \cdot \lVert \boldsymbol{\varrho}(\mathbf{x}) \rVert\_{\mathbb{R}^{N}} dt d\mathbf{x} \leq \delta^{\frac{1}{2}} \cdot \mu^{\frac{1}{2}}(Q\_{T}) \cdot \lVert \boldsymbol{\varrho} \rVert\_{H}^{\frac{1}{2}}.\tag{33}$$

Combining (30)–(33), we obtain ∀*m* ≥ *m*0, ∀*n* ≥ *n*2(*m*)

$$\|I\_2^{(n)} \le 2 \cdot \delta^{\frac{1}{2}} \cdot \mu^{\frac{1}{2}}(Q\_T) \|\varphi\|\_{H}^{\frac{1}{2}} + \delta \le C\_8(T) \cdot \delta^{\frac{1}{2}}.\tag{34}$$

Inequalities (29), (34) imply (25). So we can pass to the limit in (24) and obtain that {*y*ˆ, *u*ˆ} ∈ F¯. Now let us prove that {*y*ˆ, *u*ˆ} is an optimal process in (5)–(7).

Fatou's lemma implies

$$
\underline{\lim} f(\mathfrak{y}^n, \mathfrak{a}^n) \ge f(\mathfrak{y}, \mathfrak{a}).
$$

On the other hand, for every *u* ∈ *U* and any *yn*–solution of (1) with control *u* and *ε* = *εn* we get

*J*(*y*¯ *n* , *u*¯ *n* ) ≤ *J*(*y*¯*n*, *u*).

Using the same arguments as in proof of the Lemma 1 for {*yn*} we derive that

*yn* → *y* in the sense of (19),

where *y* is a unique solution of (5) with control *u*.

Let us prove that

$$\int\_{Q} e^{-\gamma \cdot t} q(\mathbf{x}, y\_n(t, \mathbf{x})) dt d\mathbf{x} \to \int\_{Q} e^{-\gamma \cdot t} q(\mathbf{x}, y(t, \mathbf{x})) dt d\mathbf{x} \tag{35}$$

Indeed due to the Assumption 5 and (13) we have

$$\left|e^{-\gamma \cdot t} q(\mathbf{x}, y\_n(t, \mathbf{x}))\right| \le \mathsf{C}\_3 e^{-\gamma \cdot t} \left\|y\_n(t, \mathbf{x})\right\|\_{\mathbb{R}^N}^2 + e^{-\gamma \cdot t} \cdot \mathsf{K}\_2(\mathbf{x}).\tag{36}$$

As *yn* → *y* in *L* 2 (0, *T*; *H*) and a.e. in *Q*, we deduce from Lebesgue's Dominated Convergence theorem:

$$\forall T > 0 \quad \int\_{\tilde{Q}\_T} e^{-\gamma \cdot t} q(\mathbf{x}, y\_n(t, \mathbf{x})) dt d\mathbf{x} \to \int\_{\tilde{Q}\_T} e^{-\gamma \cdot t} q(\mathbf{x}, y(t, \mathbf{x})) dt d\mathbf{x}, n \to \infty. \tag{37}$$

On the other hand, from (12) and (36)

$$\begin{split} &\int\_{T}^{+\infty} \int\_{\Omega} e^{-\gamma \cdot t} |q(\mathbf{x}, y\_{\mathrm{n}}(t, \mathbf{x}))| dt d\mathbf{x} \leq \int\_{T}^{+\infty} e^{-\gamma \cdot t} \left[ \mathbb{C}\_{3} \cdot ||y\_{\mathrm{n}}(t)||\_{H}^{2} + ||K\_{2}||\_{L^{1}} \right] dt \\ &\leq \int\_{T}^{+\infty} e^{-\gamma \cdot t} \left[ \mathbb{C}\_{3} e^{-\delta \cdot t} \cdot ||y\_{0}(t)||\_{H}^{2} + \frac{\mathbb{C}\_{3} \cdot \mathbb{C}\_{5}}{\delta} + \mathbb{C}\_{3} \cdot \mathbb{C}\_{5} \cdot ||u||\_{H}^{2} + ||K\_{2}||\_{L^{1}} \right] dt \\ &\leq \mathbb{C}\_{9} \cdot e^{-\gamma \cdot T} , \end{split} \tag{38}$$

where *C*9 does not depend on *T* and *n*. The last inequality together with with (37) leads to (35).

From (35) we conclude the following inequality: ∀{*y*, *u*} ∈ F¯

$$I(\mathcal{Y}, \mathcal{U}) \le \underline{\lim} I(\mathcal{Y}', \mathfrak{a}'') \le \underline{\lim} I(y\_{n\prime}\mathfrak{a}) = I(\mathcal{Y}, \mathfrak{a}).\tag{39}$$

This means that {*y*ˆ, *u*ˆ} is a solution of (5)–(7).

Now we substitute *u* = *u*ˆ in previous arguments. Then *y* = *y*ˆ due to uniqueness. So from (39), we obtain

$$J(\mathfrak{y}, \mathfrak{a}) \le \underline{\lim} J(\mathfrak{y}^n, \mathfrak{a}^n) \le J(\mathfrak{y}, \mathfrak{a}).$$

These inequalities mean that up to subsequence

$$J(\mathfrak{J}^{\mathfrak{n}}, \mathfrak{a}^{\mathfrak{n}}) \to J(\mathfrak{J}, \mathfrak{a}), \mathfrak{n} \to \infty. \tag{40}$$

Since *J*(*y*ˆ, *u*ˆ) = inf{*y*,*u*}∈F¯ *J*(*y*, *u*), then convergence in (40) holds for the whole sequence. Therefore (21) is proved.

Moreover, up to subsequence *y*¯ *n* tends to *y*ˆ in *L* 2 *loc*(0, +∞; *H*). So, repeating arguments (37) and (38) for *y*¯ *n* , and using boundness of {*u*¯ *n*}, we have

$$\int\_{Q} e^{-\gamma \cdot t} q(\mathbf{x}, \mathfrak{z}^{\eta}(t, \mathfrak{x})) dt d\mathfrak{x} \to \int\_{Q} e^{-\gamma \cdot t} q(\mathbf{x}, \mathfrak{z}(t, \mathfrak{x})) dt d\mathfrak{x}.$$

Then from (40) and weak convergence we deduce (22)

**Corollary 1.** *An optimal control u*¯ ∈ *U of the averaged problem (5)–(7) can serve as an "approximate" optimal control in the initial problem (1), that is:*

$$J(\mathfrak{J}^{\varepsilon}, \mathfrak{u}^{\varepsilon}) - J(\mathfrak{y}^{\varepsilon}, \mathfrak{u}) \to 0, \varepsilon \to 0,\tag{41}$$

*where yε is a solution of (1) with control u* = *u.*¯

Indeed, for *y εn* , *εn* → 0, we can repeat arguments of the proof of the Theorem, and due to the uniqueness of the solution of (5) for *u* = *u*¯ we have up to subsequence

> *y εn* → *y*¯ in the sense of (19)

Then (35) holds and taking into account strong convergence (22), we obtain (41).

#### **5. Conclusions and Future Research**

We sought to obtain a theoretical result that demonstrates the effectiveness of the averaging method of finding an approximate solution of the optimal control problem for a non-linear parabolic system with fast-oscillating coefficients with respect to a time variable. We proved that the optimal control of the problem with averaging coefficients can be considered as an "approximately" optimal for the initial perturbed system. To demonstrate effectiveness of the method we plan to continue research focusing on the practical applications and simulation results using in particular genetic algorithms.

**Author Contributions:** Conceptualization, O.A.K., O.V.K., A.R. and V.S.; methodology, O.A.K., O.V.K., A.R. and V.S.; formal analysis, O.A.K., O.V.K., A.R. and V.S.; investigation, O.A.K., O.V.K., A.R. and V.S.; writing—original draft preparation, O.A.K., O.V.K., A.R. and V.S.; writing—review and editing, O.A.K., O.V.K., A.R. and V.S. All authors have read and agreed to the published version of the manuscript.

**Funding:** This research received no external funding.

**Institutional Review Board Statement:** Not applicable.

**Informed Consent Statement:** Not applicable.

**Data Availability Statement:** Not applicable.

**Conflicts of Interest:** The authors declare no conflict of interest.

#### **References**

## *Article* **Some Korovkin-Type Approximation Theorems Associated with a Certain Deferred Weighted Statistical Riemann-Integrable Sequence of Functions**

**Hari Mohan Srivastava 1,2,3,4,\* , Bidu Bhusan Jena 5 and Susanta Kumar Paikray 5**

**Abstract:** Here, in this article, we introduce and systematically investigate the ideas of deferred weighted statistical Riemann integrability and statistical deferred weighted Riemann summability for sequences of functions. We begin by proving an inclusion theorem that establishes a relation between these two potentially useful concepts. We also state and prove two Korovkin-type approximation theorems involving algebraic test functions by using our proposed concepts and methodologies. Furthermore, in order to demonstrate the usefulness of our findings, we consider an illustrative example involving a sequence of positive linear operators in conjunction with the familiar Bernstein polynomials. Finally, in the concluding section, we propose some directions for future research on this topic, which are based upon the core concept of statistical Lebesgue-measurable sequences of functions.

**Keywords:** Riemann and Lebesgue integrals; statistical Riemann and Lebesgue integral; deferred weighted Riemann summability; Banach space; Bernstein polynomials; positive linear operators; Korovkin-type approximation theorems; Lebesgue-measurable sequences of functions

**MSC:** 40A05; 40G15; 33C45; 41A36

#### **1. Introduction and Motivation**

The relatively more familiar theory of ordinary convergence is one of the most important topics of study of sequence spaces. It has indeed gradually progressed to a very high level of development. Two prominent researchers, Fast [1] and Steinhaus [2], independently created a new idea in the theory of sequence spaces, which is known as *statistical* convergence. This fruitful concept is extremely valuable for studies in various areas of pure and applied mathematical sciences. It is remarkably more powerful than the traditional convergence and has provided a vital area of research in recent years. Furthermore, such a concept is closely related to the study of Real Analysis, Analytic Probability theory and Number theory, and so on. For some recent related developments on this subject, the reader can see, for example, the works in [3–18].

Suppose that E ⊆ N. Moreover, let

$$\mathfrak{E}\_k = \{ \eta : \eta \le k \quad \text{and} \quad \eta \in \mathfrak{E} \}. $$

**Citation:** Srivastava, H.M.; Jena, B.B.; Paikray, S.K. Some Korovkin-Type Approximation Theorems Associated with a Certain Deferred Weighted Statistical Riemann-Integrable Sequence of Functions. *Axioms* **2022**, *11*, 128. https://doi.org/10.3390/ axioms11030128

Academic Editor: Yurii Kharkevych

Received: 7 February 2022 Accepted: 11 March 2022 Published: 12 March 2022

**Publisher's Note:** MDPI stays neutral with regard to jurisdictional claims in published maps and institutional affiliations.

**Copyright:** © 2022 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (https:// creativecommons.org/licenses/by/ 4.0/).

Then, the natural (or asymptotic) density *d*(E) of E is

$$d(\mathfrak{E}) = \lim\_{k \to \infty} \frac{|\mathfrak{E}\_k|}{k} = \tau\_\prime$$

where *τ* is a real and finite number, and |E*k* | is the cardinality of E*k* .

A sequence (*un*) is said to be statistically convergent to *α* if, for each *e* > 0,

E*e* = {*η* : *η* ∈ N and |*uη* − *α*| = *e*}

has zero natural density (see [1,2]). Thus, for every *e* > 0,

$$d(\mathfrak{E}\_{\varepsilon}) = \lim\_{k \to \infty} \frac{|\mathfrak{E}\_{\varepsilon}|}{k} = 0.$$

We write

$$\text{stat}\lim\_{k\to\infty}\mathfrak{u}\_k=\mathfrak{a}.$$

For a closed and bounded interval I := [*a*, *b*] ⊂ R, we define the partition of [*a*, *b*] as an ordered set that is finite and we denote it as follows:

$$\mathfrak{P} := \{ (r\_0, r\_1, \dots, r\_k) : a = r\_0 < r\_1 < \dots < r\_k = b \}.$$

We now divide the interval [*a*, *b*] into the following non-overlapping subintervals:

$$\mathcal{T}\_1 := [r\_0, r\_1], \quad \mathcal{T}\_2 := [r\_1, r\_2], \quad \cdots \quad , \quad \mathcal{T}\_k := [r\_{k-1}, r\_k].$$

The resulting partition P is then given by

$$\mathfrak{P} := \{ [r\_{i-1}, r\_i] : i = 1, 2, 3, \dots, k \}.$$

Next, in order to find the norm of the partition P, we have

$$\|\mathfrak{P}\|\!:=\max\{r\_1-r\_0, r\_2-r\_1, r\_3-r\_2, \dots, r\_k-r\_{k-1}\}.$$

Let *γi* (*i* = 1, 2, 3, · · · , *k*) be a point that is chosen arbitrarily from each of the subintervals (I) *k i*=1 . We refer to these points as the tags of the subintervals. We also call the subintervals associated with the tags the tagged partitions of I. We denote it as follows:

$$\mathcal{P} := \{ ([r\_{i-1}, r\_i]; \gamma\_i) : i = 1, 2, 3, \dots, k \}.$$

Let [*a*, *b*] ⊂ R. Suppose that, for each *i* ∈ N, there is a function *hi* : [*a*, *b*] → R. We thus construct the sequence (*hi*)*i*∈N of functions over the closed interval [*a*, *b*].

We now define a subsequence (*hi*) *k i* of functions with respect to the Riemann sum associated with a tagged partition P as follows:

$$\delta(h\_i; \mathcal{P}) := \sum\_{i=1}^k h(\gamma\_i)(r\_i - r\_{i-1}).$$

We next recall the definition of the Riemann integrability.

A sequence (*hk* )*k*∈N of functions is Riemann-integrable to *h* on [*a*, *b*] if, for each *e* > 0, there exists *σe* > 0 such that, for any tagged partition P of [*a*, *b*] with kP k < *σe*, we have

$$|\delta(h\_k; \mathcal{P}) - h| < \epsilon.$$

The definition of statistically Riemann-integrable functions is given as follows.

**Definition 1.** *A sequence* (*hk* )*k*∈N *of functions is statistically Riemann-integrable to h on* [*a*, *b*] *if, for every e* > 0 *and for each x* ∈ [*a*, *b*], *there exists σe* > 0, *and for any tagged partition* P *of* [*a*, *b*] *with* kP k < *σe*, *the set*

$$\mathfrak{E}\_{\mathfrak{e}} = \{ \eta : \eta \in \mathbb{N} \quad \text{and} \quad |\delta(h\_{\eta}; \mathcal{P}) - h| \ge \epsilon \}$$

*has zero natural density. That is, for every e* > 0,

$$d(\mathfrak{E}\_{\mathfrak{e}}) = \lim\_{k \to \infty} \frac{|\mathfrak{E}\_{\mathfrak{e}}|}{k} = 0.$$

*We write*

$$\operatorname{stat}\_{\text{Rie}} \lim\_{\mathbf{k} \to \infty} \operatorname{ffi}(\mathbf{h\_{k}}; \mathcal{P}) = \mathbf{h}.$$

By making use of Definition 1, we first establish an inclusion theorem as Theorem 1 below.

**Theorem 1.** *If a sequence of functions* (*hk* ) *is Riemann-integrable to h over* [*a*, *b*], *then* (*hk* ) *is statistically Riemann-integrable to the same function h over* [*a*, *b*]*.*

**Proof.** Given *e* > 0, there exists *σe* > 0. Suppose that P is any tagged partition of [*a*, *b*] such that kP k < *σe*. Then

$$|\delta(h\_k; \mathcal{P}) - h| < \epsilon.$$

Since, for each *e* > 0, P is any tagged partition of [*a*, *b*] such that kP k < *σe*, so we have

$$\lim\_{k \to \infty} \frac{1}{k} |\{\eta : \eta \in \mathbb{N} \quad \text{and} \quad |\delta(h\_k; \mathcal{P}) - h| \ge \epsilon\}| \le \lim\_{k \to \infty} |\delta(h\_k; \mathcal{P}) - h| < \epsilon.$$

Consequently, by Definition 1, we get

$$\texttt{statRe}\lim\_{k\to\infty}\delta(h\_k;\mathcal{P}) = h\_r$$

which completes the proof of Theorem 1.

**Remark 1.** *In order to demonstrate that the converse of Theorem* 1 *is not true, we consider Example* 1 *below.*

**Example 1.** *Let hk* : [0, 1] → R *be a sequence of functions defined by*

$$h\_k(\mathbf{x}) = \begin{cases} \frac{1}{2} & (\mathbf{x} \in \mathbb{Q} \cap [0, 1]; \, k = j^2, \, j \in \mathbb{N}) \\\\ \frac{1}{n} & (\text{otherwise}). \end{cases} \tag{1}$$

*It is easily seen that the sequence* (*hk* ) *of functions is statistically Riemann-integrable to* 0 *over the closed interval* [0, 1]*, but it is not Riemann-integrable (in the usual sense) over* [0, 1]*.*

Motivated mainly by the above-mentioned investigations and developments, we introduce and study the ideas of deferred weighted statistical Riemann integrability and statistical deferred weighted Riemann summability of sequences of real-valued functions. We first prove an inclusion theorem connecting these two potentially useful concepts. We then state and prove two Korovkin-type approximation theorems with algebraic test functions based on the methodologies and techniques that we have adopted here. Furthermore, we consider an illustrative example involving a positive linear operator in conjunction with the familiar Bernstein polynomials, which shows the effectiveness of our findings. Finally, based upon the core concept of statistical Lebesgue-measurable sequences of functions, we

suggest some possible directions for future research on this topic in the concluding section of our study.

#### **2. Deferred Weighted Statistical Riemann Integrability**

Let (*φk* ) and (*ϕk* ) be sequences of non-negative integers with the regularity conditions given

$$
\phi\_k < \phi\_k \qquad \text{and} \qquad \lim\_{k \to \infty} \phi\_k = +\infty.
$$

Moreover, let (*pi*) be a sequence of non-negative real numbers with

$$P\_k = \sum\_{i=\phi\_k+1}^{\varphi\_k} p\_i.$$

We then define the deferred weighted summability mean for *σ*(*hk* ;P) associated with tagged partition P as follows:

$$\mathcal{W}(\delta(h\_k; \mathcal{P})) = \frac{1}{P\_k} \sum\_{\varrho = \varrho\_k + 1}^{\varrho\_k} p\_\varrho \delta(h\_\varrho; \mathcal{P}). \tag{2}$$

We now present the following definitions for our proposed study.

**Definition 2.** *A sequence* (*hk* )*k*∈N *of functions is said to be deferred weighted statistically Riemannintegrable to h on* [*a*, *b*] *if, for all e* > 0, *there exists σe* > 0, *and for any tagged partition* P *of* [*a*, *b*] *with* kP k < *σe*, *the following set*

$$\{\eta : \eta \le \mathcal{P}\_k \quad \text{and} \quad p\_\eta |\delta(h\_\eta; \mathcal{P}) - h| \ge \epsilon\}$$

*has zero natural density. Thus, for every e* > 0*, we have*

$$\lim\_{k \to \infty} \frac{|\{\eta : \eta \le P\_k \quad \text{and} \quad p\_\eta |\delta(h\_\eta; \mathcal{P}) - k| \ge \epsilon\}|}{P\_k} = 0.$$

*We write*

$$DWR\_{\text{stat}} \lim\_{k \to \infty} \delta(h\_k; \mathcal{P}) = h.$$

**Definition 3.** *A sequence* (*hk* )*k*∈N *of functions is said to statistically deferred weighted Riemann summable to h on* [*a*, *b*] *if, for all e* > 0 ∃ *σe* > 0 *and for any tagged partition* P *of* [*a*, *b*] *with* kP k < *σe*, *the set*

$$\{\eta : \eta \le k \quad \text{and} \quad |\mathcal{W}(\delta(h\_{\eta}; \mathcal{P})) - h| \ge \epsilon\}$$

*has zero natural density. Thus, for all e* > 0, *we have*

$$\lim\_{k \to \infty} \frac{|\{\eta : \eta \le k \quad \text{and} \quad |\mathcal{W}(\delta(h\_{\eta}; \mathcal{P})) - h| \ge \epsilon\}|}{k} = 0.$$

*We write*

$$\operatorname{statDWR}\_{\mathbf{k}\to\infty} \lim\_{\mathbf{k}\to\infty} \operatorname{ffi}(\mathbf{h}\_{\mathbf{k}}; \mathcal{P}) = \mathbf{h}.$$

An inclusion theorem between the two new potentially useful notions in Definitions 2 and 3 is now given by Theorem 2 below.

**Theorem 2.** *If the sequence* (*hk* )*k*∈N *of functions is deferred weighted statistically Riemannintegrable to a function h over* [*a*, *b*], *then it is statistically deferred weighted Riemann summable to the same function h over* [*a*, *b*], *but not conversely.*

**Proof.** Suppose that the sequence (*hk* )*k*∈N is deferred weighted statistically Riemannintegrable to a function *h* on [*a*, *b*]. Then, by Definition 2, we have

$$\lim\_{k \to \infty} \frac{|\{\eta : \eta \le P\_k \quad \text{and} \quad p\_\eta | \delta(h\_\eta; \mathcal{P}) - h | \ge \epsilon\}|}{P\_k} = 0.$$

Now, if we choose the two sets as follows,

$$\mathcal{O}\_{\epsilon} = \{ \eta : \eta \le P\_k \qquad \text{and} \qquad p\_\eta |\delta(h\_\eta; \mathcal{P}) - h| \ge \epsilon \}$$

and

$$\mathcal{O}^c\_\varepsilon = \{ \eta : \eta \le P\_k \qquad \text{and} \qquad p\_\eta |\delta(h\_\eta; \mathcal{P}) - h| < \varepsilon \},$$

then we have

$$\begin{split} |\mathcal{W}(\delta(h\_{k};\mathcal{P}))-h| &= \left| \frac{1}{P\_{k}} \sum\_{\varrho=\Phi\_{k}+1}^{\Phi\_{k}} p\_{\varrho} \delta(h\_{\varrho};\mathcal{P}) - h \right| \\ &\leq \left| \frac{1}{P\_{k}} \sum\_{\varrho=\Phi\_{k}+1}^{\Phi\_{k}} p\_{\varrho} [\delta(h\_{\varrho};\mathcal{P}) - h] \right| + \left| \frac{1}{P\_{k}} \sum\_{\varrho=\Phi\_{k}+1}^{\Phi\_{k}} p\_{\varrho} h - h \right| \\ &\leq \frac{1}{P\_{k}} \sum\_{\varrho=\Phi\_{k}+1}^{\Phi\_{k}} p\_{\varrho} |\delta(h\_{\varrho};\mathcal{P}) - h| + \frac{1}{P\_{k}} \sum\_{\begin{subarray}{c} \varrho=\Phi\_{k}+1 \\ (\eta \in \mathcal{O}\_{\varepsilon}^{c}) \end{subarray}}^{\Phi\_{k}} p\_{\varrho} |\delta(h\_{\varrho};\mathcal{P}) - h| \\ &+ |h| \left| \frac{1}{P\_{k}} \sum\_{\varrho=\Phi\_{k}+1}^{\Phi\_{k}} p\_{\varrho} - 1 \right| \\ &\leq \frac{1}{P\_{k}} |\mathcal{O}\_{\varepsilon}| + \frac{1}{P\_{k}} |\mathcal{O}\_{\varepsilon}^{c}|. \end{split}$$

We thus obtain

$$|\mathcal{W}(\delta(h\_k; \mathcal{P})) - h| < \epsilon.$$

Hence, clearly, the sequence of functions (*hk* ) is statistically deferred weighted Riemannsummable to *h* over [*a*, *b*].

The following example shows that the converse statement of Theorem 2 is not true.

**Example 2.** *Let hk* : [0, 1] → R *be a sequence of functions of the form given by*

$$h\_k(\mathbf{x}) = \begin{cases} 0 & (\mathbf{x} \in \mathbb{Q} \cap [0, 1]; \ k \text{ is even}) \\\\ 1 & (\mathbf{x} \in \mathbb{R} - \mathbb{Q} \cap [0, 1]; \ k \text{ is odd}), \end{cases} \tag{3}$$

*where*

*φk* = 2*k ϕk* = 4*k* and *pk* = 1.

*The above-specified sequence* (*hk* ) *of functions trivially indicates that it is neither Riemannintegrable nor deferred weighted statistically Riemann-integrable. However, as per our proposed mean (2), it is easy to see that*

$$\begin{aligned} \mathcal{W}(\delta(h\_k; \mathcal{P})) &= \frac{1}{\varphi\_k - \phi\_k} \sum\_{\varrho = \phi\_k + 1}^{\varphi\_k} \delta(h\_\varrho; \mathcal{P}) \\ &= \frac{1}{2k} \sum\_{m = 2k + 1}^{4k} \delta(h\_\varrho; \mathcal{P}) = \frac{1}{2} .\end{aligned}$$

*Thus, clearly, the sequence* (*hk* ) *of functions has deferred weighted Riemann sum* 1 2 *under the tagged partition* P*. Therefore, the sequence* (*hk* ) *of functions is statistically deferred weighted Riemann-summable to* 1 2 *over* [0, 1]*, but it is not deferred weighted statistically Riemann-integrable over* [0, 1]*.*

### **3. Korovkin-Type Approximation Theorems via the** W(*δ*(*hk***;** P))**-Mean**

Many researchers have worked toward extending (or generalizing) the approximationtheoretic aspects of the Korovkin-type approximation theorems in several different areas of mathematics, such as (for example) probability space, measurable space, sequence spaces, and so on. In Real Analysis, Harmonic Analysis and other related fields, this notion is immensely useful. In this regard, we have chosen to refer the interested reader to the recent works (see, for example, [19–28]).

Let *C*[0, 1] be the space of all continuous real-valued functions defined on [0, 1]. Suppose also that it is a Banach space with the norm k.k∞. Then, for *h* ∈ *C*[0, 1], the norm of *h* is given by

$$||h||\_{\infty} = \sup \{ |h(\rho)| \, : \, 0 \le \rho \le 1 \}.$$

We say that G*j* : *C*[0, 1] → *C*[0, 1] is a sequence of positive linear operators, if

$$\mathfrak{G}\_j(h;\rho) \ge 0 \quad \text{as} \quad h \ge 0.$$

Now, in view of our above-proposed definitions, we state and prove the following Korovkin-type approximation theorems.

**Theorem 3.** *Let* G*j* : *C*[0, 1] → *C*[0, 1] *be a sequence of positive linear operators. Then*, *for h* ∈ *C*[0, 1],

$$\text{DWR}\_{\text{stat}} \lim\_{j \to \infty} ||\mathfrak{G}\_j(h; \rho) - h(\rho)||\_{\infty} = 0 \tag{4}$$

*if and only if*

$$\text{DWR}\_{\text{stat}} \lim\_{j \to \infty} ||\mathfrak{G}\_j(1;\rho) - 1||\_{\infty} = 0,\tag{5}$$

$$\text{DWR}\_{\text{stat}} \lim\_{j \to \infty} \|\mathfrak{G}\_j(\rho; \rho) - \rho\|\_{\infty} = 0 \tag{6}$$

*and*

$$\text{DWR}\_{\text{stat}} \lim\_{j \to \infty} \||\mathfrak{G}\_j(\rho^2; \rho) - \rho^2\|\_{\infty} = 0. \tag{7}$$

**Proof.** Since each of the following functions

$$h\_0(\rho) = 1, \quad h\_1(\rho) = 2\rho \quad \text{and} \quad h\_2(\rho) = 3\rho^2.$$

belongs to *C*[0, 1] and is continuous on [0, 1], the implication given by (4) obviously implies (5) to (7).

In order to complete the proof of Theorem 3, we first assume that the conditions (5) to (7) hold true. If *h* ∈ *C*[0, 1], then there exists a constant L > 0 such that

$$|h(\rho)| \le \mathcal{L} \quad (\forall \, \rho \in [0, 1]).$$

We thus find that

$$|h(r) - h(\rho)| \le 2\mathcal{L} \quad (r, \rho \in [0, 1]).\tag{8}$$

Clearly, for given *e* > 0, there exists *δ* > 0 such that

$$|f(r) - f(\rho)| < \epsilon \tag{9}$$

whenever

If

If we now choose

$$\text{The first-order coupling between the two-dimensional } \mathcal{N} \text{-matrices is the only possible } \mathcal{N} \text{-matrices with } \mathcal{N} = \{0, 1, 2, \dots, N\} \text{ and } \mathcal{N} = \{0, 1, 2, \dots, N\}.$$

$$
\mu\_1 = \mu\_1(r\_\prime \rho) = (2r - 2\rho)^2
$$

then we obtain

$$|h(r) - h(\rho)| < \frac{2\mathcal{L}}{\theta^2} \mu\_1(r\_\prime \rho). \tag{10}$$

2 .

Thus, from Equations (9) and (10), we get

$$|h(r) - h(\rho)| < \varepsilon + \frac{2\mathcal{L}}{\theta^2} \mu\_1(r, \rho)\_{\prime\prime}$$

which implies that

$$-\varepsilon - \frac{2\mathcal{L}}{\theta^2}\mu\_1(r,\rho) \le h(r) - h(\rho) \le \varepsilon + \frac{2\mathcal{L}}{\theta^2}\mu\_1(r,\rho). \tag{11}$$

Now, since G*m*(1; *ρ*) is monotone and linear, by applying the operator G*m*(1; *ρ*) to the inequality (11), we get

$$\begin{split} \mathfrak{G}\_{m}(\mathbf{1};\rho) \Big( -\varepsilon - \frac{2\mathcal{L}}{\theta^{2}} \mu\_{1}(r,\rho) \Big) &\leq \mathfrak{G}\_{m}(\mathbf{1};\rho) \Big( h(r) - h(\rho) \Big) \\ &\leq \mathfrak{G}\_{m}(\mathbf{1};\rho) \Big( \varepsilon + \frac{2\mathcal{L}}{\theta^{2}} \mu\_{1}(r,\rho) \Big). \end{split}$$

We note that *ρ* is fixed, and so *h*(*ρ*) is a constant number. Therefore, we have

$$\begin{split} -\varepsilon \mathfrak{G}\_{m}(1;\rho) - \frac{2\mathcal{L}}{\theta^{2}} \mathfrak{G}\_{m}(\mu\_{1};\rho) &\leq \mathfrak{G}\_{m}(\mathbb{H};\rho) - h(\rho)\mathfrak{G}\_{m}(1;\rho) \\ &\leq \varepsilon \mathfrak{G}\_{m}(1;\rho) + \frac{2\mathcal{L}}{\theta^{2}} \mathfrak{G}\_{m}(\mu\_{1};\rho) .\end{split} \tag{12}$$

We also know that

$$\mathfrak{G}\_{\mathfrak{m}}(h;\rho) - h(\rho) = [\mathfrak{G}\_{\mathfrak{m}}(h;\rho) - h(\rho)\mathfrak{G}\_{\mathfrak{m}}(1;\rho)] + h(\rho)[\mathfrak{G}\_{\mathfrak{m}}(1;\rho) - 1]. \tag{13}$$

Thus, by using (12) and (13), we obtain

$$\mathfrak{G}\_{\mathfrak{m}}(h;\rho) - h(\rho) < \epsilon \mathfrak{G}\_{\mathfrak{m}}(1;\rho) + \frac{2\mathcal{L}}{\theta^2} \mathfrak{G}\_{\mathfrak{m}}(\mu\_1;\rho) + h(\rho)[\mathfrak{G}\_{\mathfrak{m}}(1;\rho) - 1]. \tag{14}$$

We now estimate G*m*(*µ*1; *ρ*) as follows:

$$\begin{split} \mathfrak{G}\_{m}(\mu\_{1};\rho) &= \mathfrak{G}\_{m}((2r-2\rho)^{2};\rho) = \mathfrak{G}\_{m}(2r^{2}-8\rho r+4\rho^{2};\rho) \\ &= \mathfrak{G}\_{m}(4r^{2};\rho) - 8t\mathfrak{G}\_{m}(r;\rho) + 4\rho^{2}\mathfrak{G}\_{m}(1;\rho) \\ &= 4\left[\mathfrak{G}\_{m}(r^{2};\rho) - \rho^{2}\right] - 8t\left[\mathfrak{G}\_{m}(r;\rho) - \rho\right] \\ &\quad + 4\rho^{2}\left[\mathfrak{G}\_{m}(1;\rho) - 1\right]. \end{split}$$

so that, in view of (14), we obtain

$$\begin{split} \mathfrak{G}\_{m}(h;\rho) - h(\rho) &< \epsilon \mathfrak{G}\_{m}(1;\rho) + \frac{2\mathcal{L}}{\theta^{2}} \{ 4[\mathfrak{G}\_{m}(r^{2};\rho) - \rho^{2}] \\ &- 8\rho[\mathfrak{G}\_{m}(r;\rho) - \rho] + 4\rho^{2}[\mathfrak{G}\_{m}(1;\rho) - 1] \} \\ &+ h(\rho)[\mathfrak{G}\_{m}(1;\rho) - 1]. \end{split}$$

$$\begin{split} &= \epsilon[\mathfrak{G}\_{m}(1;\rho) - 1] + \epsilon + \frac{2\mathcal{L}}{\theta^{2}} \{ 4[\mathfrak{G}\_{m}(r^{2};\rho) - \rho^{2}] \} \\ &- 8\rho[\mathfrak{G}\_{m}(r;\rho) - \rho] + 4\rho^{2}[\mathfrak{G}\_{m}(1;\rho) - 1] \{ \\ &+ h(\rho)[\mathfrak{G}\_{m}(1;\rho) - 1]. \end{split}$$

Furthermore, since *e* > 0 is arbitrary, we can write

$$\begin{split} |\mathfrak{G}\_{\mathfrak{m}}(h;\rho) - h(\rho)| &\leq \varepsilon + \left(\varepsilon + \frac{8\mathcal{L}}{\theta^{2}} + \mathcal{L}\right) |\mathfrak{G}\_{\mathfrak{m}}(1;\rho) - 1| \\ &+ \frac{16\mathcal{L}}{\theta^{2}} |\mathfrak{G}\_{\mathfrak{m}}(r;\rho) - \rho| + \frac{8\mathcal{L}}{\theta^{2}} |\mathfrak{G}\_{\mathfrak{m}}(r^{2};\rho) - \rho^{2}| \\ &\leq \mathcal{A}(|\mathfrak{G}\_{\mathfrak{m}}(1;\rho) - 1| + |\mathfrak{G}\_{\mathfrak{m}}(r;\rho) - \rho| \\ &+ |\mathfrak{G}\_{\mathfrak{m}}(r^{2};\rho) - \rho^{2}|), \end{split} \tag{15}$$

where

$$\mathcal{A} = \max \left( \epsilon + \frac{8\mathcal{L}}{\theta^2} + \mathcal{L}, \frac{16\mathcal{L}}{\theta^2}, \frac{8\mathcal{L}}{\theta^2} \right).$$

Now, for a given *ω* > 0, there exists *e* > 0 (*e* < *ω*) such that

$$\mathfrak{T}\_{\mathfrak{m}}(\rho;\omega) = \{ m : m \le P\_k \quad \text{and} \quad p\_m |\mathfrak{G}\_m(h;\rho) - h(\rho)| \ge \omega \}.$$

Furthermore, for *ν* = 0, 1, 2, we have

$$\mathfrak{T}\_{\mathsf{V},\mathsf{m}}(\rho;\omega) = \left\{ m : m \le P\_{\mathsf{k}} \quad \text{and} \quad p\_{\mathsf{m}} \left| \mathfrak{G}\_{\mathsf{m}}(h;\rho) - h\_{\mathsf{V}}(\rho) \right| \ge \frac{\omega - \varepsilon}{\mathfrak{Z}\mathcal{A}} \right\},$$

so that

$$\mathfrak{T}\_m(\rho;\omega) \le \sum\_{\nu=0}^2 \mathfrak{T}\_{\nu,m}(\rho;\omega).$$

Clearly, we obtain

$$\frac{||\mathfrak{T}\_{\mathfrak{m}}(\rho;\omega)||\_{\mathcal{C}[0,1]}}{P\_{k}} \le \sum\_{\nu=0}^{2} \frac{||\mathfrak{T}\_{\nu,\mathfrak{m}}(\rho;\omega)||\_{\mathcal{C}[0,1]}}{P\_{k}}.\tag{16}$$

Now, using the above assumption about the implications in (5) to (7) and by Definition 2, the right-hand side of (16) tends to zero as *n* → ∞. Consequently, we get

$$\lim\_{k \to \infty} \frac{\|\mathfrak{T}\_m(\rho; \omega)\|\_{\mathcal{C}[0,1]}}{P\_k} = 0 \ (\delta, \omega > 0).$$

Therefore, the implication (4) holds true.

**Theorem 4.** *Let* G*j* : *C*[0, 1] → *C*[0, 1] *be a sequence of positive linear operators. Then*, *for h* ∈ *C*[0, 1],

$$\text{statp}\_{\text{DWR}} \lim\_{j \to \infty} ||\mathfrak{G}\_j(h; \rho) - h(\rho)||\_{\infty} = 0 \tag{17}$$

*if and only if*

$$\text{stat}\_{\text{DWR}} \lim\_{j \to \infty} ||\mathfrak{G}\_j(1;\rho) - 1||\_{\infty} = 0,\tag{18}$$

$$\text{stat}\_{\text{DWR}} \lim\_{j \to \infty} ||\mathfrak{G}\_j(\rho; \rho) - \rho||\_{\infty} = 0 \tag{19}$$

*and*

$$\text{stat}\_{\text{DWR}} \lim\_{j \to \infty} ||\mathfrak{G}\_j(\rho^2; \rho) - \rho^2||\_{\infty} = 0. \tag{20}$$

**Proof.** The proof of Theorem 4 is similar to the proof of Theorem 3. Therefore, we choose to skip the details involved.

In view of Theorem 4, here, we consider an illustrative example. In this connection, we now recall the following operator:

$$
\rho(1+\rho D) \qquad \left(D = \frac{d}{d\rho}\right),
\tag{21}
$$

which was used by Al-Salam [29] and, more recently, by Viskov and Srivastava [30].

**Example 3.** *Consider the Bernstein polynomials* B*n*(*h*; *β*) *on C*[0, 1] *given by*

$$\mathfrak{B}\_{k}(h;\beta) = \sum\_{\varrho=0}^{k} f\left(\frac{\varrho}{k}\right) \binom{k}{\varrho} \beta^{\varrho} (1-\beta)^{k-\varrho} \quad (\beta \in [0,1]; k = 0, 1, \cdots). \tag{22}$$

*Here, in this example, we introduce the positive linear operators on C*[0, 1] *under the composition of the Bernstein polynomials and the operators given by (21) as follows:*

$$\mathfrak{G}\_{\varrho}(h;\beta) = [1 + h\_{\varrho}]\beta(1 + \beta D)\mathfrak{B}\_{\varrho}(h;\beta) \quad (\forall \ h \in \mathbb{C}[0,1]),\tag{23}$$

*where* (*h\$*) *is the same as mentioned in Example 2.*

*We now estimate the values of each of the testing functions* 1*, β and β* 2 *by using our proposed operators (23) as follows:*

$$\mathfrak{G}\_{\varrho}(1;\beta) = [1 + h\_{\varrho}]\beta (1 + \beta D)1 = [1 + h\_{\varrho}]\beta\_{\varkappa}$$

$$\mathfrak{G}\_{\varrho}(t;\beta) = [1 + h\_{\varrho}]\beta(1 + \beta D)\beta = [1 + h\_{\varrho}]\beta(1 + \beta).$$

*and*

$$\begin{split} \mathfrak{G}\_{\varrho}(t^{2};\beta) &= [1+h\_{\varrho}]\beta(1+\beta D)\left\{\beta^{2}+\frac{\beta(1-\beta)}{\varrho}\right\}, \\ &= [1+h\_{\varrho}]\left\{\beta^{2}\left(2-\frac{3\beta}{\varrho}\right)\right\}. \end{split}$$

*Consequently, we have*

$$\text{statp}\_{\text{DWR}} \lim\_{\varrho \to \infty} ||\mathfrak{G}\_{\varrho}(1; \beta) - 1||\_{\infty} = 0,\tag{24}$$

$$\text{stat}\_{\text{DWR}} \lim\_{\boldsymbol{\theta} \to \infty} ||\mathfrak{G}\_{\boldsymbol{\theta}}(\boldsymbol{\beta}; \boldsymbol{\beta}) - \boldsymbol{\beta}||\_{\infty} = \mathbf{0} \tag{25}$$

*and*

$$\text{statDWR} \lim\_{\varrho \to \infty} ||\mathfrak{G}\_{\varrho}(\mathfrak{f}^2; \mathfrak{\boldsymbol{\beta}}) - \mathfrak{\boldsymbol{\beta}}^2||\_{\infty} = \mathbf{0},\tag{26}$$

*that is, the sequence* G*\$*(*h*; *β*) *satisfies the conditions (18) to (20). Therefore, by Theorem 4, we have*

$$\operatorname{statDWR}\_{\varrho \to \infty} \lim\_{\theta \to \infty} ||\mathfrak{G}\_{\varrho}(h; \beta) - h||\_{\infty} = 0.$$

*Hence, the given sequence* (*hk* ) *of functions mentioned in Example 2 is statistically deferred weighted Riemann-summable, but not deferred weighted statistically Riemann-integrable. Therefore, our above-proposed operators defined by (23) satisfy Theorem 4. However, they do not satisfy for statistical versions of deferred weighted Riemann-integrable sequence of functions (see Theorem 3).*

#### **4. Concluding Remarks and Directions for Further Research**

In this concluding section of our present investigation, we further observe the potential usefulness of our Theorem 4 over Theorem 3 as well as over the classical versions of the Korovkin-type approximation theorems.

**Remark 2.** *Let us consider the sequence* (*h\$*)*\$*∈N *of functions in Example 2. Suppose also that* (*h\$*) *is statistically deferred weighted Riemann-summable, so that*

$$\text{stat}\_{\text{DWR}} \lim\_{\% \to \infty} \text{ffi}(\text{h}\_{\%}; \mathcal{P}) = \frac{1}{2} \text{ on } [0, 1].$$

*We then find that*

$$\text{stat}\_{\text{DWR}} \lim\_{\mathbf{k} \to \infty} \|\mathfrak{G}\_{\mathbf{k}}(\mathbf{h} \cdot ; \mathfrak{e}) - \mathbf{f} \cdot (\mathfrak{e} \mathbf{e})\|\_{\infty} = 0 \quad (\text{}^{\circ} = 0, 1, 2). \tag{27}$$

*Thus, by Theorem 4, we immediately get*

$$\text{stat}\_{\text{DWR}} \lim\_{j \to \infty} ||\mathfrak{G}\_{\mathbf{k}}(\mathbf{h}; \mathfrak{a}) - \mathbf{h}(\mathfrak{a})||\_{\infty} = 0,\tag{28}$$

*where*

$$h\_0(\rho) = 1, \quad h\_1(\rho) = \rho \quad \text{and} \quad h\_2(\rho) = \rho^2.$$

*Now, the given sequence* (*hk* ) *of functions is statistically deferred weighted Riemann-summable, but neither deferred weighted statistically Riemann-integrable nor classically Riemann-integrable. Therefore, our Korovkin-type approximation Theorem* 4 *properly works under the operators defined in the Equation* (23)*, but the classical as well as statistical versions of the deferred weighted Riemannintegrable sequence of functions do not work for the same operators. Clearly, this observation leads us to the fact that our Theorem* 4 *is a non-trivial extension of Theorem* 3 *as well as the classical Korovkin-type approximation theorem [31].*

**Remark 3.** *Motivated by some recently published results by Jena et al. [32] and Srivastava et al. [33], we choose to draw the attention of the interested readers toward the potential for further research associated with the analogous notion of statistical Lebesgue-measurable sequences of functions.*

**Author Contributions:** Formal analysis, H.M.S. and S.K.P.; Investigation, B.B.J.; Methodology, S.K.P.; Supervision, H.M.S. and S.K.P.; Writing-original draft, B.B.J.; Writing—review & editing, H.M.S. and S.K.P. All authors have read and agreed to the published version of the manuscript.

**Funding:** This research received no external funding.

**Data Availability Statement:** Not applicable.

**Conflicts of Interest:** The authors declare that they have no conflicts of interest.

#### **References**

## *Article* **On Some Generalizations of Reverse Dynamic Hardy Type Inequalities on Time Scales**

**Ahmed A. El-Deeb 1,\* and Clemente Cesarano 2,\***

**Abstract:** In the present paper, we prove some new reverse type dynamic inequalities on T. Our main inequalities are proved by using the chain rule and Fubini's theorem on time scales T. Our results extend some existing results in the literature. As special cases, we obtain some new discrete inequalities, quantum inequalities and integral inequalities.

**Keywords:** reverse Hardy's inequality; dynamic inequality; time scale

**MSC:** 26D10; 26D15; 34N05; 26E70

#### **1. Introduction**

In 1920, the renowned English mathematician Godfrey Harold Hardy [1] proved the following result.

**Theorem 1.** *Assume that* { *fn*} ∞ *n*=1 *is a sequence of nonnegative real numbers. If r* > 1*, then*

$$\sum\_{n=1}^{\infty} \frac{1}{n^r} \left( \sum\_{k=1}^n f\_k \right)^r \le \left( \frac{r}{r-1} \right)^r \sum\_{n=1}^\infty f\_n^r. \tag{1}$$

Inequality (1) is known in the literature as discrete Hardy' inequality.

In 1925, Hardy himself [2] gave the integral analogous of inequality (1) in the following form.

**Theorem 2.** *Suppose that f is a nonnegative continuous function defined on* [0, ∞)*. If r* > 1*, then*

$$\int\_0^\infty \frac{1}{\lambda^r} \left( \int\_0^\lambda f(\zeta) d\zeta \right)^r d\lambda \le \left( \frac{r}{r-1} \right)^r \int\_0^\infty f^r(\lambda) d\lambda. \tag{2}$$

In 1927, Littlewood and Hardy [3] proved the reversed version of inequality (2) in the following manner:

**Theorem 3.** *Let f be a nonnegative function on* [0, ∞)*. If* 0 < *r* < 1*, then*

$$\int\_0^\infty \frac{1}{\lambda^r} \left( \int\_\lambda^\infty f(\zeta) d\zeta \right)^r d\lambda \ge \left( \frac{r}{1-r} \right)^r \int\_0^\infty f^r(\lambda) d\lambda. \tag{3}$$

In 1928, Hardy [4] established a generalization of inequality (2). He proved that:

**Theorem 4.** *Suppose that f is a nonnegative continuous function defined on* [0, ∞)*. Then,*

$$\int\_0^\infty \frac{1}{\lambda^\gamma} \left( \int\_0^\lambda f(\zeta) d\zeta \right)^r d\lambda \le \left( \frac{r}{\gamma - 1} \right)^r \int\_0^\infty \lambda^{r - \gamma} f^r(\lambda) d\lambda, \qquad \text{for} \quad r \ge \gamma > 1,\tag{4}$$

**Citation:** El-Deeb, A.A.; Cesarano, C. On Some Generalizations of Reverse Dynamic Hardy Type Inequalities on Time Scales. *Axioms* **2022**, *11*, 336. https://doi.org/10.3390/ axioms11070336

Academic Editor: Yurii Kharkevych

Received: 19 May 2022 Accepted: 8 July 2022 Published: 11 July 2022

**Publisher's Note:** MDPI stays neutral with regard to jurisdictional claims in published maps and institutional affiliations.

**Copyright:** © 2022 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (https:// creativecommons.org/licenses/by/ 4.0/).

*and*

$$\int\_0^\infty \frac{1}{\lambda^\gamma} \left( \int\_\lambda^\infty f(\zeta) d\zeta \right)^r d\lambda \le \left( \frac{r}{1-\gamma} \right)^r \int\_0^\infty \lambda^{r-\gamma} f^r(\lambda) d\lambda, \quad \text{for} \quad r > 1 > \gamma \ge 0. \tag{5}$$

In 1928, Copson [5] gave the next two discrete inequalities as generalizations of inequality (1).

**Theorem 5.** *Let* { *fn*} ∞ *n*=1 *and* {*θn*} ∞ *n*=1 *be sequences of nonnegative real numbers. Then,*

$$\sum\_{n=1}^{\infty} \frac{\theta\_n \left(\sum\_{k=1}^n \theta\_k f\_k\right)^r}{\left(\sum\_{k=1}^n \theta\_k\right)^\gamma} \le \left(\frac{r}{\gamma - 1}\right)^r \sum\_{n=1}^\infty \theta\_n f\_n^r \left(\sum\_{k=1}^n \theta\_k\right)^{r-\gamma}, \qquad \text{for} \quad r \ge \gamma > 1,\tag{6}$$

*and*

$$\sum\_{n=1}^{\infty} \frac{\theta\_n \left(\sum\_{k=n}^{\infty} \theta\_k f\_k\right)^r}{\left(\sum\_{k=1}^n \theta\_k\right)^{\gamma}} \le \left(\frac{r}{1-\gamma}\right)^r \sum\_{n=1}^{\infty} \theta\_n f\_n^r \left(\sum\_{k=1}^n \theta\_k\right)^{r-\gamma}, \qquad \text{for} \quad r > 1 > \gamma \ge 0. \tag{7}$$

In 1970, Leindler [6] explored some discrete Hardy inequality versions (1) and was able to demonstrate that:

**Theorem 6.** *Let* { *fn*} ∞ *n*=1 *and* {*θn*} ∞ *n*=1 *be sequences of real numbers that are not negative and r* > 1*, then*

$$\sum\_{n=1}^{\infty} \theta\_n \left( \sum\_{k=1}^n f\_k \right)^r \le r^r \sum\_{n=1}^{\infty} \theta\_n^{1-r} f\_n^r \left( \sum\_{k=n}^{\infty} \theta\_k \right)^r,\tag{8}$$

*and*

$$\sum\_{n=1}^{\infty} \theta\_n \left( \sum\_{k=n}^{\infty} f\_k \right)^r \le r^r \sum\_{n=1}^{\infty} \theta\_n^{1-r} f\_n^r \left( \sum\_{k=1}^n \theta\_k \right)^r. \tag{9}$$

In 1976, Copson [7] gave the inequalities' continuous versions (6) and (7). He arrived at the following conclusion specifically:.

**Theorem 7.** *Let f and θ be continuous functions that are not negative on* [0, ∞)*. Then,*

$$\int\_{0}^{\infty} \frac{\theta(\lambda) \left( \int\_{0}^{\lambda} \theta(\zeta) f(\zeta) d\zeta \right)^{r}}{\left( \int\_{0}^{\lambda} \theta(\zeta) d\zeta \right)^{\gamma}} d\lambda \le \left( \frac{r}{\gamma - 1} \right)^{r} \int\_{0}^{\infty} \theta(\lambda) f'(\lambda) \left( \int\_{0}^{\lambda} \theta(\zeta) d\zeta \right)^{r - \gamma} d\lambda, \qquad \text{for} \quad r \ge \gamma > 1,\tag{10}$$

$$\int\_{0}^{\infty} \frac{\theta(\lambda) \left( \int\_{\lambda}^{\infty} \theta(\zeta) f(\zeta) d\zeta \right)^{r}}{\left( \int\_{0}^{\lambda} \theta(\zeta) d\zeta \right)^{\gamma}} d\lambda \le \left( \frac{r}{1 - \gamma} \right)^{r} \int\_{0}^{\infty} \theta(\lambda) f'(\lambda) \left( \int\_{0}^{\lambda} \theta(\zeta) d\zeta \right)^{r - \gamma} d\lambda, \quad \text{for} \quad r > 1 > \gamma \ge 0. \tag{11}$$

In 1982, Lyon [8] discovered a reverse version of the discrete Hardy inequality (1) for the special case when *r* = 2. According to his conclusion:

**Theorem 8.** *Let* { *fn*} ∞ *n*=0 *be a nonincreasing sequence of real numbers that are nonnegative. Then,*

$$\sum\_{n=0}^{\infty} \left( \frac{1}{n+1} \sum\_{k=0}^{n} f\_k \right)^2 \ge \frac{\pi^2}{6} \sum\_{n=0}^{\infty} f\_n^2. \tag{12}$$

In 1986, Renaud [9] proved the following two results.

**Theorem 9.** *Assume that* { *fn*} ∞ *n*=1 *is a nonincreasing sequence of nonnegative real numbers. If r* > 1*, then,*

$$\sum\_{n=1}^{\infty} \left(\sum\_{k=n}^{\infty} f\_k\right)^r \ge \sum\_{n=1}^{\infty} n^r f\_n^r. \tag{13}$$

**Theorem 10.** *Assume that f is a nonincreasing nonnegative function defined on* [0, ∞)*. If r* > 1*, then,*

$$\int\_0^\infty \left( \int\_\lambda^\infty f(\zeta) d\zeta \right)^p d\mathbf{x} \ge \int\_0^\infty \lambda^p f^r(\lambda) d\lambda. \tag{14}$$

In 1990, the reverses of inequalities (8) and (9) were demonstrated by Leindler in [10] as the following:

**Theorem 11.** *If* { *fn*} ∞ *n*=1 *and* {*θn*} ∞ *n*=1 *are sequences of nonnegative real numbers and* 0 < *r* ≤ 1*, then,*

$$\sum\_{n=1}^{\infty} \theta\_n \left( \sum\_{k=1}^n f\_k \right)^r \ge r^r \sum\_{n=1}^{\infty} \theta^{1-r}(n) f\_n^r \left( \sum\_{k=n}^{\infty} \theta\_k \right)^r,\tag{15}$$

*and*

$$\sum\_{n=1}^{\infty} \theta\_n \left( \sum\_{k=n}^{\infty} f\_k \right)^r \ge r^r \sum\_{n=1}^{\infty} \theta\_n^{1-r} f\_n^r \left( \sum\_{k=1}^n \theta\_k \right)^r. \tag{16}$$

Hilger, in his Ph.D. thesis [11], was the first one to accomplish the unification and extension of differential equations, difference equations, *q*-difference equations, and so on to the encompassing theory of dynamic equations on time scales.

Throughout this work, a knowledge and understanding of time scales and time-scale notation is assumed; for an excellent introduction to the calculus on time scales, see Bohner and Peterson [12,13].

In 2005, Rehák [ ˇ 14] was a forerunner in extending Hardy-type inequalities to time scales. He expanded the original Hardy inequalities (1) and (2) to a time scale of our choosing, and so, he combined them into a single form, as illustrated below.

**Theorem 12.** *Suppose* T *is a time scale, and f* ∈ *Crd*([*a*, ∞)T, [0, ∞))*. If r* ≥ 1*, then,*

$$\int\_{a}^{\infty} \left( \frac{\int\_{a}^{\sigma(\eta)} f(\zeta) \Delta \zeta}{\sigma(\eta) - a} \right)^{r} \Delta \eta < \left( \frac{r}{r - 1} \right)^{r} \int\_{a}^{\infty} f^{r}(\eta) \Delta \eta \,\,\tag{17}$$

*unless f* ≡ 0*.*

In 2017, Agarwal et al. [15] presented the next dynamic inequality.

**Theorem 13.** *Let* T *be a time scale such that* 0 ∈ T*. Moreover, assume f is a nonincreasing nonnegative function on* [0, ∞)T*. If r* > 1*, then,*

$$\int\_0^\infty \frac{1}{\eta^r} \left( \int\_0^\eta f(\zeta) \Delta \zeta \right)^r \Delta \eta \ge \frac{r}{r-1} \int\_0^\infty f^r(\eta) \Delta \eta. \tag{18}$$

Very recently, El-Deeb et al. [16] established the next dynamic inequalities.

**Theorem 14.** *Suppose* T *is a time scale with a* ∈ [0, ∞)T*. Additionally, suppose that f* > 0 *and θ* > 0 *are rd-continuous functions on* [*a*, ∞)T *and f is nonincreasing.*

(*i*) *If r* ≥ 1 *and γ* ≥ 0*, then*

$$\int\_{a}^{\infty} \frac{\theta(\eta) \left( \int\_{\eta}^{\infty} \theta(\zeta) f(\zeta) \Delta \zeta \right)^{r}}{\left( \int\_{a}^{\sigma(\eta)} \theta(\zeta) \Delta \zeta \right)^{\gamma}} \Delta \eta \geq \int\_{a}^{\infty} \frac{\theta(\eta) \left( \int\_{a}^{\eta} \theta(\zeta) \Delta \zeta \right)^{r} f^{r}(\eta)}{\left( \int\_{a}^{\sigma(\eta)} \theta(\zeta) \Delta \zeta \right)^{\gamma}} \Delta \eta. \tag{19}$$

(*ii*) *If r* ≥ 1 *and γ* > 1*, then*

$$\int\_{a}^{\infty} \frac{\theta(\eta) \left( \int\_{\eta}^{\infty} \theta(\zeta) f(\zeta) \Delta \zeta \right)^{r}}{\left( \int\_{\eta}^{\infty} \theta(\zeta) \Delta \zeta \right)^{\gamma}} \Delta \eta \geq \frac{r}{\gamma - 1} \int\_{a}^{\infty} \theta(\eta) \left( \int\_{\eta}^{\infty} \theta(\zeta) \Delta \zeta \right)^{r - \gamma} f^{r}(\eta) \Delta \eta. \tag{20}$$

(*iii*) *If r* ≥ 1 *and γ* > 1*, then*

$$\int\_{a}^{\infty} \frac{\theta(\eta) \left( \int\_{a}^{\eta} \theta(\zeta) f(\zeta) \Delta \zeta \right)^{r}}{\left( \int\_{a}^{\eta} \theta(\zeta) \Delta \zeta \right)^{\gamma}} \Delta \eta \geq \frac{r}{\gamma - 1} \int\_{a}^{\infty} \theta(\eta) \left( \int\_{a}^{\eta} \theta(\zeta) \Delta \zeta \right)^{r - \gamma} f^{r}(\eta) \Delta \eta. \tag{21}$$

(*iv*) *If r* ≥ 1 *and* 0 ≤ *γ* < 1*, then*

$$\int\_{a}^{\infty} \frac{\theta(\eta) \left( \int\_{a}^{\eta} \theta(\zeta) f(\zeta) \Delta \zeta \right)'}{\left( \int\_{\sigma(\eta)}^{\infty} \theta(\zeta) \Delta \zeta \right)^{\gamma}} \Delta \eta \geq \frac{r}{1-\gamma} \int\_{a}^{\infty} \theta(\eta) \left( \int\_{a}^{\eta} \theta(\zeta) \Delta \zeta \right)^{\gamma-1} \left( \int\_{\eta}^{\infty} \theta(\zeta) \Delta \zeta \right)^{1-\gamma} f'(\eta) \Delta \eta. \tag{22}$$

For more details on Hardy-type inequalities and other types on time scales, we suggest [17–29] for the reader.

**Theorem 15** (**Fubini's Theorem, see [Theorem 1.1, Page 300] [30]**)**.** *Assume that* (*λ*, Σ1, *µ*∆) *and* (*Y*, Σ2, *ν*∆) *are two finite-dimensional time scales measure spaces. Moreover, suppose that f* : *λ* × *Y* → R *is a delta integrable function and define the functions*

$$\Phi(y) = \int\_{\lambda} f(\lambda, y) d\mu\_{\Delta}(\lambda), \quad y \in \Upsilon\_{\nu}$$

*and*

$$\hat{\Psi}(\lambda) = \int\_Y f(\lambda, y) d\nu\_\Delta(y), \quad \lambda \in \lambda.$$

*Then,* Φ *is delta integrable on Y and* Ψˆ *is delta integrable on λ and*

$$\int\_{\lambda} d\mu\_{\Delta}(\lambda) \int\_{Y} f(\lambda, y) d\nu\_{\Delta}(y) = \int\_{Y} d\nu\_{\Delta}(y) \int\_{\lambda} f(\lambda, y) d\mu\_{\Delta}(\lambda).$$

The basic theorems that will be required in the proof of our results are presented next.

**Theorem 16** (**Chain rule on time scales, see [Theorem 1.87, Page 31] [12]**)**.** *Assume g* : R → R*, g* : T → R *is delta differentiable on* T *κ , and f* : R → R *is continuously differentiable. Then, there exists c* ∈ [*η*, *σ*(*η*)] *with*

$$(f \circ \mathfrak{g})^{\Delta}(\eta) = f'(\mathfrak{g}(\mathfrak{c})) \mathfrak{g}^{\Delta}(\eta). \tag{23}$$

**Theorem 17** (**Chain rule on time scales, see [Theorem 1.90, Page 32] [12]**)**.** *Let f* : R → R *be continuously differentiable and suppose g* : T → R *is delta differentiable. Then, f* ◦ *g* : T → R *is delta differentiable and the formula*

$$(f \circ g)^{\Delta}(\eta) = \left\{ \int\_0^1 \left[ f'(\hbar g^{\sigma}(\eta) + (1 - h)g(\eta)) \right] d\hbar \right\} \mathfrak{g}^{\Delta}(\eta).$$

*holds.*

In this manuscript, we show and prove some new dynamic Hardy-type which are reverse inequalities on time scales. The dynamic Hardy-type inequalities we obtained are entirely original, and as a result, we could obtain some integral and discrete inequalities of Hardy-type that are new. Furthermore, our findings generalize inequities (19)–(22). This paper is organized in the following way: Some basic concepts of the calculus on time scales and useful lemmas are introduced in Section 1. In Section 2, we state and prove the main results. In Section 3, we state the conclusion.

#### **2. Main Results**

The version of inequality (14) on time scales is given as a special case of the following theorem.

**Theorem 18.** *Assume that* T *is a time scale with* 0 ≤ *a* ∈ T*. Additionally, let f , g,* ˇ*ξ and θ be nonnegative functions defined on* [0, ∞)T *such that f and g are nonincreasing. Moreover, let* Ψˇ : R+ → R+ *be a differentiable function such that* Ψˇ 0 *is nondecreasing and* Ψˇ 0 (*xy*) = Ψˇ 0 (*x*)Ψˇ 0 (*y*) *for all x*, *y* ∈ R+*. If γ* ≥ 0*, then*

$$\int\_{a}^{\infty} \frac{\tilde{\xi}(\eta)g(\eta)\Psi\left(\int\_{\eta}^{\infty} \tilde{\xi}(\zeta)f(\zeta)\Delta\zeta\right)}{\left(\int\_{a}^{\sigma(\eta)}\theta(\zeta)\Delta\zeta\right)^{\gamma}}d\eta \geq \int\_{a}^{\infty} \frac{\tilde{\xi}(\eta)g(\eta)\Psi\left(\int\_{a}^{\eta}\tilde{\xi}(\zeta)\Delta\zeta\right)\Psi'(f(\eta))f(\eta)}{\left(\int\_{a}^{\sigma(\eta)}\theta(\zeta)\Delta\zeta\right)^{\gamma}}d\eta. \tag{24}$$

**Proof.** Owing to nonincreasity of *f* , we have for *λ* ≥ *η* ≥ *a*

$$\int\_{\eta}^{\lambda} \widetilde{\xi}(\zeta) f(\zeta) \Delta \zeta \ge f(\lambda) \int\_{\eta}^{\lambda} \widetilde{\xi}(\zeta) \Delta \zeta \nu$$

then, since Ψˇ 0 is nondecreasing,

$$\Psi'\left(\int\_{\eta}^{\lambda} \xi(\zeta)f(\zeta)d\zeta\right) \ge \Psi'\left(f(\lambda)\int\_{\eta}^{\lambda} \xi(\zeta)d\zeta\right) = \Psi'(f(\lambda))\Psi'\left(\int\_{\eta}^{\lambda} \xi(\zeta)d\zeta\right). \tag{25}$$

Applying the chain rule (23), there exists *c* ∈ [*λ*, *σ*(*λ*)] such that

$$\mathbb{E}\left[\Psi\left(\int\_{\eta}^{\lambda}\stackrel{\mathsf{x}}{\xi}(\zeta)f(\zeta)\Delta\zeta\right)\right]^{\Delta\_{\lambda}} = \Psi'\left(\int\_{\eta}^{c}\stackrel{\mathsf{x}}{\xi}(\zeta)f(\zeta)\Delta\zeta\right)\left(\int\_{\eta}^{\lambda}\stackrel{\mathsf{x}}{\xi}(\zeta)f(\zeta)\Delta\zeta\right)^{\Delta\_{\lambda}}.$$

Since *c* ≥ *λ*, Ψˇ 0 is nondecreasing, and Z *λ η* ˇ*ξ*(*ζ*)*f*(*ζ*)∆*ζ* ∆*λ* = ˇ*ξ*(*λ*)*f*(*λ*) ≥ 0, we have

$$\left[\Psi\left(\int\_{\eta}^{\lambda}\check{\xi}(\zeta)f(\zeta)\Delta\zeta\right)\right]^{\Delta\_{\lambda}}\geq\check{\xi}(\lambda)f(\lambda)\Psi'\left(\int\_{\eta}^{\lambda}\check{\xi}(\zeta)f(\zeta)\Delta\zeta\right).\tag{26}$$

Combining (25) with (26) yields

$$\left[\Psi\left(\int\_{\eta}^{\lambda}\xi(\zeta)f(\zeta)d\zeta\right)\right]^{\Delta\_{\lambda}}\geq\xi(\lambda)\Psi'\left(\int\_{\eta}^{\lambda}\xi(\zeta)d\zeta\right)\Psi'(f(\lambda))f(\lambda).$$

and so

$$\frac{\left(\widetilde{\xi}(\eta)\mathcal{g}(\eta)\left[\Psi\left(\int\_{\eta}^{\lambda}\widetilde{\xi}(\zeta)f(\zeta)\Delta\zeta\right)\right]\right)^{\Delta\_{\lambda}}}{\left(\int\_{a}^{\sigma(\eta)}\theta(\zeta)\Delta\zeta\right)^{\gamma}} \geq \frac{\widetilde{\xi}(\eta)\mathcal{g}(\eta)\widetilde{\xi}(\lambda)\Psi'\left(\int\_{\eta}^{\lambda}\widetilde{\xi}(\zeta)\Delta\zeta\right)\Psi'(f(\lambda))f(\lambda)}{\left(\int\_{a}^{\sigma(\eta)}\theta(\zeta)\Delta\zeta\right)^{\gamma}}.$$

Considering that *λ* ≥ *η* implies: (*i*) *σ*(*λ*) ≥ *σ*(*η*) and hence Z *σ*(*λ*) *a θ*(*ζ*)∆*ζ* ≥ Z *σ*(*η*) *a θ*(*ζ*)∆*ζ*; (*ii*) *g*(*λ*) ≤ *g*(*η*), we obtain

$$\frac{\left(\mathop{\xi}(\eta)\mathbbm{g}(\eta)\left[\Psi\left(\int\_{\eta}^{\lambda}\widetilde{\xi}(\zeta)f(\zeta)d\zeta\right)\right]\right)^{\Delta\_{\lambda}}}{\left(\int\_{a}^{\sigma(\eta)}\theta(\zeta)d\zeta\right)^{\gamma}} \geq \frac{\left(\xi(\eta)\widetilde{\xi}(\lambda)\mathbbm{g}(\lambda)\Psi'\left(\int\_{\eta}^{\lambda}\widetilde{\xi}(\zeta)d\zeta\right)\Psi'\left(f(\lambda)\right)f(\lambda)\right)^{\gamma}}{\left(\int\_{a}^{\sigma(\lambda)}\theta(\zeta)d\zeta\right)^{\gamma}}.$$

If we integrate both sides with respect to *λ* over [*η*, ∞)T, we obtain

$$\frac{\left(\xi(\eta)\,\mathfrak{g}(\eta)\,\Psi\left(\int\_{\eta}^{\infty}\frac{\mathfrak{x}}{\xi}(\zeta)f(\zeta)\,\Delta\zeta\right)\right)}{\left(\int\_{a}^{\sigma(\eta)}\theta(\zeta)\,\Delta\zeta\right)^{\gamma}} \geq \int\_{\eta}^{\infty} \frac{\,\frac{\mathfrak{x}}{\xi}(\eta)\,\mathfrak{f}(\lambda)\,\mathfrak{g}(\lambda)\,\Psi'\left(\int\_{\eta}^{\lambda}\frac{\mathfrak{x}}{\xi}(\zeta)\,\Delta\zeta\right)\Psi'\left(f(\lambda)\right)f(\lambda)}{\left(\int\_{a}^{\sigma(\lambda)}\theta(\zeta)\,\Delta\zeta\right)^{\gamma}}d\lambda.$$

If we integrate both sides once more, but with respect to *η* over [*a*, ∞)T, we obtain

$$\int\_{a}^{\infty} \frac{\xi(\eta)\varrho(\eta)\Psi\left(\int\_{\eta}^{\infty}\xi(\zeta)f(\zeta)\Delta\zeta\right)}{\left(\int\_{a}^{\tau(\eta)}\theta(\zeta)\Delta\zeta\right)^{\gamma}}d\eta \geq \int\_{a}^{\infty} \tilde{\xi}(\eta) \left(\int\_{\eta}^{\infty} \frac{\xi(\lambda)g(\lambda)\Psi'\left(\int\_{\eta}^{\lambda}\xi(\zeta)\Delta\zeta\right)\Psi'(f(\lambda))f(\lambda)}{\left(\int\_{a}^{\tau(\lambda)}\theta(\zeta)\Delta\zeta\right)^{\gamma}}d\lambda\right)d\eta. \tag{27}$$

By Using Fubini's theorem on time scales, (27) can be rewritten as

$$\int\_{a}^{\infty} \frac{\breve{\xi}(\eta)\varrho(\eta)\Psi\Big(\int\_{\eta}^{\infty} \breve{\xi}(\zeta)f(\zeta)\Delta\zeta\Big)}{\left(\int\_{a}^{\tau(\eta)}\theta(\zeta)\Delta\zeta\right)^{\gamma}}d\eta \geq \int\_{a}^{\infty} \frac{\breve{\xi}(\lambda)g(\lambda)\Psi'(f(\lambda))f(\lambda)\left(\int\_{a}^{\lambda}\breve{\xi}(\eta)\Psi'\left(\int\_{\eta}^{\lambda}\breve{\xi}(\zeta)\Delta\zeta\right)d\eta\right)}{\left(\int\_{a}^{\tau(\lambda)}\theta(\zeta)\Delta\zeta\right)^{\gamma}}d\lambda. \tag{28}$$

Now, from the chain rule (23), one can see that there exists *c* ∈ [*η*, *σ*(*η*)] with

$$\left[-\Psi\left(\int\_{\eta}^{\lambda}\not\xi(\zeta)\Delta\zeta\right)\right]^{\Delta\_{\eta}}=-\Psi'\left(\int\_{c}^{\lambda}\not\xi(\zeta)\Delta\zeta\right)\left(\int\_{\eta}^{\lambda}\not\xi(\zeta)\Delta\zeta\right)^{\Delta\_{\eta}}.$$

$$\left.\ldots\right.\quad\ldots\right.\quad\left.\ldots\right.$$

Since *c* ≥ *η*, Ψˇ 0 is nondecreasing, *r* ≥ 1 and Z *λ η* ˇ*ξ*(*ζ*)∆*ζ* ∆*η* = − ˇ*ξ*(*η*) ≤ 0, we have

$$\left[-\,\Psi\left(\int\_{\eta}^{\lambda}\xi(\zeta)\,\Delta\zeta\right)\right]^{\Delta\eta}\leq\!\left(\eta\right)\Psi'\left(\int\_{\eta}^{\lambda}\xi(\zeta)\,\Delta\zeta\right).\tag{29}$$

Substituting (29) into (28) leads to

$$\begin{split} &\int\_{a}^{\infty} \frac{\breve{\xi}(\eta)g(\eta)\Psi\left(\int\_{\eta}^{\infty} \breve{\xi}(\xi)f(\xi)\Delta\zeta\right)}{\left(\int\_{a}^{\sigma(\eta)} \theta(\xi)\Delta\zeta\right)^{\gamma}}d\eta \\ &\geq \int\_{a}^{\infty} \frac{\breve{\xi}(\lambda)g(\lambda)\Psi'(f(\lambda))f(\lambda)\left(\int\_{a}^{\lambda} \left[-\varPsi\left(\int\_{\eta}^{\lambda} \breve{\xi}(\xi)\Delta\zeta\right)\right]^{\Delta\_{\eta}}\Delta\eta\right)}{\left(\int\_{a}^{\sigma(\lambda)} \theta(\xi)\Delta\zeta\right)^{\gamma}}d\Delta\lambda \\ &= \int\_{a}^{\infty} \frac{\breve{\xi}(\lambda)g(\lambda)\Psi\left(\int\_{a}^{\lambda} \breve{\xi}(\xi)\Delta\zeta\right)\Psi'(f(\lambda))f(\lambda)}{\left(\int\_{a}^{\sigma(\lambda)} \theta(\xi)\Delta\zeta\right)^{\gamma}}d\Delta\lambda. \end{split}$$

This shows the validity of (24).

**Remark 1.** *In Theorem 18, if we take* Ψˇ (*η*) = *η r , r* ≥ 1*,* ˇ*ξ*(*η*) = *θ*(*η*) *and g*(*η*) = 1*, then inequality* (24) *reduces to inequality* (19)*.*

**Corollary 1.** *In Theorem 18, if we take* Ψˇ (*η*) = *η r ,* ˇ*ξ*(*η*) = *g*(*η*) = 1 *and a* = *γ* = 0*, then inequality* (24) *reduces to*

$$\int\_0^\infty \left( \int\_\eta^\infty f(\zeta) \, \Delta \zeta \right)' \Delta \eta \ge \int\_0^\infty \left( \int\_0^\eta \check{\xi}(\zeta) \, \Delta \zeta \right)' f'(\eta) \, \Delta \eta \, \zeta$$

*which is the time scales version of* (14)*.*

**Corollary 2.** *If* T = R *in Theorem 18, then inequality* (24) *reduces to*

$$\int\_{a}^{\infty} \frac{\tilde{\xi}(\eta)g(\eta)\Psi\left(\int\_{\eta}^{\infty} \tilde{\xi}(\zeta)f(\zeta)d\zeta\right)}{\left(\int\_{a}^{\eta}\theta(\zeta)d\zeta\right)^{\gamma}}d\eta \geq \int\_{a}^{\infty} \frac{\tilde{\xi}(\eta)g(\eta)\Psi\left(\int\_{a}^{\eta}\tilde{\xi}(\zeta)d\zeta\right)\Psi'(f(\eta))f(\eta)}{\left(\int\_{a}^{\eta}\theta(\zeta)d\zeta\right)^{\gamma}}d\eta.$$

**Remark 2.** *In Corollary 2, if we take* Ψˇ (*η*) = *η r ,* ˇ*ξ*(*η*) = *g*(*η*) = 1*, a* = *γ* = 0*, then we reclaim inequality* (14)*.*

**Corollary 3.** *If* T = *h*Z *in Theorem 18, then inequality* (24) *is reduced to*

$$\sum\_{n=\frac{\mathfrak{s}}{\mathfrak{h}}}^{\infty} \frac{\mathfrak{z}'(nh)\mathfrak{g}(nh)\Psi\left(h\sum\_{m=\frac{\mathfrak{s}}{\mathfrak{h}}}^{\infty}\mathfrak{z}(mh)f(mh)\right)}{\left(\sum\_{m=\frac{\mathfrak{s}}{\mathfrak{h}}}^{\frac{\mathfrak{s}}{\mathfrak{h}}}\theta(mh)\right)^{\gamma}} \geq \sum\_{n=\frac{\mathfrak{s}}{\mathfrak{s}}}^{\infty} \frac{\mathfrak{z}'(nh)\mathfrak{g}(nh)\Psi\left(h\sum\_{m=\frac{\mathfrak{s}}{\mathfrak{h}}}^{\frac{\mathfrak{s}}{\mathfrak{s}}-1}\xi(mh)\right)\Psi'\left(f(nh)\right)f(nh)}{\left(\sum\_{m=\frac{\mathfrak{s}}{\mathfrak{h}}}^{\frac{\mathfrak{s}}{\mathfrak{s}}}\theta(mh)\right)^{\gamma}}.$$

**Corollary 4.** *In Corollary 3, if we take h* = 1*, then, inequality* (24) *will be reduced to*

$$\sum\_{n=a}^{\infty} \frac{\xi(n)\varrho(n)\Psi\left(\sum\_{m=n}^{\infty}\xi(m)f(m)\right)}{\left(\sum\_{m=a}^{n}\theta(m)\right)^{\gamma}} \geq \sum\_{n=a}^{\infty} \frac{\xi(n)\varrho(n)\Psi\left(\sum\_{m=a}^{n-1}\xi(m)\right)\Psi'(f(n))f(n)}{\left(\sum\_{m=a}^{n}\theta(m)\right)^{\gamma}}.$$

**Remark 3.** *In Corollary 4, if we take* Ψˇ (*η*) = *η r ,* ˇ*ξ*(*η*) = *g*(*η*) = 1*, a* = 1 *and γ* = 0*, then we reclaim inequality* (13)*.*

**Corollary 5.** *If* T = *q* Z *in Theorem 18, then*

$$\begin{split} &\sum\_{n=(\log\_q a)}^{\infty} \frac{q^n \breve{\xi}(q^n) \mathcal{g}(q^n) \Psi\left((q-1) \sum\_{m=(\log\_q a)}^{\infty} q^m \breve{\xi}(q^m) f(q^m)\right)}{\left(\sum\_{m=(\log\_q a)}^{\infty} q^m h(q^m)\right)^{\gamma}} \\ &\geq \sum\_{n=(\log\_q a)}^{\infty} \frac{q^n \breve{\xi}(q^n) \mathcal{g}(q^n) \Psi\left((q-1) \sum\_{m=(\log\_q a)}^{\infty} q^m \breve{\xi}(q^m)\right) \Psi'\left(f(q^n)\right) f(q^n)}{\left(\sum\_{m=(\log\_q a)}^{\infty} q^m h(q^m)\right)^{\gamma}}. \end{split}$$

Now, as a new result, we are interested in discussing the inequality (24) in the case of the extrema of integration R *η a θ*(*s*)∆*s* being replaced to be from *η* to ∞. In fact, that is what we will do in the following theorem.

**Theorem 19.** *Assume that* T *is a time scale with* 0 ≤ *a* ∈ T*. Additionally, let f , g, θ and* ˇ*ξ be nonnegative functions defined on* [0, ∞)T *such that f and g are nonincreasing. Furthermore, let* Ψˇ : R+ → R+ *be a differentiable function such that* Ψˇ 0 *is nondecreasing and* Ψˇ 0 (*xy*) = Ψˇ 0 (*x*)Ψˇ 0 (*y*) *for all x*, *y* ∈ R+*. If γ* > 1*, then*

$$\int\_{a}^{\infty} \frac{\tilde{\xi}(\eta)\varrho(\eta)\Psi\Big(\int\_{\eta}^{\infty}\theta(\zeta)f(\zeta)\Delta\zeta\Big)}{\left(\int\_{\eta}^{\infty}\tilde{\xi}(\zeta)\Delta\zeta\right)^{\gamma}}d\eta \geq \frac{1}{\gamma-1}\int\_{a}^{\infty} \frac{\theta(\eta)\varrho(\eta)\Psi^{\prime}\Big(\int\_{\eta}^{\infty}\theta(\zeta)\Delta\zeta\Big)\Psi^{\prime}(f(\eta))f(\eta)}{\left(\int\_{\eta}^{\infty}\tilde{\xi}(\zeta)\Delta\zeta\right)^{\gamma-1}}d\eta. \tag{30}$$

**Proof.** Because of nonincreasity of *f* , we have for *η* ≥ *λ* ≥ *a*

$$\int\_{\lambda}^{\infty} \theta(\zeta) f(\zeta) \Delta \zeta \le f(\lambda) \int\_{\lambda}^{\infty} \theta(\zeta) \Delta \zeta \nu$$

therefore, because Ψˇ 0 is nondecreasing,

$$\Psi'\left(\int\_{\lambda}^{\infty}\theta(\zeta)f(\zeta)d\zeta\right) \ge \Psi'\left(f(\lambda)\int\_{\lambda}^{\infty}\theta(\zeta)d\zeta\right) = \Psi'(f(\lambda))\Psi'\left(\int\_{\lambda}^{\infty}\theta(\zeta)d\zeta\right). \tag{31}$$

From the chain rule (23), we see that there is *c* ∈ [*λ*, *σ*(*λ*)] with

$$\left[\Psi\left(\int\_{\lambda}^{\infty}\theta(\zeta)f(\zeta)\Delta\zeta\right)\right]^\Delta = \Psi'\left(\int\_{\varepsilon}^{\infty}\theta(\zeta)f(\zeta)\Delta\zeta\right)\left(\int\_{\lambda}^{\infty}\theta(\zeta)f(\zeta)\Delta\zeta\right)^\Delta.$$

Since *c* ≥ *λ*, Ψˇ 0 is nondecreasing, *r* ≥ 1 and Z ∞ *λ θ*(*ζ*)*f*(*ζ*)∆*ζ* ∆ = −*θ*(*λ*)*f*(*λ*) ≤ 0, we have

$$\left[\Psi\left(\int\_{\lambda}^{\infty}\theta(\zeta)f(\zeta)\,\Delta\zeta\right)\right]^{\Delta}\geq-\theta(\lambda)f(\lambda)\Psi'\left(\int\_{\lambda}^{\infty}\theta(\zeta)f(\zeta)\,\Delta\zeta\right)\tag{32}$$

Combining (31) with (32) yields

$$\left[\Psi\left(\int\_{\lambda}^{\infty}\theta(\zeta)f(\zeta)\Delta\zeta\right)\right]^\Delta \geq -\theta(\lambda)\Psi'\left(\int\_{\lambda}^{\infty}\theta(\zeta)\Delta\zeta\right)\Psi'\left(f(\lambda)\right)f(\lambda)\lambda$$

which implies

$$\frac{\left(\widetilde{\xi}(\eta)\underline{\boldsymbol{g}}(\eta)\left[\boldsymbol{\Psi}\left(\int\_{\lambda}^{\infty}\theta(\boldsymbol{\zeta})\boldsymbol{f}(\boldsymbol{\zeta})\,\Delta\boldsymbol{\zeta}\right)\right]\right)^{\Lambda}}{\left(\int\_{\eta}^{\infty}\widetilde{\xi}(\boldsymbol{\zeta})\,\Delta\boldsymbol{\zeta}\right)^{\gamma}} \geq \frac{-\widetilde{\xi}(\eta)\underline{\boldsymbol{g}}(\eta)\theta(\boldsymbol{\lambda})\boldsymbol{\Psi}'\left(\int\_{\lambda}^{\infty}\theta(\boldsymbol{\zeta})\,\Delta\boldsymbol{\zeta}\right)\boldsymbol{\Psi}'(\boldsymbol{f}(\boldsymbol{\lambda}))\boldsymbol{f}(\boldsymbol{\lambda})}{\left(\int\_{\eta}^{\infty}\widetilde{\xi}(\boldsymbol{\zeta})\,\Delta\boldsymbol{\zeta}\right)^{\gamma}}.$$

As *g* is nonincreasing and *λ* ≤ *η*, we have *g*(*λ*) ≥ *g*(*η*) and hence,

$$\frac{\left[\tilde{\xi}(\eta)\varrho(\eta)\left[\Psi\left(\int\_{\lambda}^{\infty}\theta(\zeta)f(\zeta)\Delta\zeta\right)\right]\right]^\Delta}{\left(\int\_{\eta}^{\infty}\tilde{\xi}(\zeta)\Delta\zeta\right)^\gamma} \geq \frac{-\check{\xi}(\eta)\theta(\lambda)\varrho(\lambda)\Psi'\left(\int\_{\lambda}^{\infty}\theta(\zeta)\Delta\zeta\right)\Psi'\left(f(\lambda)\right)f(\lambda)}{\left(\int\_{\eta}^{\infty}\check{\xi}(\zeta)\Delta\zeta\right)^\gamma}.$$

Now, after both sides are integrated with respect to *λ* over [*a*, *η*]T, we could have

$$\begin{split} & \frac{\widetilde{\xi}(\eta)\mathcal{g}(\eta)\left[\Psi\left(\int\_{\eta}^{\infty}\theta(\zeta)f(\zeta)\Delta\zeta\right)-\Psi\left(\int\_{a}^{\infty}\theta(\zeta)f(\zeta)\Delta\zeta\right)\right]}{\left(\int\_{\eta}^{\infty}\widetilde{\xi}(\zeta)\Delta\zeta\right)^{\gamma}} \\ & \geq \int\_{a}^{\eta}\frac{-\widetilde{\xi}(\eta)\theta(\lambda)\mathcal{g}(\lambda)\Psi'\left(\int\_{\lambda}^{\infty}\theta(\zeta)\Delta\zeta\right)\Psi'(f(\lambda))f(\lambda)}{\left(\int\_{\eta}^{\infty}\widetilde{\xi}(\zeta)\Delta\zeta\right)^{\gamma}}d\lambda. \end{split}$$

$$\begin{split} \text{Since } & \Psi\left(\int\_{\eta}^{\infty} \theta(\zeta)f(\zeta)d\zeta\right) \geq \Psi\left(\int\_{\eta}^{\infty} \theta(\zeta)f(\zeta)d\zeta\right) - \Psi\left(\int\_{a}^{\infty} \theta(\zeta)f(\zeta)d\zeta\right), \text{ we have} \\ \frac{\int\_{\eta}^{\infty} (\eta)g(\eta)\Psi\left(\int\_{\eta}^{\infty} \theta(\zeta)f(\zeta)d\zeta\right)}{\left(\int\_{\eta}^{\infty} \check{\xi}(\zeta)\Delta\zeta\right)^{\gamma}} \geq -\int\_{a}^{\eta} \check{\xi}(\eta)\theta(\lambda)g(\lambda)\Psi'\left(\int\_{\lambda}^{\infty} \theta(\zeta)\Delta\zeta\right) \left(\int\_{\eta}^{\infty} \check{\xi}(\zeta)\Delta\zeta\right)^{-\gamma}\Psi'(f(\lambda))f(\lambda)d\lambda. \end{split}$$

Afterwards, if both sides are integrated with respect to *η* over [*a*, ∞)T, we obtain

$$\begin{split} &\int\_{a}^{\infty} \frac{\mathop{\!}^{\sharp}\_{\xi}(\eta)\varrho(\eta)\Psi\Big(\int\_{\eta}^{\infty}\theta(\zeta)f(\zeta)\Delta\zeta\Big)}{\left(\int\_{\eta}^{\infty}\xi(\zeta)\Delta\zeta\right)^{\gamma}}d\eta\\ &\geq -\int\_{a}^{\infty}\mathop{\!}^{\sharp}\_{\xi}(\eta)\left(\int\_{\eta}^{\infty}\xi(\zeta)\Delta\zeta\right)^{-\gamma}\left(\int\_{a}^{\eta}\theta(\lambda)g(\lambda)\Psi'\Big(\int\_{\lambda}^{\infty}\theta(\zeta)\Delta\zeta\Big)\Psi'(f(\lambda))f(\lambda)d\lambda\right)d\eta. \end{split} \tag{33}$$

Using Fubini's theorem on time scales, (33) can be rewritten as

$$\begin{split} &\int\_{a}^{\infty} \frac{\tilde{\xi}(\eta)g(\eta)\Psi\left(\int\_{\eta}^{\infty}\theta(\zeta)f(\zeta)\Delta\zeta\right)}{\left(\int\_{\eta}^{\infty}\tilde{\xi}(\zeta)\Delta\zeta\right)^{\gamma}}d\eta\\ &\geq -\int\_{a}^{\infty}\theta(\lambda)g(\lambda)\tilde{\Psi}'\left(\int\_{\lambda}^{\infty}\theta(\zeta)\Delta\zeta\right)\tilde{\Psi}'(f(\lambda))f(\lambda)\left(\int\_{\lambda}^{\infty}\tilde{\xi}(\eta)\left(\int\_{\eta}^{\infty}\tilde{\xi}(\zeta)\Delta\zeta\right)^{-\gamma}d\eta\right)d\lambda. \end{split} \tag{34}$$

If we take a look at the chain rule, (23), we could say that there exists *c* ∈ [*η*, *σ*(*η*)] such that

$$\left[-\left(\int\_{\eta}^{\infty} \frac{\mathfrak{z}}{\xi}(\zeta) \Delta \zeta\right)^{1-\gamma}\right]^{\Delta} = -(1-\gamma) \left(\int\_{c}^{\infty} \widetilde{\xi}(\zeta) \Delta \zeta\right)^{-\gamma} \left(\int\_{\eta}^{\infty} \widetilde{\xi}(\zeta) \Delta \zeta\right)^{\Delta}.$$

$$\text{Since } c \ge \eta, \gamma > 1 \text{ and } \left(\int\_{\eta}^{\infty} \widetilde{\xi}(\zeta) \Delta \zeta\right)^{\Delta} = -\widetilde{\xi}(\eta) \le 0, \text{ we get}$$

$$\left[-\left(\int\_{\eta}^{\infty} \widetilde{\xi}(\zeta) \Delta \zeta\right)^{1-\gamma}\right]^{\Delta} \ge -(\gamma-1)\widetilde{\xi}(\eta) \left(\int\_{\eta}^{\infty} \widetilde{\xi}(\zeta) \Delta \zeta\right)^{-\gamma}.\tag{35}$$

Substituting (35) into (34) leads to

$$\begin{split} &\int\_{a}^{\infty} \frac{\widetilde{\xi}(\eta)g(\eta)\Psi\left(\int\_{\eta}^{\infty}\theta(\zeta)f(\zeta)\Delta\zeta\right)}{\left(\int\_{\eta}^{\infty}\widetilde{\xi}(\zeta)\Delta\zeta\right)^{\gamma}}d\eta \\ &\geq \frac{1}{\gamma-1}\int\_{a}^{\infty}\theta(\lambda)g(\lambda)\Psi'\left(\int\_{\lambda}^{\infty}\theta(\zeta)\Delta\zeta\right)\Psi'(f(\lambda))f(\lambda)\left(\int\_{\lambda}^{\infty}\left[-\left(\int\_{\eta}^{\infty}\widetilde{\xi}(\zeta)\Delta\zeta\right)^{1-\gamma}\right]^{\Delta}\Delta\eta\right)d\lambda \\ &=\frac{1}{\gamma-1}\int\_{a}^{\infty}\theta(\lambda)g(\lambda)\Psi'\left(\int\_{\lambda}^{\infty}\theta(\zeta)\Delta\zeta\right)\left(\int\_{\lambda}^{\infty}\widetilde{\xi}(\zeta)\Delta\zeta\right)^{1-\gamma}\Psi'(f(\lambda))f(\lambda)\Delta\lambda. \end{split}$$

from which inequality (30) follows.

**Remark 4.** *In Theorem 19, if we take* Ψˇ (*η*) = *η r ,* ˇ*ξ*(*η*) = *θ*(*η*) *and g*(*η*) = 1*, then inequality* (30) *reduces to inequality* (20)*.*

**Corollary 6.** *If* T = R *in Theorem 19, then, inequality* (30) *will be reduced to*

$$\int\_{a}^{\infty} \frac{\xi(\eta)\varrho(\eta)\Psi\left(\int\_{\eta}^{\infty}\theta(\zeta)f(\zeta)d\zeta\right)}{\left(\int\_{\eta}^{\infty}\check{\xi}(\zeta)d\zeta\right)^{\gamma}}d\eta \geq \frac{1}{\gamma-1}\int\_{a}^{\infty} \frac{\theta(\eta)\varrho(\eta)\Psi'\left(\int\_{\eta}^{\infty}\theta(\zeta)d\zeta\right)\Psi'(f(\eta))f(\eta)}{\left(\int\_{\eta}^{\infty}\check{\xi}(\zeta)d\zeta\right)^{\gamma-1}}d\eta.$$

**Corollary 7.** *If* T = *h*Z *in Theorem 19, then inequality* (30) *is reduced to*

$$\sum\_{n=\frac{\mathfrak{s}}{\mathfrak{s}}}^{\infty} \frac{\widetilde{\xi}(nh)\mathfrak{g}(nh)\Psi\left(h\sum\_{m=\frac{\mathfrak{s}}{\mathfrak{s}}}^{\infty}\theta(mh)f(mh)\right)}{\left(h\sum\_{m=\frac{\mathfrak{s}}{\mathfrak{s}}}^{\infty}\widetilde{\xi}(mh)\right)^{\gamma}} \geq \frac{1}{\gamma-1} \sum\_{n=\frac{\mathfrak{s}}{\mathfrak{s}}}^{\infty} \frac{\theta(nh)\mathfrak{g}(nh)\Psi'\left(h\sum\_{m=\frac{\mathfrak{s}}{\mathfrak{s}}}^{\infty}\theta(mh)\right)\Psi'\left(f(nh)\right)f(nh)}{\left(h\sum\_{m=\frac{\mathfrak{s}}{\mathfrak{s}}}^{\infty}\widetilde{\xi}(mh)\right)^{\gamma-1}}.$$

**Corollary 8.** *In Corollary 7, if we take h* = 1*, then inequality* (30) *reduces to*

$$\sum\_{n=a}^{\infty} \frac{\tilde{\xi}(n)\varrho(n)\Psi\left(\sum\_{m=n}^{\infty}\theta(m)f(m)\right)}{\left(\sum\_{m=n}^{\infty}\tilde{\xi}(m)\right)^{\gamma}} \geq \frac{1}{\gamma-1} \sum\_{n=a}^{\infty} \frac{\theta(n)\varrho(n)\Psi'\left(\sum\_{m=n}^{\infty}\theta(m)\right)\Psi'(f(n))f(n)}{\left(\sum\_{m=n}^{\infty}\tilde{\xi}(m)\right)^{\gamma-1}}.$$

**Corollary 9.** *If* T = *q* Z *in Theorem 19, then inequality* (30) *will be reduced to*

$$\begin{split} &\sum\_{n=(\log\_q a)}^{\infty} \frac{q^n \breve{\xi}(q^n) \mathfrak{g}(q^n) \Psi\left((q-1) \sum\_{m=(\log\_q n)}^{\infty} q^m h(q^m) f(q^m)\right)}{\left((q-1) \sum\_{m=(\log\_q n)}^{\infty} q^m \breve{\xi}(q^m)\right)^{\gamma}} \\ &\geq \frac{1}{\gamma-1} \sum\_{n=(\log\_q a)}^{\infty} \frac{q^n h(q^n) \mathfrak{g}(q^n) \breve{\Psi}'\left((q-1) \sum\_{m=(\log\_q n)}^{\infty} q^m h(q^m)\right) \breve{\Psi}'\left(f(q^n)\right) f(q^n)}{\left((q-1) \sum\_{m=(\log\_q n)}^{\infty} q^m \breve{\xi}(q^m)\right)^{\gamma-1}}. \end{split}$$

In the next theorem, we make a broad popularization of Theorem 13.

**Theorem 20.** *Let* T *be a time scale with* 0 ≤ *a* ∈ T*. Moreover, suppose that f , g, θ and* ˇ*ξ are nonnegative functions defined on* [0, ∞)T *such that f is nonincreasing and g is nondecreasing. In addition, let* Ψˇ : R+ → R+ *be a differentiable function such that* Ψˇ 0 *is nondecreasing and* Ψˇ 0 (*xy*) = Ψˇ 0 (*x*)Ψˇ 0 (*y*) *for all x*, *y* ∈ R+*. If γ* > 1*, then*

$$\int\_{a}^{\infty} \frac{\xi(\eta)\varrho(\eta)\Psi\left(\int\_{0}^{\eta} \theta(\zeta)f(\zeta)d\zeta\right)}{\left(\int\_{0}^{\eta}\xi(\zeta)d\zeta\right)^{\gamma}}d\eta \geq \frac{1}{\gamma-1} \int\_{a}^{\infty} \frac{\theta(\eta)g(\eta)\Psi'\left(\int\_{0}^{\eta}\theta(\zeta)d\zeta\right)\Psi'(f(\eta))f(\eta)}{\left(\int\_{0}^{\eta}\xi(\zeta)d\zeta\right)^{\gamma-1}}d\eta. \tag{36}$$

**Proof.** As a result of of the nonincreasity of *f* , we have for *η* ≥ *λ* ≥ 0

$$\int\_0^\lambda \theta(\zeta) f(\zeta) \Delta \zeta \ge f(\lambda) \int\_0^\lambda \theta(\zeta) \Delta \zeta \nu$$

then, since Ψˇ 0 is nondecreasing,

$$\Psi'\left(\int\_0^\lambda \theta(\zeta)f(\zeta)\Delta\zeta\right) \ge \Psi'\left(f(\lambda)\int\_0^\lambda \theta(\zeta)\Delta\zeta\right) = \Psi'(f(\lambda))\Psi'\left(\int\_0^\lambda \theta(\zeta)\Delta\zeta\right). \tag{37}$$

Using the chain rule (23), there exists *c* ∈ [*λ*, *σ*(*λ*)] such that

$$\left[\Psi\left(\int\_0^\lambda \theta(\zeta)f(\zeta)\Delta\zeta\right)\right]^\Delta = \Psi'\left(\int\_0^c \theta(\zeta)f(\zeta)\Delta\zeta\right)\left(\int\_0^\lambda \theta(\zeta)f(\zeta)\Delta\zeta\right)^\Delta.$$

Since *c* ≥ *λ*, Ψˇ 0 is nondecreasing, *r* ≥ 1 and Z *λ* 0 *θ*(*ζ*)*f*(*ζ*)∆*ζ* ∆ = *θ*(*λ*)*f*(*λ*) ≥ 0, we have

$$\left[\Psi\left(\int\_0^\lambda \theta(\zeta)f(\zeta)d\zeta\right)\right]^\Delta \ge \theta(\lambda)f(\lambda)\Psi'\left(\int\_0^\lambda \theta(\zeta)f(\zeta)d\zeta\right). \tag{38}$$

By using (37) and (38) together we could have

$$\left[\Psi'\left(\int\_0^\lambda \theta(\zeta)f(\zeta)\Delta\zeta\right)\right]^\Delta \ge \theta(\lambda)\Psi'\left(\int\_0^\lambda \theta(\zeta)\Delta\zeta\right)\Psi'\left(f(\lambda)\right)f(\lambda).$$

and thus

$$\frac{\left(\mathop{\xi}(\eta)\operatorname{g}(\eta)\left[\Psi\left(\int\_{0}^{\lambda}\theta(\zeta)f(\zeta)\Delta\zeta\right)\right]\right)^{\Delta}}{\left(\int\_{0}^{\eta}\xi(\zeta)\Delta\zeta\right)^{\gamma}} \geq \frac{\left(\operatorname{\widetilde{\xi}}(\eta)\operatorname{g}(\eta)\theta(\lambda)\Psi'\left(\int\_{0}^{\lambda}\theta(\zeta)\Delta\zeta\right)\Psi'(f(\lambda))f(\lambda)\right)^{\gamma}}{\left(\int\_{0}^{\eta}\xi(\zeta)\Delta\zeta\right)^{\gamma}}.$$

As *g* is nondecreasing and *λ* ≤ *η*, we have *g*(*λ*) ≤ *g*(*η*) and hence,

$$\frac{\left(\widetilde{\xi}(\eta)g(\eta)\left[\Psi\left(\int\_0^\lambda \theta(\zeta)f(\zeta)d\zeta\right)\right]\right)^\Delta}{\left(\int\_0^\eta \widetilde{\xi}(\zeta)d\zeta\right)^\gamma} \ge \frac{\widetilde{\xi}(\eta)g(\lambda)\theta(\lambda)\Psi'\left(\int\_0^\lambda \theta(\zeta)d\zeta\right)\Psi'(f(\lambda))f(\lambda)}{\left(\int\_0^\eta \widetilde{\xi}(\xi)d\zeta\right)^\gamma}.$$

Integrating both sides of the last inequality with respect to *λ* over [0, *η*]T gives

$$\begin{split} \frac{\xi(\eta)\varrho(\eta)\left[\Psi\left(\int\_{0}^{\eta}\theta(\xi)f(\xi)\Delta\xi\right)-\Psi\left(\int\_{0}^{\eta}\theta(\xi)f(\xi)\Delta\xi\right)\right]}{\left(\int\_{0}^{\eta}\xi(\xi)\Delta\xi\right)^{\gamma}} \\ &\geq \int\_{a}^{\eta} \frac{\xi(\eta)\theta(\lambda)\varrho(\lambda)\Psi'\left(\int\_{0}^{\lambda}\theta(\xi)\Delta\xi\right)\Psi'\left(f(\lambda)\right)f(\lambda)}{\left(\int\_{0}^{\eta}\xi(\xi)\Delta\xi\right)^{\gamma}}d\lambda. \\ \text{where} \quad \eta\left(\int\_{0}^{\eta}\theta(\xi)\Delta\xi\right)\leq\_{\mathsf{X}}\xi\left(\int\_{0}^{\eta}\theta(\xi)f(\xi)\Delta\xi\right). \end{split}$$

$$\begin{aligned} \text{Since } & \Psi \left( \int\_0^\eta \theta(\zeta) f(\zeta) \, d\zeta \right) \ge \Psi \left( \int\_0^\eta \theta(\zeta) f(\zeta) \, d\zeta \right) - \Psi \left( \int\_0^a \theta(\zeta) f(\zeta) \, d\zeta \right), \text{ we obtain} \\ \text{then } & \dots \end{aligned}$$

$$\frac{\int\_{\mathfrak{a}} \breve{\xi}(\eta) g(\eta) \Psi \Big( \int\_{0}^{\eta} \theta(\zeta) f(\zeta) \Delta \zeta \Big)}{\left( \int\_{0}^{\eta} \breve{\xi}(\zeta) \Delta \zeta \right)^{\gamma}} \geq \int\_{a}^{\eta} \breve{\xi}(\eta) \theta(\lambda) g(\lambda) \Psi' \Big( \int\_{0}^{\lambda} \theta(\zeta) \Delta \zeta \Big) \Big( \int\_{0}^{\eta} \breve{\xi}(\zeta) \Delta \zeta \Big)^{-\gamma} \Psi'(f(\lambda)) f(\lambda) \Delta \zeta$$

After integrating both sides with respect to *η* over [*a*, ∞)T,

$$\begin{split} &\int\_{a}^{\infty} \frac{\tilde{\xi}(\eta)\varrho(\eta)\Psi\Big(\int\_{0}^{\eta}\theta(\zeta)f(\zeta)\Delta\zeta\Big)}{\left(\int\_{0}^{\eta}\tilde{\xi}(\zeta)\Delta\zeta\right)^{\gamma}}d\eta\\ &\geq\int\_{a}^{\infty}\tilde{\xi}(\eta)\left(\int\_{0}^{\eta}\tilde{\xi}(\zeta)\Delta\zeta\right)^{-\gamma}\left(\int\_{a}^{\eta}\theta(\lambda)g(\lambda)\Psi'\Big(\int\_{0}^{\lambda}\theta(\zeta)\Delta\zeta\Big)\Psi'(f(\lambda))f(\lambda)d\lambda\right)d\eta. \end{split} \tag{39}$$

Employing Fubini's theorem on time scales, (39) can be rewritten as

$$\begin{split} &\int\_{a}^{\infty} \frac{\tilde{\xi}(\eta)g(\eta)\Psi\left(\int\_{0}^{\eta}\theta(\zeta)f(\zeta)\Delta\zeta\right)}{\left(\int\_{0}^{\eta}\tilde{\xi}(\zeta)\Delta\zeta\right)^{\gamma}}d\eta\\ &\geq\int\_{a}^{\infty}\theta(\lambda)g(\lambda)\Psi'\left(\int\_{0}^{\lambda}\theta(\zeta)\Delta\zeta\right)\Psi'(f(\lambda))f(\lambda)\left(\int\_{\lambda}^{\infty}\tilde{\xi}(\eta)\left(\int\_{0}^{\eta}\tilde{\xi}(\zeta)\Delta\zeta\right)^{-\gamma}\Delta\eta\right)d\lambda. \end{split} \tag{40}$$

Additionally, by taking a look at the chain rule (23), we can say that there exists *c* ∈ [*η*, *σ*(*η*)] such that

$$\left[-\left(\int\_0^\eta \tilde{\xi}(\zeta)\Delta\zeta\right)^{1-\gamma}\right]^\Delta = -(1-\gamma)\left(\int\_0^c \tilde{\xi}(\zeta)\Delta\zeta\right)^{-\gamma}\left(\int\_0^\eta \tilde{\xi}(\zeta)\Delta\zeta\right)^\Delta.$$

$$\text{Since } c \ge \eta, \gamma > 1 \text{ and } \left(\int\_0^\eta \underline{\xi}(\zeta) \, \Delta\zeta \right)^\Delta = \underline{\xi}(\eta) \ge 0 \text{, we get}$$

$$\left[ - \left( \int\_0^\eta \underline{\xi}(\zeta) \, \Delta\zeta \right)^{1-\gamma} \right]^\Delta \le (\gamma - 1)\underline{\xi}(\eta) \left( \int\_0^\eta \underline{\xi}(\zeta) \, \Delta\zeta \right)^{-\gamma}.\tag{41}$$

Substituting (41) into (40) leads to

$$\begin{split} &\int\_{a}^{\infty} \frac{\breve{\xi}(\eta)g(\eta)\Psi\Big(\int\_{0}^{\eta}\theta(\zeta)f(\zeta)\Delta\zeta\Big)}{\left(\int\_{0}^{\eta}\breve{\xi}(\zeta)\Delta\zeta\right)^{\gamma}}d\eta\\ &\geq \frac{1}{\gamma-1}\int\_{a}^{\infty}\theta(\lambda)g(\lambda)\Psi'\Big(\int\_{0}^{\lambda}\theta(\zeta)\Delta\zeta\Big)\Psi'(f(\lambda))f(\lambda)\left(\int\_{\lambda}^{\infty}\left[-\left(\int\_{0}^{\eta}\breve{\xi}(\zeta)\Delta\zeta\right)^{1-\gamma}\right]^{\Delta}\Delta\eta\right)d\lambda.\\ &=\frac{1}{\gamma-1}\int\_{a}^{\infty}\theta(\lambda)g(\lambda)\Psi'\Big(\int\_{0}^{\lambda}\theta(\zeta)\Delta\zeta\Big)\left(\int\_{0}^{\lambda}\breve{\xi}(\zeta)\Delta\zeta\right)^{1-\gamma}\Psi'(f(\lambda))f(\lambda)d\lambda. \end{split}$$

This concludes the proof.

**Remark 5.** *In Theorem 20, if we make* Ψˇ (*η*) = *η r ,* ˇ*ξ*(*η*) = *θ*(*η*) *and g*(*η*) = 1*, then inequality* (36) *reduces to inequality* (21)*.*

**Remark 6.** *In Theorem 20, if we make* Ψˇ (*η*) = *η r ,* ˇ*ξ*(*η*) = *θ*(*η*) = *g*(*η*) = 1*, r* = *γ and a* = 0*, then we reclaim Theorem 13.*

**Corollary 10.** *If* T = R *in Theorem 20, then, inequality* (36) *boils down to*

$$\int\_{a}^{\infty} \frac{\xi(\eta)\varrho(\eta)\Psi\left(\int\_{0}^{\eta} \theta(\zeta)f(\zeta)d\zeta\right)}{\left(\int\_{0}^{\eta} \tilde{\xi}(\zeta)d\zeta\right)^{\gamma}} d\eta \geq \frac{1}{\gamma - 1} \int\_{a}^{\infty} \frac{\theta(\eta)g(\eta)\Psi'\left(\int\_{0}^{\eta} \theta(\zeta)d\zeta\right)\Psi'(f(\eta))f(\eta)}{\left(\int\_{0}^{\eta} \tilde{\xi}(\zeta)d\zeta\right)^{\gamma - 1}} d\eta.$$

**Corollary 11.** *If* T = *h*Z *in Theorem 20, then, inequality* (36) *boils down to*

$$\sum\_{n=\frac{\mathfrak{s}}{\mathfrak{h}}}^{\infty} \frac{\mathfrak{xi}(nh)g(nh)\Psi\left(h\sum\_{m=0}^{\frac{n}{\mathfrak{h}}-1}\theta(mh)f(mh)\right)}{\left(h\sum\_{m=0}^{\frac{n}{\mathfrak{h}}-1}\tilde{\xi}(mh)\right)^{\gamma}} \geq \frac{1}{\gamma-1} \sum\_{n=\frac{\mathfrak{s}}{\mathfrak{h}}}^{\infty} \frac{\theta(nh)g(nh)\Psi'\left(h\sum\_{m=0}^{\frac{n}{\mathfrak{h}}-1}\theta(mh)\right)\Psi'\left(f(nh)\right)f(nh)}{\left(h\sum\_{m=0}^{\frac{n}{\mathfrak{h}}-1}\tilde{\xi}(mh)\right)^{\gamma-1}}.$$

**Corollary 12.** *In Corollary 11, if we take* T = Z*, and inequality* (36) *abbreviates to*

$$\sum\_{n=a}^{\infty} \frac{\tilde{\xi}(n)g(n)\Psi\left(\sum\_{m=0}^{n-1} \theta(m)f(m)\right)}{\left(\sum\_{m=0}^{n-1} \tilde{\xi}(m)\right)^{\gamma}} \ge \frac{1}{\gamma - 1} \sum\_{n=a}^{\infty} \frac{\theta(n)g(n)\Psi'\left(\sum\_{m=0}^{n-1} \theta(m)\right)\Psi'(f(n))f(n)}{\left(\sum\_{m=0}^{n-1} \tilde{\xi}(m)\right)^{\gamma - 1}}.$$

**Corollary 13.** *If* T = *q* Z *in Theorem 20, and inequality* (36) *abbreviates to*

$$\begin{split} &\sum\_{n=(\log\_q a)}^{\infty} \frac{q^n \widetilde{\xi}(q^n) \mathfrak{g}(q^n) \Psi\left((q-1) \sum\_{m=0}^{(\log\_q n)-1} q^m h(q^m) f(q^m)\right)}{\left((q-1) \sum\_{m=0}^{(\log\_q n)-1} q^m \widetilde{\xi}(q^m)\right)^{\gamma}} \\ &\geq \frac{1}{\gamma-1} \sum\_{n=(\log\_q a)}^{\infty} \frac{q^n h(q^n) \mathfrak{g}(q^n) \Psi'\left((q-1) \sum\_{m=0}^{(\log\_q n)-1} q^m h(q^m)\right) \Psi'(f(q^n)) f(q^n)}{\left((q-1) \sum\_{m=0}^{(\log\_q n)-1} q^m \widetilde{\xi}(q^m)\right)^{\gamma-1}}. \end{split}$$

Now, as a new result, we are interested in discussing the results in Theorem (20) in the case of the extrema of integration R *η a* ˇ*ξ*∆*s* being replaced to be from *η* to ∞. In fact, that is exactly what we shall accomplish in the next theorem.

**Theorem 21.** *Suppose that* T *is a time scale with* 0 ≤ *a* ∈ T*. Moreover, assume that f , g, θ and* ˇ*ξ are nonnegative functions defined on* [0, ∞)T *such that f is nonincreasing and g is nondecreasing. Moreover, let* Ψˇ : R+ → R+ *be a differentiable function such that* Ψˇ 0 *is nondecreasing and* Ψˇ 0 (*xy*) = Ψˇ 0 (*x*)Ψˇ 0 (*y*) *for all x*, *y* ∈ R+*. If* 0 ≤ *γ* < 1*, then*

$$\int\_{a}^{\infty} \frac{\xi(\eta)\varrho(\eta)\Psi\left(\int\_{0}^{\eta} \theta(\zeta)f(\zeta)\Delta\zeta\right)}{\left(\int\_{\sigma(\eta)}^{\infty} \xi(\zeta)\Delta\zeta\right)^{\gamma}} \Delta\eta \geq \frac{1}{1-\gamma} \int\_{a}^{\infty} \frac{\theta(\eta)g(\eta)\Psi'\left(\int\_{0}^{\eta} \theta(\zeta)\Delta\zeta\right)\Psi'(f(\eta))f(\eta)}{\left(\int\_{\eta}^{\infty} \xi(\zeta)\Delta\zeta\right)^{\gamma-1}} \Delta\eta. \tag{42}$$

**Proof.** Due to nonincreasity of *f* , we have for *η* ≥ *λ* ≥ 0

$$\int\_0^\lambda \theta(\zeta) f(\zeta) \Delta \zeta \ge f(\lambda) \int\_0^\lambda \theta(\zeta) \Delta \zeta \, \, \zeta$$

and thus,

$$\Psi\left(\int\_0^\lambda \theta(\zeta)f(\zeta)\Delta\zeta\right) \ge \Psi\left(f(\lambda)\int\_0^\lambda \theta(\zeta)\Delta\zeta\right) = \Psi(f(\lambda))\Psi\left(f(\lambda)\int\_0^\lambda \theta(\zeta)\Delta\zeta\right). \tag{43}$$

Applying the chain rule (23), there exists *c* ∈ [*λ*, *σ*(*λ*)] such that

$$\mathbb{E}\left[\Psi\left(\int\_0^\lambda \theta(\zeta)f(\zeta)\Delta\zeta\right)\right]^\Delta = \check{\Psi}'\left(\int\_0^c \theta(\zeta)f(\zeta)\Delta\zeta\right)\left(\int\_0^\lambda \theta(\zeta)f(\zeta)\Delta\zeta\right)^\Delta.$$

Since *c* ≥ *λ*, Ψˇ 0 is nondecreasing, *r* ≥ 1 and Z *λ* 0 *θ*(*ζ*)*f*(*ζ*)∆*ζ* ∆ = *θ*(*λ*)*f*(*λ*) ≥ 0, we get

$$\left[\Psi\left(\int\_0^\lambda \theta(\zeta)f(\zeta)d\zeta\right)\right]^\Delta \ge \theta(\lambda)f(\lambda)\Psi'\left(\int\_0^\lambda \theta(\zeta)f(\zeta)d\zeta\right).\tag{44}$$

Combining (43) with (44) gives