$$| (\delta + \upsilon)(\gamma - \upsilon) | = \sqrt{[(\delta + \mathbf{x})^2 + y^2][(+y^2 + \gamma - \mathbf{x})^2]} \ge | (\gamma - \mathbf{x})(\delta + \mathbf{x}) |,\tag{11}$$

where *γ* ≥ *δ* > *ρ* ≥ p *x* 2 + *y* 2 .

On the other hand, a simple calculation results in

$$(\delta + \mathbf{x})(\gamma - \mathbf{x}) \ge (\delta - \rho)(\gamma + \rho), \ge \in [-\rho, \rho]. \tag{12}$$

It is easily seen that (11) and (12) imply the second inequality of (10). Finally, multiplying the inequality *ρ* 2 ≥ |*v*| 2 by the first inequality of (10), we obtain (9) and the proof is complete.

**Lemma 6.** *If v* ∈ C*, δ*, *γ* ∈ R*, and γ* ≥ *δ* > *ρ* ≥ |*v*|*, then*

$$\left|\frac{2v^2[2\gamma\delta + (\gamma - \delta)v]}{(\gamma - v)^2(\delta + v)^2}\right| \le \frac{2r^2[2\gamma\delta - (\gamma - \delta)\rho]}{(\gamma + \rho)^2(\delta - \rho)^2}.\tag{13}$$

**Proof.** The inequality obviously holds provided that *γ* = *δ* (see (10)), thus, we have to prove it in the case that *γ* > *δ*.

We can then prove the following inequality:

$$\left|\frac{2\gamma\delta + (\gamma - \delta)v}{(\delta + v)(\gamma - v)}\right| \le \frac{2\gamma\delta - (\gamma - \delta)\rho}{(\delta - \rho)(\gamma + \rho)}, \quad \gamma \ge \delta > \rho \ge |v|. \tag{14}$$

We define *z* = *x* + *iy* and define the mapping

$$\phi : [ -\rho, \rho ] \to \mathbb{R}, \ \phi(y) = \frac{(\omega + \mathfrak{x})^2 + y^2}{[(\delta + \mathfrak{x})^2 + y^2][(\gamma - \mathfrak{x})^2 + y^2]}, \ \omega = \frac{2\gamma\delta}{\gamma - \delta}.$$

Then, we have

$$\phi'(y) = 2y \frac{[(\delta + \mathbf{x})^2 + y^2][(\gamma - \mathbf{x})^2 + y^2] - [(\delta + \mathbf{x})^2 + (\gamma - \mathbf{x})^2 + 2y^2][(\omega + \mathbf{x})^2 + y^2]}{[(\delta + \mathbf{x})^2 + y^2]^2[(\gamma - \mathbf{x})^2 + y^2]^2}.$$

As *φ* 0 (*y*) < 0, *y* ∈ (0, *ρ*) and *φ* 0 (*y*) > 0, *y* ∈ (−*ρ*, 0), it follows that

$$\phi(y) \le \phi(0) = \frac{(\omega + \mathfrak{x})^2}{[(\delta + \mathfrak{x})^2][(\gamma - \mathfrak{x})^2]}, \ y \in [-\rho, \rho]. \tag{15}$$

We can determine the maximum of the function

$$\varphi: [ -\rho, \rho ] \to \mathbb{R}, \ \rho(\mathfrak{x}) = \frac{\omega + \mathfrak{x}}{(\delta + \mathfrak{x})(\gamma - \mathfrak{x})}.$$

We have

$$\phi'(x) = \frac{x^2 + 2\omega x - \gamma \delta}{(\delta + x)^2 (\gamma - x)^2}$$

The derivative *ϕ* 0 (*x*) = 0 has one positive root, *x*1 = p *ω*2 + *γδ* − *ω*, and one negative root, *x*2 = − p *ω*2 + *γδ* − *ω*. As *x*2 < −*r* and *x*1 ∈ (−*ρ*, *ρ*), it follows the inequality

$$\frac{\omega + \mathbf{x}}{(\delta + \mathbf{x})(\gamma - \mathbf{x})} = \varrho(\mathbf{x}) \le \max\{\varrho(-\rho), \varrho(\rho)\} = \varrho(-\rho) = \frac{\omega - \rho}{(\delta - \rho)(\gamma + \rho)}\tag{16}$$

for every *x* ∈ [−*ρ*, *ρ*]. From (15) and (16), we have (14). Finally, multiplying the inequalities (14), |*v* 2 | ≤ *ρ* 2 and the first inequality of (10), we infer (13).

**Lemma 7.** *If the functions gν and hν are defined by (4) and (5), respectively, then*

$$\frac{z g\_{\nu}^{\prime\prime}(z)}{g\_{\nu}^{\prime}(z)} = z \frac{z \mathbf{J}\_{\nu+2}(z) - \mathbf{3} \mathbf{J}\_{\nu+1}(z)}{\mathbf{J}\_{\nu}(z) - z \mathbf{J}\_{\nu+1}(z)}. \tag{17}$$

$$\frac{zh\_{\nu}^{\prime\prime}(z)}{h\_{\nu}^{\prime}(z)} = \frac{zf\_{\nu+2}(z^{\frac{1}{2}}) - 4z^{\frac{1}{2}}f\_{\nu+1}(z^{\frac{1}{2}})}{4f\_{\nu}(z^{\frac{1}{2}}) - 2z^{\frac{1}{2}}f\_{\nu+1}(z^{\frac{1}{2}})}.\tag{18}$$

.

**Proof.** We differentiate the equality (4), and at the second time we differentiate it logarithmically. After multiplying by *z*, we obtain the following equality:

$$\frac{z g\_{\nu}^{\prime\prime}(z)}{g\_{\nu}^{\prime}(z)} = \frac{z^2 f\_{\nu}^{\prime\prime}(z) + 2z(1-\nu)f\_{\nu}^{\prime}(z) + \nu(\nu-1)f\_{\nu}(z)}{z f\_{\nu}^{\prime}(z) + (1-\nu)f\_{\nu}(z)}.$$

The function *Jν* is a solution of the Bessel differential equation; thus, we can replace the function *z* 2 *J* 00 *ν* using the equality *z* 2 *J* 00 *ν* (*z*) = (*ν* 2 − *z* 2 )*Jν*(*z*) − *zJ*0 *ν* (*z*), and it follows that

$$\frac{z g\_{\nu}^{\prime\prime}(z)}{g\_{\nu}^{\prime}(z)} = \frac{z(1 - 2\nu) f\_{\nu}^{\prime}(z) + (2\nu^2 - \nu - z^2) f\_{\nu}(z)}{z f\_{\nu}^{\prime}(z) + (1 - \nu) f\_{\nu}(z)}$$

In the second step, we use the following well-known equality: *zJ*0 *ν* (*z*) = *νJν*(*z*) − *zJν*+1(*z*), and infer

$$\frac{z\mathbf{g}\_{\nu}^{\prime\prime}(z)}{\mathbf{g}\_{\nu}^{\prime}(z)} = \frac{z(2\nu - 1)J\_{\nu+1}(z) - z^2 J\_{\nu}(z)}{J\_{\nu}(z) - zJ\_{\nu+1}(z)}.$$

Finally, we replace *zJν*(*z*) in the numerator by *zJν*(*z*) = 2(*ν* + 1)*Jν*+1(*z*) − *zJν*+2(*z*), and obtain (17).

We differentiate equality (5) twice, similarly to the case of the function *gν*, and obtain

$$\frac{zh\_{\nu}^{\prime\prime}(z)}{h\_{\nu}^{\prime}(z)} = \frac{\nu(\nu-2)f\_{\nu}(z^{\frac{1}{2}}) + (3-2\nu)z^{\frac{1}{2}}f\_{\nu}^{\prime}(z^{\frac{1}{2}}) + zf\_{\nu}^{\prime\prime}(z^{\frac{1}{2}})}{2(2-\nu)f\_{\nu}(z^{\frac{1}{2}}) + 2z^{\frac{1}{2}}f\_{\nu}^{\prime}(z^{\frac{1}{2}})}.$$

We use the equality *zJ*00 *ν* (*z* 1 2 ) = (*ν* 2 − *z*)*Jν*(*z* 1 2 ) − *z* 1 2 *J* 0 *ν* (*z* 1 2 ), and obtain

$$\frac{zh\_{\nu}^{\prime\prime}(z)}{h\_{\nu}^{\prime}(z)} = \frac{(2\nu^2 - 2\nu - z)f\_{\nu}(z^{\frac{1}{2}}) + (2 - 2\nu)z^{\frac{1}{2}}f\_{\nu}^{\prime}(z^{\frac{1}{2}})}{2(2 - \nu)f\_{\nu}(z^{\frac{1}{2}}) + 2z^{\frac{1}{2}}f\_{\nu}^{\prime}(z^{\frac{1}{2}})}.$$

Now, using the equality *z* 1 2 *J* 0 *ν* (*z* 1 2 ) = *νJν*(*z* 1 2 ) − *z* 1 2 *Jν*+1(*z* 1 2 ), we infer

$$\frac{zh\_{\nu}^{\prime\prime}(z)}{h\_{\nu}^{\prime}(z)} = \frac{(2\nu - 2)z^{\frac{1}{2}}J\_{\nu+1}(z^{\frac{1}{2}}) - zJ\_{\nu}(z^{\frac{1}{2}})}{4J\_{\nu}(z^{\frac{1}{2}}) - 2z^{\frac{1}{2}}J\_{\nu+1}(z^{\frac{1}{2}})} \; .$$

and combining this with the equality *z* 1 2 *Jν*(*z* 1 2 ) = 2(*ν* + 1)*Jν*+1(*z* 1 2 ) − *z* 1 2 *Jν*+2(*z* 1 2 ), (18) follows.

#### **3. Main Results**

**Theorem 1.** *If α* ∈ [0, 1) *and ν* ∈ (−2, −1)*, then the radius of convexity of order α for the mapping gν is rc ν* (*α*) = *r*1, *where r*1 *is the unique root of the equation*

$$1 + r \frac{I\_{\nu+2}(r) + 3I\_{\nu+1}(r)}{I\_{\nu+1}(r) + rI\_{\nu}(r)} = \alpha \tag{19}$$

*in the interval* (0, *a*).

**Proof.** According to the proof of Theorem 1 [2], the equalities

$$\frac{zg\_{\nu}'(z)}{g\_{\nu}(z)} = 1 - 2\sum\_{n=1}^{\infty} \frac{z^2}{j\_{\nu,n}^2 - z^2}, \\ \sum\_{n=1}^{\infty} \frac{1}{j\_{\nu,n}^2} = \frac{1}{4(\nu+1)}.$$

imply

$$\frac{z g\_{\nu}'(z)}{g\_{\nu}(z)} = 1 - \frac{a^2}{2(1+\nu)} \frac{z^2}{a^2 + z^2} - 2 \sum\_{n=2}^{\infty} \frac{a^2 + f\_{\nu,n}^2}{f\_{\nu,n}^2} \frac{z^4}{(a^2 + z^2)(f\_{\nu,n}^2 - z^2)}.$$

The logarithmic differentation of this equality leads to

$$\begin{split} 1 + \frac{z g\_{\nu}^{\prime\prime}(z)}{g\_{\nu}^{\prime}(z)} &= \\ 1 - \frac{a^2}{2(1+\nu)} \frac{z^2}{a^2 + z^2} - 2 \sum\_{n=2}^{\infty} \frac{a^2 + j\_{\nu,n}^2}{j\_{\nu,n}^2} \frac{z^4}{(a^2 + z^2)(j\_{\nu,n}^2 - z^2)} - \\ \frac{\frac{a^2}{1+\nu} \frac{a^2 z^2}{(a^2 + z^2)^2} + 2 \sum\_{n=2}^{\infty} \frac{a^2 + j\_{\nu,n}^2}{j\_{\nu,n}^2} \frac{2z^4 [2a^2 j\_{\nu,n}^2 + z^2 (j\_{\nu,n}^2 - a^2)]}{(a^2 + z^2)^2 (j\_{\nu,n}^2 - z^2)^2}}{1 - \frac{a^2}{2(1+\nu)} \frac{z^2}{a^2 + z^2} - 2 \sum\_{n=2}^{\infty} \frac{a^2 + j\_{\nu,n}^2}{j\_{\nu,n}^2} \frac{z^4}{(a^2 + z^2)(j\_{\nu,n}^2 - z^2)}}. \end{split} \tag{20}$$

It is proven in Theorem 1 [2] that the radius of starlikeness, *r* ∗ *gν* , for the function *gν* is the smallest root of the equation

$$\begin{aligned} 1 + \frac{a^2}{2(1+\nu)} \frac{r^2}{a^2 - r^2} \\ -2 \sum\_{n=2}^\infty \frac{a^2 + j\_{\nu,n}^2}{j\_{\nu,n}^2} \frac{r^4}{(a^2 - r^2)(j\_{\nu,n}^2 + r^2)} = ir \frac{g\_\nu'(ir)}{g\_\nu(ir)} = 0, \end{aligned}$$

in the interval (0, *a*). Thus, we have

∗

).

$$0 < r\_{\mathcal{g}\_{\mathcal{V}}}^{\*} < a < j\_{\nu,2} < j\_{\nu,3} < \dots < j\_{\nu,\mathcal{U}} < \dots < \dots \tag{21}$$

Taking into account that *ν* + 1 < 0, the equality (20) implies the following inequality:

$$\begin{split} \text{Re}\left(1+\frac{z\mathbf{g}\_{\nu}^{\prime\prime}(z)}{\mathbf{g}\_{\nu}^{\prime}(z)}\right) &\geq \\ 1+\left.\frac{a^{2}}{2(1+\nu)}\right|\frac{z^{2}}{a^{2}+z^{2}}\left|-2\sum\_{n=2}^{\infty}\frac{a^{2}+j\_{\nu,n}^{2}}{j\_{\nu,n}^{2}}\right|\frac{z^{4}}{(a^{2}+z^{2})(j\_{\nu,n}^{2}-z^{2})}\right|-\text{(22)} \\ \leq \frac{-\frac{a^{2}}{1+\nu}\left|\frac{a^{2}z^{2}}{(a^{2}+z^{2})^{2}}\right|+2\sum\_{n=2}^{\infty}\frac{a^{2}+j\_{\nu,n}^{2}}{j\_{\nu,n}^{2}}\left|\frac{2z^{4}[2a^{2}j\_{\nu,n}^{2}+z^{2}(j\_{\nu,n}^{2}-a^{2})]}{(a^{2}+z^{2})(j\_{\nu,n}^{2}-z^{2})^{2}}\right|}{1+\frac{a^{2}}{2(1+\nu)}\left|\frac{z^{2}}{a^{2}+z^{2}}\right|-2\sum\_{n=2}^{\infty}\frac{a^{2}+j\_{\nu,n}^{2}}{j\_{\nu,n}^{2}}\left|\frac{z^{4}}{(a^{2}+z^{2})(j\_{\nu,n}^{2}-z^{2})}\right|} \end{split} \tag{22}$$

for every *z* ∈ *U*(*r*

*ν* Using *δ* = *a* 2 , *ρ* = *r* 2 and *v* = *z* 2 in Lemma 4, we obtain

$$\frac{a^2}{2(1+\nu)} \left| \frac{z^2}{a^2 + z^2} \right| \ge \frac{a^2}{2(1+\nu)} \frac{r^2}{a^2 - r^2} \text{ and} \frac{a^2}{2(1+\nu)} \left| \frac{z^2}{(a^2 + z^2)^2} \right| \ge \tag{23}$$

$$\frac{a^2}{(a^2 - r^2)^2} \frac{r^2}{2(1+\nu)}.$$

In a similar manner, Lemma 5 and Lemma 6 imply that

$$\begin{array}{c|c} \left| \frac{z^4}{(a^2+z^2)(j\_{\nu,n}^2-z^2)} \right| \le \frac{r^4}{(a^2-r^2)(j\_{\nu,n}^2+r^2)}\\ \left| \frac{2z^4[2a^2j\_{\nu,n}^2+z^2(j\_{\nu,n}^2-a^2)]}{(a^2+z^2)^2(j\_{\nu,n}^2-z^2)^2} \right| \le \frac{2r^4[2a^2j\_{\nu,n}^2-r^2(j\_{\nu,n}^2-a^2)]}{(a^2-r^2)^2(j\_{\nu,n}^2+r^2)^2}. \end{array} \tag{24}$$

Now, inequalities (22)–(24) imply the following inequality:

$$\begin{split} \text{Re}\left(1+\frac{zg\_{\nu}^{\mu}(z)}{g\_{\nu}^{\nu}(z)}\right) &\geq \\ 1+\frac{a^{2}}{2(1+\nu)}\frac{r^{2}}{a^{2}-r^{2}}-2\sum\_{n=2}^{\infty}\frac{a^{2}+j\_{\nu,n}^{2}}{j\_{\nu,n}^{2}}\frac{r^{4}}{(a^{2}-r^{2})(j\_{\nu,n}^{2}+r^{2})}-\\ -\frac{a^{2}}{1+\nu}\frac{a^{2}r^{2}}{(a^{2}-r^{2})^{2}}+2\sum\_{n=2}^{\infty}\frac{a^{2}+j\_{\nu,n}^{2}}{j\_{\nu,n}^{2}}\frac{2r^{4}[2r^{2}\_{\mu,n}-r^{2}(j\_{\nu,n}^{2}-a^{2})]}{(a^{2}-r^{2})^{2}(j\_{\nu,n}^{2}+r^{2})^{2}} &= 1+\frac{irg\_{\nu}^{\mu\prime}(ir)}{g\_{\nu}^{\prime}(ir)}=\Phi(r),\end{split} \tag{25}$$

provided that *a* > *r* ∗ *gν* > |*z*|, where *r* ∗ *gν* verifies the inequalities (21).

The following equalities hold: Φ(0) = 1 and lim*r*%*r* ∗ *gν* Φ(*r*) = −∞. Consequently, equation 1 + *irg*00 *ν* (*ir*) *g* 0 *ν*(*ir*) = *α* has a real root in the interval (0,*r* ∗ *gν* ). The smallest positive real root of the equation 1 + *irg*00 *ν* (*ir*) *g* 0 *ν*(*ir*) = *α* is denoted by *r c gν* (*α*), and this root is the radius of convexity of order *α* of the function *gν*. The first equality of Lemma 7 and the equality *Jν*(*iz*) = *i ν Iν*(*z*) imply that the equation 1 + *irg*00 *ν* (*ir*) *g* 0 *ν*(*ir*) = *α* is equivalent to (19).

We determine the radius of uniform convexity of the mapping *gν* in the next theorem.

**Theorem 2.** *If ν* ∈ (−2, −1), *then the radius of uniform convexity for the mapping gν is r* ∗ *ν* (*α*) = *r*2, *where r*2 *is the smallest positive root of the equation*

$$\frac{1}{2} + r \frac{I\_{\nu+2}(r) + 3I\_{\nu+1}(r)}{I\_{\nu+1}(r) + rI\_{\nu}(r)} = 0 \tag{26}$$

*in the interval* (0,*r* ∗ *ν* ).

**Proof.** Equality (20) implies the following inequality:

$$\begin{array}{c|c} \left| \frac{zg\_{\nu}^{\mu}(z)}{g\_{\nu}^{\mu}(z)} \right| \leq \\ -\frac{a^{2}}{2(1+\nu)} \left| \frac{z^{2}}{a^{2}+z^{2}} \right| + 2\sum\_{n=2}^{\infty} \frac{a^{2}+j\_{\nu,n}^{2}}{j\_{\nu,n}^{2}} \left| \frac{z^{4}}{(a^{2}+z^{2})(j\_{\nu,n}^{2}-z^{2})} \right| + \\ -\frac{a^{2}}{1+\nu} \left| \frac{a^{2}z^{2}}{(a^{2}+z^{2})^{2}} \right| + 2\sum\_{n=2}^{\infty} \frac{a^{2}+j\_{\nu,n}^{2}}{j\_{\nu,n}^{2}} \left| \frac{2z^{4}[2a^{2}j\_{\nu,n}^{2}+z^{2}(j\_{\nu,n}^{2}-a^{2})]}{(a^{2}+z^{2})^{2}(j\_{\nu,n}^{2}-z^{2})^{2}} \right| \\ \phantom{\left| \frac{a^{2}}{1+\frac{a^{2}}{2(1+\nu)}} \right|} \frac{z^{2}}{a^{2}+z^{2}} \left| -2\sum\_{n=2}^{\infty} \frac{a^{2}+j\_{\nu,n}^{2}}{j\_{\nu,n}^{2}} \right| \frac{z^{4}}{(a^{2}+z^{2})(j\_{\nu,n}^{2}-z^{2})} \end{array} . \tag{27}$$

We can again use inequalities (22) and (23), and in combination with (27), we have

$$\begin{split} \left| \frac{zg\_{\nu}^{\prime\prime}(z)}{g\_{\nu}^{\prime}(z)} \right| &\leq -\frac{a^{2}}{2(1+\nu)} \frac{r^{2}}{a^{2}-r^{2}} + 2\sum\_{n=2}^{\infty} \frac{a^{2}+j\_{\nu,n}^{2}}{j\_{\nu,n}^{2}} \frac{r^{4}}{(a^{2}-r^{2})(j\_{\nu,n}^{2}+r^{2})} + 1 \\ &\frac{-\frac{a^{2}}{1+\nu}\frac{a^{2}r^{2}}{(a^{2}-r^{2})^{2}} + 2\sum\_{n=2}^{\infty} \frac{a^{2}+j\_{\nu,n}^{2}}{j\_{\nu,n}^{2}} \frac{2r^{4}[2a^{2}j\_{\nu,n}^{2}-r^{2}(j\_{\nu,n}^{2}-a^{2})]}{(a^{2}-r^{2})^{2}(j\_{\nu,n}^{2}+r^{2})^{2}} = -\frac{i\mathbf{r}g\_{\nu}^{\prime\prime}(ir)}{g\_{\nu}^{\prime\prime}(ir)}. \end{split}$$

Inequalities (25) and (27) imply

$$\left| \mathrm{Re} \left( 1 + \frac{z g\_{\nu}^{\prime\prime}(z)}{g\_{\nu}^{\prime}(z)} \right) - \left| \frac{z g\_{\nu}^{\prime\prime}(z)}{g\_{\nu}^{\prime}(z)} \right| \geq 1 + 2 \frac{i r g\_{\nu}^{\prime\prime}(ir)}{g\_{\nu}^{\prime}(ir)}, z \in \mathsf{U}(r\_{\nu}^{\*}). \tag{28}$$

The smallest positive root of the equation 1 + 2 *irg*00 *ν* (*ir*) *g* 0 *ν*(*ir*) = 0 in the interval (0,*r* ∗ *ν* ) is denoted by *r uc ν* . According to (28), the value *r uc ν* is the biggest with the property that

$$\left| \mathrm{Re} \left( 1 + \frac{z \mathcal{g}\_{\nu}^{\prime\prime}(z)}{\mathcal{g}\_{\nu}^{\prime}(z)} \right) - \left| \frac{z \mathcal{g}\_{\nu}^{\prime\prime}(z)}{\mathcal{g}\_{\nu}^{\prime}(z)} \right| > 0, \; z \in \mathcal{U}(r\_{\nu}^{\mu c}). \right.$$

Lemma 7 and the equality *Jν*(*iz*) = *i ν Iν*(*z*) imply that the equation 1 + 2 *irg*00 *ν* (*ir*) *g* 0 *ν*(*ir*) = 0 is equivalent to (26), completing the proof.

Theorems 1 and 2 imply the following result.

**Corollary 1.** *The mapping gν is uniformly convex in the disk U*(*r*) *if and only if it is convex of order* 1 2 .

**Theorem 3.** *If α* ∈ [0, 1) *and ν* ∈ (−2, −1), *then the radius of convexity of order α for the mapping hν is rc hν* (*α*) = *r*3, *where r*3 *is the smallest real root of the equation*

$$1 + \frac{rI\_{\nu+2}(r^{\frac{1}{2}}) + 4r^{\frac{1}{2}}I\_{\nu+1}(r^{\frac{1}{2}})}{4I\_{\nu}(r^{\frac{1}{2}}) + 2r^{\frac{1}{2}}I\_{\nu+1}(r^{\frac{1}{2}})} = a \tag{29}$$

*in the interval* (0,*r* ∗ *hν* ). **Proof.** According to the proof of Theorem 2 [2], the equalities

$$\frac{zh\_{\nu}'(z)}{h\_{\nu}(z)} = 1 - \sum\_{n=1}^{\infty} \frac{z}{j\_{\nu,n}^2 - z}, \quad \sum\_{n=1}^{\infty} \frac{1}{j\_{\nu,n}^2} = \frac{1}{4(\nu+1)}$$

imply

$$\frac{zh\_{\nu}'(z)}{h\_{\nu}(z)} = 1 - \frac{a^2}{4(\nu+1)} \cdot \frac{z}{a^2+z} - \sum\_{n=2}^{\infty} \frac{a^2 + f\_{\nu,n}^2}{f\_{\nu,n}^2} \cdot \frac{z^2}{(a^2+z)\left(f\_{\nu,n}^2 - z\right)}.$$

where *z* ∈ *U*(0,*r*).

The logarithmic differentiation of the equality leads to

$$1 + \frac{zh\_{\nu}^{\prime\prime}(z)}{h\_{\nu}^{\prime}(z)} = 1 - \frac{a^2}{4(\nu+1)} \cdot \frac{z}{a^2 + z} - \sum\_{n=2}^{\infty} \frac{a^2 + j\_{\nu,n}^2}{j\_{\nu,n}^2} \cdot \frac{z^2}{(a^2 + z)\left(j\_{\nu,n}^2 - z\right)} -$$

$$-\frac{\frac{a^2}{4(1+\nu)} \cdot \frac{a^2 z}{(a^2 + z)^2} + \sum\_{n=2}^{\infty} \frac{a^2 + j\_{\nu,n}^2}{j\_{\nu,n}^2} \cdot \frac{z^2 \left[2a^2 j\_{\nu,n}^2 + z \left(j\_{\nu,n}^2 - a^2\right)\right]}{\left(j\_{\nu,n}^2 - z\right)^2 \left(a^2 + z\right)^2}}{1 - \frac{a^2}{4a^2 + z} \cdot \frac{z}{(\nu+1)} - \sum\_{n=2}^{\infty} \frac{a^2 + j\_{\nu,n}^2}{j\_{\nu,n}^2} \cdot \frac{z^2}{\left(j\_{\nu,n}^2 - z\right) \left(a^2 + z\right)}}. \tag{30}$$

It is proven in [2] that the radius of starlikeness, *r* ∗ *hν* , for function *hν* is the smallest root of the equation

$$\frac{-rh\_\nu'(-r)}{h\_\nu(-r)} = 0, \ r \in \left(0, a^2\right), \ z \in \mathcal{U}(0, r).$$

However,

$$\frac{-rh\_{\nu}^{\prime}(-r)}{h\_{\nu}(-r)} = 1 + \frac{a^2}{4(\nu+1)} \cdot \frac{r}{a^2 - r}$$

$$-\sum\_{n=2}^{\infty} \frac{a^2 + j\_{\nu,n}^2}{j\_{\nu,n}^2} \cdot \frac{r^2}{(a^2 - r)\left(j\_{\nu,n}^2 + r\right)} = 0, \ r \in \left(0, a^2\right).$$

Taking into the account that *ν* + 1 < 0, we obtain from relation (30)

$$\begin{split} \mathrm{Re}\left(1+\frac{zh\_{\nu}^{\mu\prime}(z)}{h\_{\nu}^{\prime}(z)}\right) &\geq \frac{a^2}{4(\nu+1)} \cdot \left|\frac{z}{a^2+z}\right| - \sum\_{n=2}^{\infty} \frac{a^2+j\_{\nu,n}^2}{j\_{\nu,n}^2} \cdot \left|\frac{z^2}{(a^2+z)\left(j\_{\nu,n}^2-z\right)}\right| - \\ &\quad -\frac{\frac{-a^2}{4(\nu+1)} \cdot \left|\frac{a^2z}{(a^2+z)^2}\right| + \sum\_{n=2}^{\infty} \frac{a^2+j\_{\nu,n}^2}{j\_{\nu,n}^2} \cdot \left|\frac{z^2\left|2a^2j\_{\nu,n}^2+z\left(j\_{\nu,n}^2-a^2\right)\right|}{(a^2+z)^2\left(j\_{\nu,n}^2-z\right)^2}\right|}{1+\frac{a^2}{4(\nu+1)} \cdot \left|\frac{z}{a^2+z}\right| - \sum\_{n} \frac{a^2+j\_{\nu,n}^2}{j\_{\nu,n}^2} \cdot \left|\frac{z^2}{(a^2+z)\left(j\_{\nu,n}^2-z\right)}\right|}\end{split} \tag{31}$$

and *z* ∈ *U*(0,*r*),*r* ∈ 0,*r* ∗ *hν* . We obtain from Lemmas 4 and 5 the following inequality:

$$\mathrm{Re}\left(1+\frac{zh\_{\nu}^{\prime\prime}(z)}{h\_{\nu}^{\prime}(z)}\right) \ge 1 + \frac{a^2}{4(\nu+1)} \cdot \frac{r}{a^2-r} - \sum\_{n=2}^{\infty} \frac{a^2+j\_{\nu,n}^2}{j\_{\nu,n}^2} \cdot \frac{r^2}{(a^2-r)(a^2+r)} - \cdots$$

$$-\frac{-\frac{a^2}{4(\nu+1)} \cdot \frac{a^2r}{(a^2-r)^2} + \sum\_{n=2}^{\infty} \frac{a^2+j\_{\nu,n}^2}{j\_{\nu,n}^2} \cdot \frac{r^2[2a^2j\_{\nu,n}^2 - r(j\_{\nu,n}^2 - a^2)]}{(a^2-r)^2(j\_{\nu,n}^2 + r)^2}}{1 + \frac{a^2}{4(\nu+1)} \cdot \frac{r}{a^2-r} - \sum\_{n=2}^{\infty} \frac{a^2+j\_{\nu,n}^2}{j\_{\nu,n}^2} \cdot \frac{r^2}{(a^2-r)(\frac{j\_{\nu,n}^2}{j\_{\nu,n}^2 - r})}} = \tag{32}$$

$$= 1 - \frac{r h\_{\nu}^{\prime\prime}(-r)}{h\_{\nu}^{\prime}(-r)} = \psi(r), \ a > r\_{h\_{\nu}}^{\ast}>|z|\_{\prime}$$

similarly to the proof of Theorem 1. The mapping

$$\psi : \left( 0, r\_{h\_{\mathcal{V}}}^{\*} \right) \to \mathbb{R}, \; \psi(r) = 1 + \frac{-rh\_{\mathcal{V}}^{\prime\prime}(-r)}{h\_{\mathcal{V}}^{\prime}(-r)} \lambda$$

is strictly decreasing, and *a* > *r* ∗ *hν* > |*z*|.

We then have lim *r*%*r* ∗ *ψ*(*r*) = −∞, *ψ*(0) = 1, and the equation

$$1 + \frac{-r h\_{\nu}^{\prime\prime}(-r)}{h\_{\nu}^{\prime}(-r)} = \alpha$$

has at least one real root in the interval 0,*r* ∗ *hν* .

*hν*

The smallest positive real root of the equation 1 − *rh*00 *ν* (−*r*) *h* 0 *ν*(−*r*) = *α* is denoted by *r c hν* (*α*), and this root is the radius of convexity of order *α* of the function *hν*. The second equality of Lemma 7 and the equality *Jν*(*iz*) = *i ν Iν*(*z*) imply that the equation 1 − *rh*00 *ν* (−*r*) *h* 0 *ν*(−*r*) = *α* is equivalent to (29).

**Theorem 4.** *If α* ∈ [0, 1) *and ν* ∈ (−2, −1), *then the radius of uniform convexity of hν is r* ∗ *hν* (*α*) = *r*4, *where r*4 *is the smallest positive root of the equation*

$$\frac{rI\_{\nu+2}(r^{\frac{1}{2}}) + 4r^{\frac{1}{2}}I\_{\nu+1}(r^{\frac{1}{2}})}{4I\_{\nu}(r^{\frac{1}{2}}) + 2r^{\frac{1}{2}}I\_{\nu+1}(r^{\frac{1}{2}})} = \frac{1}{2} \tag{33}$$

*in the interval* (0,*r* ∗ *hν* ).

>

**Proof.** Equality (30) implies the following inequality:

$$\begin{split} \left| \frac{zh\_{\nu}^{\mu\prime}(z)}{h\_{\nu}^{\prime}(z)} \right| &\leq -\frac{a^{2}}{4(\nu+1)} \cdot \left| \frac{z}{a^{2}+z} \right| + \sum\_{n=2}^{\infty} \frac{a^{2} + j\_{\nu,n}^{2}}{j\_{\nu,n}^{2}} \cdot \left| \frac{z^{2}}{(a^{2}+z)\left(j\_{\nu,n}^{2}-z\right)} \right| + \\ &+ \frac{\frac{-a^{2}}{4(\nu+1)} \cdot \left| \frac{a^{2}z}{(a^{2}+z)^{2}} \right| + \sum\_{n=2}^{\infty} \frac{a^{2} + j\_{\nu,n}^{2}}{j\_{\nu,n}^{2}} \cdot \left| \frac{z^{2}\left[2a^{2}j\_{\nu,n}^{2}+z\left(j\_{\nu,n}^{2}-a^{2}\right)\right]}{(a^{2}+z)^{2}\left(j\_{\nu,n}^{2}-z\right)^{2}} \right|} \right|. \end{split} \tag{34}$$

We obtain the following from the relation (31), Lemma 4, and the relation (34):

$$\left|\frac{zh\_{\nu}^{\prime\prime}(z)}{h\_{\nu}^{\prime}(z)}\right| \leq -\frac{a^2}{4(\nu+1)} \cdot \frac{r}{a^2 - r} + \sum\_{n=2}^{\infty} \frac{a^2 + \hat{f}\_{\nu,n}^2}{\hat{f}\_{\nu,n}^2} \cdot \frac{r^2}{(a^2 - r)\left(\hat{f}\_{\nu,n}^2 + r\right)} +$$

$$+\frac{\frac{a^2}{4(\nu+1)} \cdot \frac{a^2r}{(a^2 - r)^2} + \sum\_{n=2}^{\infty} \frac{a^2 + \hat{f}\_{\nu,n}^2}{\hat{f}\_{\nu,n}^2} \cdot \frac{r^2 \left[2a^2\hat{f}\_{\nu,n}^2 - r\left(\hat{f}\_{\nu,n}^2 - a^2\right)\right]}{\left(a^2 - r\right)^2 \left(\hat{f}\_{\nu,n}^2 + r\right)^2}}{1 - \frac{a^2}{4(a^2 - r)} \cdot \frac{r}{\nu+1} - \sum\_{n=2}^{\infty} \frac{a^2 + \hat{f}\_{\nu,n}^2}{\hat{f}\_{\nu,n}^2} \cdot \frac{r^2}{(a^2 - r)\left(\hat{f}\_{\nu,n}^2 + r\right)}} = \frac{r h\_{\nu}^{\prime\prime}(-r)}{h\_{\nu}^{\prime}(-r)}, |z| \leq r < a^2.$$

Inequalities (32) and (34) imply

$$\left| \operatorname{Re} \left( 1 + \frac{z h\_{\nu}^{\prime\prime}(z)}{h\_{\nu}^{\prime}(z)} \right) - \left| \frac{z h\_{\nu}^{\prime\prime}(z)}{h\_{\nu}^{\prime}(z)} \right| \geq 1 - \frac{2 r h\_{\nu}^{\prime\prime}(-r)}{h\_{\nu}^{\prime}(-r)}, \; z \in \mathcal{U}(r\_{h\_{\nu}}^{\*}). \tag{35}$$

The smallest positive root of the equation 1 − 2*rh*00 *ν* (−*r*) *h* 0 *ν*(−*r*) = 0 in the interval (0,*r* ∗ *hν* ) is denoted by *r uc hν* .

According to (35), the value *r uc hν* is the biggest with the property that

$$\left| \mathrm{Re} \left( 1 + \frac{z h\_{\nu}^{\prime\prime}(z)}{h\_{\nu}^{\prime}(z)} \right) - \left| \frac{z h\_{\nu}^{\prime\prime}(z)}{h\_{\nu}^{\prime}(z)} \right| > 0, \; z \in \mathcal{U}(r\_{h\_{\nu}}^{uc}).$$

The equation 1 − 2*rh*00 *ν* (−*r*) *h* 0 *ν*(−*r*) = 0 is equivalent to (33), completing the proof. Lemma 7 and the equality *Jν*(*iz*) = *i ν Iν*(*z*) imply that the equation 1 − 2*rh*00 *ν* (−*r*) *h* 0 *ν*(−*r*) = 0 is equivalent to (33).

From Theorems 3 and 4, we obtain the following corollary.

**Corollary 2.** *The function hν is uniformly convex in the disk U*(*r*) *if and only if it is convex of order* 1 2 .

**Author Contributions:** Conceptualization, L.-I.C., R.S. and P.A.K.; methodology, R.S.; software, L.- I.C., R.S. and P.A.K.; validation, R.S.; formal analysis, L.-I.C., R.S. and P.A.K.; investigation, L.-I.C., R.S. and P.A.K.; resources, L.-I.C., R.S. and P.A.K.; data curation, L.-I.C., R.S. and P.A.K.; writing—original draft preparation, L.-I.C., R.S. and P.A.K.; writing—review and editing, L.-I.C., R.S. and P.A.K.; visualization, L.-I.C., R.S. and P.A.K.; supervision, R.S.; project administration, L.-I.C., R.S. and P.A.K.; funding acquisition, L.-I.C., R.S. and P.A.K. 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.

**Acknowledgments:** The authors would like to express their sincere thanks to the referees for their valuable suggestions.

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

#### **References**

### *Article* **Cauchy Integral and Boundary Value for Vector-Valued Tempered Distributions**

**Richard D. Carmichael**

Department of Mathematics, Wake Forest University, Winston-Salem, NC 27109, USA<EMAIL_ADDRESS>
**Abstract:** Using the historically general growth condition on scalar-valued analytic functions, which have tempered distributions as boundary values, we show that vector-valued analytic functions in tubes *T C* = R*n* + *iC* obtain vector-valued tempered distributions as boundary values. In a certain vector-valued case, we study the structure of this boundary value, which is shown to be the Fourier transform of the distributional derivative of a vector-valued continuous function of polynomial growth. A set of vector-valued functions used to show the structure of the boundary value is shown to have a one–one and onto relationship with a set of vector-valued distributions, which generalize the Schwartz space D0 *L* 2 (R*n* ); the tempered distribution Fourier transform defines the relationship between these two sets. By combining the previously stated results, we obtain a Cauchy integral representation of the vector-valued analytic functions in terms of the boundary value.

**Keywords:** analytic functions; vector-valued tempered distributions; boundary value; Cauchy integral

**MSC:** 32A26; 32A40; 46F12; 46F20

#### **1. Introduction**

Tillmann [1] introduced the analysis of analytic functions, which obtain tempered distributional boundary values in S 0 (R*n* ). In [1], Tillmann worked with scalar-valued analytic functions in tubes *T Cµ* = R*n* + *iCµ*, where the *Cµ* = {*y* ∈ R*n* : (−1) *µjyj* > 0, *j* = 1, . . . , *n*} with *µ* = (*µ*1, *µ*2, . . . , *µn*) being any of the 2 *n* n-tuples, whose components are either 0 or 1 and characterize the growth conditions on the analytic functions, which obtain the S 0 (R*n* ) boundary values. This analysis by Tillmann was motivated by the work by K*o*¨the in [2,3].

Using a more restrictive growth on the analytic functions, we showed in [4] that vectorvalued analytic functions in tubes *T C* = R*n* + *iC*, where *C* is an open convex cone, having this more restrictive growth obtain vector-valued tempered distributions in S 0 (R*n* , X ), with X being a specified topological vector space. In this paper, our first objective is to generalize this result of [4] to the general growth form of Tillmann for the vector-valued analytic functions. We obtain this boundary value generalization in Section 4 of this paper.

Moreover, in Section 4, we study the structure of this boundary value in S 0 (R*n* , X ). To do this, we first restrict the topological vector space X by imposing certain conditions on it to ensure that the boundary value is the Fourier transform of a distributional derivative of a continuous vector-valued function **g**, which has polynomial growth in the norm of the space X . By further restricting X to be a Hilbert space, we show that function **g** is in a set that has a one–one and onto relationship with a set of vector-valued distributions, which generalize the D0 *L* 2 (R*n* ) distributions of Schwartz. The relationship between these two sets is obtained using the tempered distribution Fourier transform; the proof of this relationship is proved in Section 3 of this paper. Using the relationships of these noted two sets, we are able to obtain an additional structure of the tempered distribution boundary value of the analytic functions in Section 4.

A few papers have been written concerning the construction of a Cauchy integral for tempered distributions. All of these papers concern scalar-valued analytic functions

**Citation:** Carmichael, R.D. Cauchy Integral and Boundary Value for Vector-Valued Tempered Distributions. *Axioms* **2022**, *11*, 392. https://doi.org/10.3390/ axioms11080392

Academic Editor: Georgia Irina Oros

Received: 12 July 2022 Accepted: 7 August 2022 Published: 10 August 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 scalar-valued tempered distributions. The first paper known to this author was by J. Sebastião e Silva [5] (Section 5) and concerned scalar-valued analytic functions and tempered distributions in one dimension. An associated analysis by Sebastião e Silva is contained in [6]. Carmichael [7] defined a Cauchy integral for tempered distributions in the C*n* setting corresponding to analytic functions in each of the 2 *n* quadrant tubes *T Cµ* ⊂ C*n* and showed that the analytic functions with growth, such as that of Tillmann in (C − R) *n* could be recovered as the defined Cauchy integral of the tempered distribution boundary value; the results of [7] can be extended to the vector-valued analytic functions in *T Cµ* and the tempered distribution setting considered in this paper by the same techniques as those of [7]. The Cauchy integrals introduced by Sebastião e Silva in [5] and by Carmichael in [7] are in fact equivalence classes of analytic functions defined by an integral involving the Cauchy kernel.

Vladimirov [8–10] defined a Cauchy integral for tempered distributions associated with analytic functions in general tubes *T C* = R*n* + *iC* ⊂ C*n* corresponding to open convex cones *C* with the functions satisfying a growth condition similar to that of Tillmann. Vladimirov has shown that the analytic functions that he has considered can be recovered by a Cauchy integral involving the tempered distribution boundary values of the analytic functions. An associated analysis by Vladimirov is contained in [11,12]. The works mentioned in this paragraph and the previous paragraph all concern scalar-valued analytic functions and scalar-valued tempered distributions.

In Section 5 of this paper, we build on our analysis of Sections 3 and 4 to obtain a Cauchy integral representation of the vector-valued analytic functions, which are shown to have tempered vector-valued distributions as the boundary values in Section 4. The proof of our result here and the form of the Cauchy integral representation are substantially different from any of the previous results concerning Cauchy integral representation of the analytic functions having tempered distribution boundary values.

#### **2. Definitions and Notation**

Throughout, X will denote a topological vector space with the stated appropriate properties corresponding to the results that we wish to prove. For X being a normed space, we denote the norm by N . Θ will denote the zero element of X ; and if X is a Hilbert space, we denote the space by H. For integration of the vector-valued functions and vector-valued analytic functions, we refer to Dunford and Schwartz [13]. For foundational information concerning vector-valued distributions, we refer to Schwartz [14,15].

The n-dimensional notation to be used in this paper will be the same as in [16,17]. Note 0 = (0, 0, . . . , 0) is the origin in R*n* . The information concerning cones *C* ⊂ R*n* needed is explicitly stated in [16] (Section 2) and [17] (Chapter 1). We do not repeat the definitions and notations concerning cones as stated in [16] (Section 2), and we ask the reader to refer to this reference.

The *L p* (R*n* , X ) functions, 1 ≤ *p* ≤ ∞, with values in a Banach space X and their norm |**h**|*p* [13] (p. 119) are noted in [13] (Chapter III). The Fourier transform on *L* 1 (R*n* ) or *L* 1 (R*n* , X ) is given in [17] (p. 3). All Fourier (inverse Fourier) transforms on scalar or vectorvalued functions will be denoted *φ*ˆ(*x*) = F[*φ*(*t*); *x*] (F −1 [*φ*(*t*); *x*]). As stated in [18,19], the Plancherel theory is not true for vector-valued functions, except when X = H, a Hilbert space. The Plancherel theory is complete in the *L* 2 (R*n* , H) setting in that the inverse Fourier transform is the inverse mapping of the Fourier transform with F −1F = *I* = F F −1 with *I* being the identity mapping.

We denote S(R*n* ) as the tempered functions with associated distributions being S 0 (R*n* ) or associated vector-valued distributions being S 0 (R*n* , X ). The Fourier (inverse Fourier) transform on S 0 (R*n* ) and S 0 (R*n* , X ) is the usual definition and is given in [14] (p. 73).

#### **3. Fourier and Inverse Fourier Transform on a Function Subset of** S 0 (R*n* **,** H)

Let X be a Banach space. We defined the space S 0 *p* (R*n* , X ), 1 ≤ *p* < ∞, in [16]. We repeat the definition here because of the importance of these functions for our results in this paper.

**Definition 1.** *For a Banach space* X , S 0 *p* (R*n* , X ), 1 ≤ *p* < ∞, *is the set of all measurable functions g*(*t*), *t* ∈ R*n , with values in* X *such that there exists a real number m* ≥ 0 *for which* (1 + |*t*| *p* ) −*mg*(*t*) ∈ *L p* (R*n* , X )*.*

Note that *m* can be taken as a nonnegative integer in Definition 1. As noted in [16], S 0 *p* (R*n* , X ) ⊂ S0 (R*n* , X ), 1 ≤ *p* < ∞. The spaces S 0 *p* (R*n* , X ) will be important in this paper. Throughout this paper, the differential operator *Dt* , *t* ∈ R*n* will take the form

$$D\_{t} = \frac{-1}{2\pi i} \left( \frac{\partial}{\partial t\_{1}}, \frac{\partial}{\partial t\_{2}}, \dots, \frac{\partial}{\partial t\_{n}} \right) \dots$$

Thus, for *α* being any n-tuple of nonnegative integers,

$$D\_t^\alpha = \begin{pmatrix} \frac{-1}{2\pi i} \end{pmatrix}^{|\alpha|} \begin{pmatrix} \frac{\partial^{\alpha\_1}}{\partial t\_1^{\alpha\_1}}, \frac{\partial^{\alpha\_2}}{\partial t\_2^{\alpha\_2}}, \dots, \frac{\partial^{\alpha\_{\text{II}}}}{\partial t\_n^{\text{III}}} \end{pmatrix}.$$

The goal of this section is to show a one–one and onto relationship between the set of functions S 0 2 (R*n* , H) and another subset of S 0 (R*n* , H), where H is a Hilbert space. This relationship is obtained by both the Fourier and inverse Fourier transforms in S 0 (R*n* , H). We define the space that has this stated relationship to S 0 2 (R*n* , H), as follows.

**Definition 2.** *Let m be any nonnegative integer. The set of Hilbert space* H*-valued generalized functions in* S 0 (R*n* , H) *of the form*

$$V\_t = \sum\_{|\alpha| \le m} D\_t^{\alpha} \mathbf{g}\_{\alpha}(t)$$

*where gα* ∈ *L* 2 (R*n* , H), |*α*| ≤ *m, will be denoted as L*2(R*n* , H)*.*

We emphasize that *L*2(R*n* , H) ⊂ S0 (R*n* , H). When H = C1 , note that *L*2(R*n* , C1 ) = D0 *L* 2 (R*n* ), the Schwartz space of distributions contained in S 0 (R*n* ) of the form of finite sums of distributional derivatives of *L* 2 (R*n* ) functions. For *φ* ∈ D*L* 2 (R*n* ), the Schwartz space that is the set of test functions for D0 *L* 2 (R*n* ), the application h*V*, *φ*i, *V* ∈ D0 *L* 2 (R*n* ), yields a complex number. In exactly the same way, for *V* ∈ *L*2(R*n* , H) and *φ* ∈ D*L* 2 (R*n* ), the application h*V*, *φ*i yields an element of H; and the algebraic and differentiation calculations on the form h*V*, *φ*i hold for *V* ∈ *L*2(R*n* , H), as usual, just as these calculations hold on the form h*V*, *φ*i for *V* ∈ S0 (R*n* , H) and *φ* ∈ S(R*n* ). This is an important note in relation to our construction of the Cauchy integral (later in this paper).

We now obtain the relationship between S 0 2 (R*n* , H) and *L*2(R*n* , H) for any Hilbert space H.

**Lemma 1.** *The* S 0 (R*n* , H) *Fourier transform maps* S 0 2 (R*n* , H) *one-one and onto L*2(R*n* , H)*. The* S 0 (R*n* , H) *inverse Fourier transform maps L*2(R*n* , H) *one-one and onto* S 0 2 (R*n* , H)*.*

**Proof.** Let the function **g** ∈ S0 2 (R*n* , H). From Definition 1, there is a real number *m* ≥ 0 for which (1 + |*t*| 2 ) −*m***g**(*t*) ∈ *L* 2 (R*n* , H), and *m* can be taken as a nonnegative integer. Since **g** ∈ S0 2 (R*n* , H) ⊂ S0 (R*n* , H), the Fourier transform of **g** in S 0 (R*n* , H) is an element of S 0 (R*n* , H); we put *Vx* = F[**g**]*x*. Let *φ* ∈ S(R*n* ), and let ∆ denote the Laplace operator in the variable *x* ∈ R*n* . Using integration by parts, we have

$$\begin{split} \langle V\_{\mathbf{x}'}\boldsymbol{\phi}(\mathbf{x})\rangle &= \langle \mathbf{g}(t), \mathcal{F}[\boldsymbol{\phi}(\mathbf{x}); t] \rangle \\ &= \langle \frac{\mathbf{g}(t)}{(1+|t|^2)^m}, \int\_{\mathbb{R}^n} \boldsymbol{\phi}(\mathbf{x})(1+|t|^2)^m e^{2\pi i \langle \mathbf{x}, t\rangle} d\mathbf{x} \rangle \\ &= \langle \frac{\mathbf{g}(t)}{(1+|t|^2)^m}, \mathcal{F}[(1-(4\pi^2)^{-1}\Delta)^m \boldsymbol{\phi}(\mathbf{x}); t] \rangle \\ &= \langle \mathcal{F}[\frac{\mathbf{g}(t)}{(1+|t|^2)^m}; \mathbf{x}], (1-(4\pi^2)^{-1}\Delta)^m \boldsymbol{\phi}(\mathbf{x}) \rangle. \end{split} \tag{1}$$

Since (1 + |*t*| 2 ) −*m***g**(*t*) ∈ *L* 2 (R*n* , H), then **h**(*x*) = F[(1 + |*t*| 2 ) −*m***g**(*t*); *x*] ∈ *L* 2 (R*n* , H). From (1), we have

$$
\langle V\_{\mathbf{x}} \varphi(\mathbf{x}) \rangle = \langle (1 - (4\pi^2)^{-1} \Delta)^m \mathbf{h}(\mathbf{x}), \phi(\mathbf{x}) \rangle,
$$

and *Vx* = F[**g**]*x* = (1 − (4*π* 2 ) −1∆) *m***h**(*x*) ∈ *L*2(R*n* , H). Thus, the S 0 (R*n* , H) Fourier transform maps S 0 2 (R*n* , H) to *L*2(R*n* , H).

We now desire to prove that any element of *L*2(R*n* , H) is the S 0 (R*n* , H) Fourier transform of an element in S 0 2 (R*n* , H). Let *V* ∈ *L*2(R*n* , H) and *φ* ∈ S(R*n* ). By Definition 2, there is a nonnegative integer *m*, such that

$$V\_t = \sum\_{|\alpha| \le m} D\_t^{\alpha} \mathbf{g}\_{\alpha}(t)$$

with **g***α* (*t*) ∈ *L* 2 (R*n* , H), |*α*| ≤ *m*. Since *L*2(R*n* , H) ⊂ S0 (R*n* , H), F −1 [*V*]*x* exists in S 0 (R*n* , H), and we have for the nonnegative integer *m*

hF −1 [*V*]*x*, *φ*(*x*)i = ∑ |*α*|≤*m* h*D α t* **g***α* (*t*), F −1 [*φ*(*x*); *t*]i = ∑ |*α*|≤*m* (−1) |*α*| h**g***α* (*t*), *D α t* Z R*n φ*(*x*)*e* −2*πi*h*x*,*t*i *dx*i = ∑ |*α*|≤*m* (−1) |*α*| h**g***α* (*t*),(−1/2*πi*) |*α*| Z R*n φ*(*x*)(−2*πi*) |*α*| *x α e* −2*πi*h*x*,*t*i *dx*i = ∑ |*α*|≤*m* h(−1) |*α*|**g***α* (*t*), Z R*n x αφ*(*x*)*e* −2*πi*h*x*,*t*i *dx*i = ∑ |*α*|≤*m* h(−1) |*α*|**g***α* (*t*), F −1 [*x αφ*(*x*); *t*]i = ∑ |*α*|≤*m* hF −1 [(−1) |*α*|**g***α* (*t*); *x*], *x αφ*(*x*)i.

For each *α*, |*α*| ≤ *m*, put **h***α*(*x*) = F −1 [(−1) |*α*|**g***α* (*t*); *x*]. We have **h***α*(*x*) ∈ *L* 2 (R*n* , H), |*α*| ≤ *m*, since each **g***α* (*t*) ∈ *L* 2 (R*n* , H); moreover, ∑|*α*|≤*m* **h***α*(*x*) ∈ *L* 2 (R*n* , H). Thus, we have

$$
\langle \mathcal{F}^{-1}[V]\_{x'} \phi(x) \rangle = \sum\_{|a| \le m} \langle \mathbf{h}\_a(\mathbf{x}), \mathbf{x}^a \phi(\mathbf{x}) \rangle
$$

$$
= \langle \sum\_{|a| \le m} \mathbf{x}^a \mathbf{h}\_a(\mathbf{x}), \phi(\mathbf{x}) \rangle,
\tag{2}
$$

and F −1 [*V*]*x* = ∑|*α*|≤*m x α***h***α*(*x*) in S 0 (R*n* , H). For the *L* 2 (R*n* , H) norm | · |2 and the order *m* of the summation defining *V*, we consider

$$|(1+|\boldsymbol{x}|^{2})^{-m-2}\sum\_{|\boldsymbol{a}|\leq m}\boldsymbol{x}^{\boldsymbol{a}}\mathbf{h}\_{\boldsymbol{a}}(\boldsymbol{x})|\_{2}.\tag{3}$$

For |*α*| ≤ *m*, note that |*x α* | ≤ |*x*| |*α*| ≤ (1 + |*x*|) |*α*| ≤ (1 + |*x*|) *m*. Since (1 + |*x*|) *m* ≤ 2 *m* if |*x*| ≤ 1 and (1 + |*x*|) *m* ≤ (1 + |*x*| 2 ) *m* if |*x*| ≥ 1, then

$$\begin{aligned} &|\mathfrak{x}^{\mathfrak{a}}(1+|\mathfrak{x}|^{2})^{-m-2}| \leq (1+|\mathfrak{x}|)^{m}(1+|\mathfrak{x}|^{2})^{-m-2} \\ &\leq \max\_{\mathfrak{x}\in\mathbb{R}^{n}}\{2^{m},(1+|\mathfrak{x}|^{2})^{m}\}(1+|\mathfrak{x}|^{2})^{-m-2} \\ &\leq \max\{2^{m},1\}=2^{m} \end{aligned}$$

for |*α*| ≤ *m* since *m* ≥ 0 is a nonnegative integer. Thus, for the *L* 2 (R*n* , H) norm in (3), we have

$$\left| (1+|\boldsymbol{x}|^{2})^{-m-2} \sum\_{|\boldsymbol{a}| \leq m} \boldsymbol{\pi}^{\boldsymbol{a}} \mathbf{h}\_{\boldsymbol{a}}(\boldsymbol{\pi}) \right|\_{2} \leq 2^{m} \left| \sum\_{|\boldsymbol{a}| \leq m} \mathbf{h}\_{\boldsymbol{a}}(\boldsymbol{\pi}) \right|\_{2} < \infty \tag{4}$$

since ∑|*α*|≤*m* **h***α*(*x*) ∈ *L* 2 (R*n* , H). Recalling (2), we have by (4) that F −1 [*V*]*x* = ∑|*α*|≤*m x α***h***α*(*x*) ∈ S 0 2 (R*n* , H) for any *V* ∈ *L*2(R*n* , H); and *Vt* = F[∑|*α*|≤*m x α***h***α*(*x*)]*t* in S 0 (R*n* , H). Thus, the S 0 (R*n* , H) Fourier transform maps S 0 2 (R*n* , H) onto *L*2(R*n* , H); the fact that this mapping is one–one follows directly from the fact that the Fourier transform is a one–one mapping on S 0 (R*n* , H). The same statements and proofs as in this proof of Lemma 1 for the Fourier transform hold in exactly the same way for the inverse Fourier transform on S 0 (R*n* , H); and we have that the S 0 (R*n* , H) inverse Fourier transform maps *L*2(R*n* , H) one–one and onto S 0 2 (R*n* , H). The proof of Lemma 1 is complete.

Let *C* be a regular cone in R*n* ; that is, *C* is an open convex cone in R*n* , which does not contain any entire straight line. *C* ∗ = {*t* ∈ R*n* : h*t*, *y*i ≥ 0 for all *y* ∈ *C*} is the dual cone of *C*. We consider now the Cauchy kernel

$$K(z - t) = \int\_{\mathbb{C}^\*} e^{2\pi i \langle z - t, \mu \rangle} du, \; z \in T^{\mathbb{C}} = \mathbb{R}^n + i\mathbb{C}, \; t \in \mathbb{R}^n.$$

The ultradistributional test function spaces D(∗, *L p* ) ⊂ D*L p* (R*n* ), 1 < *p* ≤ ∞, where ∗ is Beurling (*Mp*) or Roumieu {*Mp*}, defined in [17] (Section 2.3, p. 21). For *C* being a regular cone, we proved in [17] (Section 4.1, Theorem 4.1.1) that *K*(*z* − ·) ∈ D(∗, *L p* ) ⊂ D*L p* (R*n* ) for *z* ∈ *T C*, 1 < *p* ≤ ∞ , under specified conditions on the sequence *Mp* of positive numbers, which we assume here. (See [17] (pp. 13–14, Theorem 4.1.1) for assumptions on the sequence *Mp*.) The Schwartz space D0 *L* 2 (R*n* ) consists of finite sums of distributional derivatives of *L* 2 (R*n* ) functions; thus, the space *L*2(R*n* , H|) is the extension of D0 *L* 2 (R*n* ) to vector-valued distributions with values in H. Thus, for *p* = 2, we emphasize that the form h*Vt* , *K*(*z* − *t*)i, *z* ∈ *T C*, is well defined for *V* ∈ *L*2(R*n* , H), and yields an element of H; the algebraic and differentiation calculations on the form h*V*, *φ*i hold for *V* ∈ *L*2(R*n* , H) and *φ* ∈ D*L* 2 (R*n* ), as usual, just as these calculations hold for the form h*V*, *φ*i for *V* ∈ S0 (R*n* , H) and *φ* ∈ S(R*n* ). We use this information in Section 5 of this paper.

#### **4. Boundary Values in** S 0 (R*n* **,** X )

Let *C* be an open convex cone in R*n* . In [4] (Theorem 8), we proved that an analytic function **f**(*z*), *z* ∈ *T C*, with values in a specified topological vector space X and satisfying a certain norm growth obtained a vector-value-tempered distributional boundary value, as *y* → 0, *y* ∈ *C* 0 ⊂⊂ *C*, for any compact subcone *C* 0 of *C*. The norm growth used in [4] (Theorem 8) was not as general as the growth of Tillmann [1] in which the original tempered distributional boundary value results in the scalar-valued case were obtained. In this section, we extend the result [4] (Theorem 8) by assuming a norm growth on the analytic function equivalent to that of Tillmann [1]; our result here also contains new information concerning the boundary value. As a corollary of our result, we obtain a precise representation of the boundary value when the conditions on the topological vector space X are restricted.

Following Vladimirov [11] (p. 230), we shall use the term "spectral function" but will extend the definition of this term to the vector-valued case. For an analytic function **f**(*z*), *z* ∈ *T C* = R*n* + *iC* ⊂ C*n* , with values in a topological vector space X , the spectral function of **f**(*z*) is that vector-valued distribution *V* ∈ D0 (R*n* , X ), such that *e* −2*π*h*y*,*t*i*Vt* ∈ S 0 (R*n* , X ), *y* ∈ *C*; and **f**(*x* + *iy*) = F[*e* −2*π*h*y*,*t*i*Vt* ]*x* in S 0 (R*n* , X ) for *z* = *x* + *iy* ∈ *T C*.

We begin by assuming that the topological vector space X is locally convex, separable, and quasi-complete where quasi-complete is in the sense of Schwartz [15] (p. 198). We further assume that X is a normed space with norm N . These stated assumptions on X were the assumptions under which we obtained [4] (Theorem 8) and are the assumptions on the topological vector space X under which we obtain Theorem 1 below.

Throughout the paper, by *y* → 0, *y* ∈ *C*, we mean that *y* → 0, *y* ∈ *C* 0 ⊂⊂ *C* for every compact subcone *C* 0 ⊂⊂ *C*.

The following theorem generalizes and extends [4] (Theorem 8) for X , satisfying the properties noted above.

**Theorem 1.** *Let C be an open convex cone. Let f*(*z*) *be analytic in TC and have values in* X *. Let*

$$\mathcal{N}(f(\mathbf{x} + i\mathbf{y})) \le M(1 + |z|)^q |y|^{-r}, \; z = \mathbf{x} + i\mathbf{y} \in T^{\mathbb{C}}.\tag{5}$$

*where M* > 0 *is a real constant, q is a nonnegative integer, r* > 1 *is an integer, and M*, *q, and r are independent of z* = *x* + *iy* ∈ *T C. There exists an element U* ∈ S0 (R*n* , X )*, such that*

$$\lim\_{y \to \overline{0}, y \in \mathbb{C}} f(x + iy) = \mathcal{U} \tag{6}$$

*in the weak and strong topologies of* S 0 (R*n* , X )*. Further, U* = F[*V*] *with V* ∈ S0 (R*n* , X ) *being the spectral function of f*(*z*), *z* ∈ *T C, , such that supp*(*V*) ⊆ *C* ∗ *.*

**Proof.** We apply the proofs of [4] (Theorems 3 and 8). Note that in the second sentence of the proof of [4] (Theorem 8) that the value of *η* ≥ 1 is arbitrary but fixed; in the present proof, we simply take *η* = 1, where it is appropriate to use *η* = 1. Let *λ* > 0; put *ρ* = *σ* + *iλ*, *σ* ∈ R1 ; and define **f** 0 (*ρ*; *x*, *y*) = **f**(*x* + *ρy*), *y* ∈ *pr*(*C*), where *pr*(*C*) denotes the projection of *C*, which is the intersection of *C* with the unit sphere in R*n* . (Thus, |*y*| = 1 if *y* ∈ *pr*(*C*).) **f** 0 (*ρ*; *x*, *y*) is an analytic function of *ρ* in the half plane *λ* = Im(*ρ*) > 0 and has values in X . We have **f** 0 (*ρ*; *x*, *y*) = **f**(*x* + *ρy*) = **f**((*x* + *σy*) + *iλy*), *λ* > 0, for *z* = *x* + *iy* with *y* ∈ *pr*(*C*); and note that *λy* ∈ *C* for *λ* > 0 and *y* ∈ *pr*(*C*). Now for *y* = Im(*z*) ∈ *pr*(*C*) and 0 < *λ* ≤ *η* = 1 we have

$$\begin{split} &\mathcal{N}(\mathbf{f}(\boldsymbol{\varrho};\boldsymbol{x},\boldsymbol{y})) \leq &\mathcal{M}(1+|(\boldsymbol{x}+\sigma\boldsymbol{y})+i\lambda\boldsymbol{y}|)^{q}|\lambda\boldsymbol{y}|^{-r} \\ &=\mathcal{M}(1+(\lambda^{2}+|\boldsymbol{x}+\sigma\boldsymbol{y}|^{2})^{1/2})^{q}\lambda^{-r} \\ &\leq &\mathcal{M}(1+(1+(|\boldsymbol{x}|+|\sigma|)^{2})^{1/2})^{q}\lambda^{-r} \\ &\leq &\mathcal{M}(1+((1+|\boldsymbol{x}|+|\sigma|)^{2})^{1/2})^{q}\lambda^{-r} \\ &=\mathcal{M}(2+|\boldsymbol{x}|+|\sigma|)^{q}\lambda^{-r} \end{split} \tag{7}$$

which is of the form, with norm N replacing the absolute value, of [4] (15), which is used in exactly the same way in the proof of [4] (Theorem 8) as in the proof of [4] (Theorem 3). Thus, for *y* = Im(*z*) ∈ *pr*(*C*) and 0 < *λ* ≤ *η* = 1 the bound on N (**f** 0 (*ρ*; *x*, *y*)) is in the proper form to proceed with the proof of this present Theorem 1 exactly as in the form of the proofs of [4] (Theorems 3 and 8). We obtain the structured function of the form Λ(−*r*−1) **f** 0 (*ρ*; *x*, *y*), *y* ∈ *pr*(*C*), which satisfies the growth (similar to [4] (37))

$$\mathcal{N}(\Lambda^{(-r-1)}\mathbf{f}'(\rho; \mathbf{x}, y)) \le M^{(r+1)}(2 + |\mathbf{x}| + |\sigma|)^q (2 + |\sigma|)^{r+1}$$

for 0 < *λ* ≤ *η* = 1 where *M*(*r*+1) is a positive constant, and obtains the representation (similar to [4] (38))

$$\mathbf{f}(\mathbf{x} + \rho y) = \mathbf{f}'(\rho; \mathbf{x}, y) = \frac{\partial^{r+1}(\Lambda^{(-r-1)}\mathbf{f}'(\rho; \mathbf{x}, y))}{\partial \sigma^{r+1}}, \; \sigma = \text{Re}(\rho).$$

Now, we proceed in our proof of Theorem 1 in exactly the same way as in [4] (Theorem 8) (p. 328) to obtain the desired boundedness properties leading to the existence of an element *V* ∈ D0 (R*n* , X ), such that *e* −2*π*h*y*,*t*i*Vt* ∈ S0 (R*n* , X ), *y* ∈ *C*, and **f**(*z*) = F[*e* −2*π*h*y*,*t*i*Vt* ]*x*, *z* = *x* + *iy* ∈ *T C* , in S 0 (R*n* , X ) from the results of Schwartz [14] (Prop. 22, p. 76). (These results of Schwartz [14] (Prop. 22, p. 76) were obtained in their original scalar-valued case in [20]; the related results were then obtained by Lions [21]). Thus, *V* ∈ D0 (R*n* , X ) is the spectral function of **f**(*z*), *z* ∈ *T C* . The remainder of the proof of [4] (Theorem 8, pp. 329–330) and the succeeding discussion after the conclusion of the proof of [4] (Theorem 8) can be applied to the present proof of Theorem 1 in the same way to yield that, in fact, *V* ∈ S0 (R*n* , X ) and that

$$\lim\_{y \to \overline{0}, y \in \mathbb{C}} \mathbf{f}(\mathbf{x} + iy) = \lim\_{y \to \overline{0}, y \in \mathbb{C}} \mathcal{F}[\mathfrak{e}^{-2\pi \langle y, t \rangle} V\_t] = \mathcal{F}[V] = \mathcal{U} \tag{8}$$

in the weak topology of S 0 (R*n* , X ). However, S(R*n* ) is a Montel space; thus, the convergence in (8) is in the strong topology of S 0 (R*n* , X ) as well. We emphasize that *V* ∈ S 0 (R*n* , X ) and that *U* = F[*V*] ∈ S0 (R*n* , X ) is the desired boundary value in (6) as obtained in (8).

We now prove that supp(*V*) ⊆ *C* ∗ . Let *to* ∈ *C*∗ = R*n* \ *C* ∗ ; *C*∗ is an open set in R*n* since *C* ∗ is a closed set. From the definition of *C* ∗ , for *to* ∈ *C*∗, there is a point *yo* ∈ *C*, such that h*yo*, *to*i < 0. Using the fact that *C*∗ is open and the continuity of h*t*, *yo*i at *to* ∈ *C*∗ as a function of *t*, there is a fixed *τ* > 0 and a fixed neighborhood *N*(*to*; *γ*) = {*t* ∈ R*n* : |*t* − *to*| < *γ*, *γ* > 0} ⊂ *C*∗, such that h*t*, *yo*i < −*τ* < 0 for all *t* ∈ *N*(*to*; *γ*). Let *φ* ∈ D(R*n* ), such that supp(*φ*) ⊂ *N*(*to*, *γ*). Recall that *V* ∈ S0 (R*n* , X ), such that *e* −2*π*h*y*,*t*i*Vt* ∈ S0 (R*n* , X ), *y* ∈ *C*, and **f**(*x* + *iy*) = F[*e* −2*π*h*y*,*t*i*Vt* ]*x*, *z* = *x* + *iy* ∈ *T C*, in S 0 (R*n* , X ). Thus *e* −2*π*h*y*,*t*i*Vt* = F −1 [**f**(*x* + *iy*)]*t* , *x* + *iy* ∈ *T C*, in S 0 (R*n* , X ); or *Vt* = *e* 2*π*h*y*,*t*iF −1 [**(***x* + *iy*)]*t* , *x* + *iy* ∈ *T C*, in S 0 (R*n* , X ). Let *y* = *βyo*, *yo* ∈ *C*, *β* > 0, now. We have *y* = *βyo* ∈ *C* and

$$\begin{split} \langle V, \Phi \rangle &= \langle e^{2\pi \langle \beta y\_o, t \rangle} \mathcal{F}^{-1} [\mathbf{f}(x + i \beta y\_o)\_{t \prime} \phi(t)] \\ &= \langle \mathcal{F}^{-1} [\mathbf{f}(x + i \beta y\_o)\_{t \prime} e^{2\pi \langle \beta y\_o, t \rangle} \phi(t)] \\ &= \langle \mathbf{f}(x + i \beta y\_o)\_{t \prime} \mathcal{F}^{-1} [e^{2\pi \langle \beta y\_o, t \rangle} \phi(t); x] \rangle \\ &= \int\_{\mathbb{R}^n} \mathbf{f}(x + i \beta y\_o) \int\_{supp(\phi)} e^{2\pi \langle \beta y\_o, t \rangle} \phi(t) e^{-2\pi i \langle x, t \rangle} dt d\mathbf{x} \end{split} \tag{9}$$

for the function *φ* ∈ D(R*n* ) chosen above. Using integration by parts and letting ∆ denote the Laplacian in the *t* ∈ R*n* variable, we have for any positive integer *m*

$$\mathcal{N}\left(\int\_{\mathbb{R}^n} \mathbf{f}(\mathbf{x} + i\beta y\_o) \int\_{\operatorname{supp}(\phi)} e^{2\pi \langle \beta y\_o, t \rangle} \phi(t) e^{-2\pi i \langle \mathbf{x}, t \rangle} dt d\mathbf{x}\right) \tag{10}$$

$$\begin{split} \mathcal{N} &= \mathcal{N}\left(\int\_{\mathbb{R}^n} \frac{\mathbf{f}(\mathbf{x} + i\beta y\_o)}{(1 + |\mathbf{x}|^2)^m} \int\_{\operatorname{supp}(\phi)} e^{2\pi \langle \beta y\_o, t \rangle} \phi(t) (1 + |\mathbf{x}|^2)^m e^{-2\pi i \langle \mathbf{x}, t \rangle} dt d\mathbf{x}\right) \\ &= \mathcal{N}\left(\int\_{\mathbb{R}^n} \frac{\mathbf{f}(\mathbf{x} + i\beta y\_o)}{(1 + |\mathbf{x}|^2)^m} \int\_{\operatorname{supp}(\phi)} (1 - \frac{\Delta}{4\pi^2})^m (e^{2\pi \langle \beta y\_o, t \rangle} \phi(t)) e^{-2\pi i \langle \mathbf{x}, t \rangle} dt d\mathbf{x}\right). \end{split}$$

(For the present, the positive integer *m* is arbitrary; later, we explicitly choose *m* to obtain the desired convergence of all integrals through Equation (15) below). For the interior integral over supp(*φ*) in (10), we note that by applying (1 − (∆/4*π* 2 ))*m* to the product *e* 2*π*h*βyo*,*t*i*φ*(*t*) and then bounding the terms in the resulting sum, including the terms involving 2*π* or it powers, we obtain a finite sum of terms involving powers of *β*(*yo*)*j* , *j* = 1, . . . , *n*, multiplied by *e* 2*π*h*βyo*,*t*i , where (*yo*)*j* is the *j*th component of *yo*, *j* = 1, . . . , *n*, and multiplied by bounds on *φ*(*t*) or one of its partial derivatives with *e* 2*π*h*βyo*,*t*i in each term of the sum. Of course, the boundedness of *φ*(*t*) and any of its partial derivatives are valid because of the compact support of *φ*(*t*). Moreover, note that |*β*(*yo*)*j* | ≤ *β*|*yo*|, *j* = 1, . . . , *n*. Thus, since the interior

integral in (10) is over supp(*φ*) ⊂ *N*(*to*; *γ*), we obtain the following bound on this interior integral:

$$\left| \int\_{\operatorname{supp}(\phi)} (1 - \frac{\Delta}{4\pi^2})^m (e^{2\pi \langle \beta y\_o, t \rangle} \phi(t)) e^{-2\pi i \langle \mathbf{x}, t \rangle} dt \right| $$

$$\leq \int\_{\operatorname{supp}(\phi)} |(1 - \frac{\Delta}{4\pi^2})^m (e^{2\pi \langle \beta y\_o, t \rangle} \phi(t))| dt \tag{11}$$

$$\leq T\_{\operatorname{supp}(\phi)} (1 + \beta |y\_o|)^{4(m+1)} \sup\_{t \in \operatorname{supp}(\phi)} e^{2\pi \langle \beta y\_o, t \rangle}$$

where *Tsupp*(*φ*) is a positive constant depending only on supp(*φ*). Using (11) in (10), we have

$$\mathcal{N}\left(\int\_{\mathbb{R}^n} \frac{\mathbf{f}(\mathbf{x} + i\beta y\_o)}{(1 + |\mathbf{x}|^2)^m} \int\_{\operatorname{supp}(\phi)} (1 - \frac{\Delta}{4\pi^2})^m (e^{2\pi \langle \beta y\_o, t\rangle} \phi(t)) e^{-2\pi i \langle \mathbf{x}, t\rangle} dt d\mathbf{x}\right)$$

$$\leq T\_{\operatorname{supp}(\phi)} (1 + \beta |y\_o|)^{4(m+1)} \sup\_{t \in \operatorname{supp}(\phi)} e^{2\pi \langle \beta y\_o, t\rangle} \int\_{\mathbb{R}^n} \frac{N(\mathbf{f}(\mathbf{x} + i\beta y\_o))}{(1 + |\mathbf{x}|^2)^m} d\mathbf{x}\tag{12}$$

where *yo* ∈ *C*, *β* > 0 is arbitrary, and supp(*φ*) ⊂ *N*(*to*; *γ*) ⊂ *C*∗, *to* ∈ *C*∗, *γ* > 0 and fixed. As noted before, since h*yo*, *to*i < 0 and *C*∗ is open, by the continuity of h*t*, *yo*i at *to* ∈ *C*∗ as a function of *t* ∈ R*n* , the fixed *τ* > 0 is chosen and the fixed *N*(*to*; *γ*) ⊂ *C*∗ is chosen, such that h*t*, *yo*i < −*τ* < 0 for all *t* ∈ *N*(*to*; *γ*) ⊂ *C*∗. Since supp(*φ*) ⊂ *N*(*to*; *γ*), we have

$$\sup\_{t \in \sup p(\phi)} e^{2\pi \langle \beta y\_\circ, t \rangle} \le e^{-2\pi \tau \beta} \rho$$

which yields from (12)

$$\mathcal{N}\left(\int\_{\mathbb{R}^n} \frac{\mathbf{f}(\mathbf{x} + i\beta y\_o)}{(1 + |\mathbf{x}|^2)^m} \int\_{\operatorname{supp}(\phi)} (1 - \frac{\Delta}{4\pi^2})^m (e^{2\pi \langle \beta y\_o, t \rangle} \phi(t)) e^{-2\pi i \langle \mathbf{x}, t \rangle} dt d\mathbf{x}\right)$$

$$\leq T\_{\operatorname{supp}(\phi)} e^{-2\pi \tau \beta} (1 + \beta |y\_o|)^{4(m+1)} \int\_{\mathbb{R}^n} \frac{\mathcal{N}(\mathbf{f}(\mathbf{x} + i\beta y\_o))}{(1 + |\mathbf{x}|^2)^m} d\mathbf{x}\tag{13}$$

where *yo* ∈ *C*, *τ* > 0, and *γ* > 0 are fixed and are independent of the arbitrary *β* > 0. We now bound the integral on the right of the inequality in (13) using the assumed growth (5) on **f**(*z*), *z* ∈ *T C*; (13) holds for all *β* > 0. To obtain the supp(*V*) containment result, we are going to let *β* → ∞ in (13); thus, we may assume that *β* > 1 in the remainder of this proof. By simple calculations and for *β* > 1 , we have

$$|1 + |\mathbf{x} + i\beta y\_o| = \beta(\frac{1}{\beta} + ((\frac{|\mathbf{x}|}{\beta})^2 + |y\_o|^2)^{1/2}) \le \beta(1 + (|\mathbf{x}|^2 + |y\_o|^2)^{1/2}),$$

and

$$(1 + |\mathbf{x} + i\beta y\_o|)^q \le \beta^q (1 + (|\mathbf{x}|^2 + |y\_o|^2)^{1/2})^q \le \beta^q (1 + |y\_o| + |\mathbf{x}|)^q$$

Hence, from (5),

$$\mathcal{N}(\mathbf{f}(\mathbf{x} + i\beta y\_o)) \le M\beta^q (1 + |y\_o| + |\mathbf{x}|)^q |\beta y\_o|^{-r}$$

and

$$\int\_{\mathbb{R}^n} \frac{\mathcal{N}(\mathbf{f}(\mathbf{x} + i\beta y\_o))}{(1 + |\mathbf{x}|^2)^m} d\mathbf{x} \le M \beta^{q-r} |y\_o|^{-r} \int\_{\mathbb{R}^n} \frac{(1 + |y\_o| + |\mathbf{x}|)^q}{(1 + |\mathbf{x}|^2)^m} d\mathbf{x}.\tag{14}$$

.

Combining (10), (12), (13), and (14) yields

$$\mathcal{N}\left(\int\_{\mathbb{R}^n} \mathbf{f}(\mathbf{x} + i\beta y\_o) \int\_{\operatorname{supp}(\phi)} e^{2\pi \langle \beta y\_o, t \rangle} \phi(t) e^{-2\pi i \langle \mathbf{x}, t \rangle} dt d\mathbf{x} \right) \tag{15}$$

$$\leq M T\_{\operatorname{supp}(\phi)} (1 + \beta |y\_o|)^{4(m+1)} \beta^{q-r} |y\_o|^{-r} e^{-2\pi \tau \beta} \int\_{\mathbb{R}^n} \frac{(1 + |y\_o| + |\mathbf{x}|)^q}{(1 + |\mathbf{x}|^2)^m} d\mathbf{x}.$$

The positive integer *m* in (15) was introduced in (10), and at that point in the proof, *m* was arbitrary. We now choose *m*, such that *m* > 2(*q* + *n* + 1). With this choice of *m*, the integral in (15) converges where *yo* ∈ *C* is a fixed point in *C*; further, with this choice of *m*, all calculations from (10) leading to (15) are valid and the integrals converge. Because of the exponential term *e* −2*πτβ*, where *τ* > 0 is fixed and now *β* > 1 is arbitrary, the right side of (15) has limit 0 as *β* → ∞. Thus, from (9) h*V*, *φ*i = Θ for *φ* ∈ D(R*n* ), such that supp(*φ*) ⊂ *N*(*to*, *γ*) ⊂ *C*∗ for *to* being an arbitrary but fixed point in the open set *C*∗ = R*n* \ *C* ∗ . That is, for each fixed point, *to* ∈ *C*∗ = R*n* \ *C* ∗ , with *C*∗ being an open set, there is a neighborhood *N*(*to*; *γ*) ⊂ *C*∗ of *to*, such that for all *φ* ∈ D(R*n* ) with supp(*φ*) ⊂ *N*(*to*; *γ*), we have h*V*, *φ*i = Θ. Thus, *V* vanishes on a neighborhood of each point of *C*∗; this proves that *V* vanishes on the open set *C*∗ = R*n* \ *C* ∗ . Thus, supp(*V*) ⊆ *C* ∗ , which is a closed set in R*n* . The proof of Theorem 1 is complete.

Yoshinaga [22] (Proposition 3) provides a representation of the tempered vectorvalued distributions in the case of the topological vector space X being a complete space of type (DF). Yoshinaga's result is as follows for X , being a complete space of type (DF): *V* ∈ S0 (R*n* , X ), if and only if there exists a continuous function **g** on R*n* with values in X , an integer *k* ≥ 0, and a n-tuple *α* of nonnegative integers, such that *V* = *Dα***g** and {**g**(*t*)/(1 + |*t*| 2 ) *kn*; *t* ∈ R*n*} is a bounded subset of X . (In fact, in Yoshinaga's symbolism, *α* = (*k*, *k*, ..., *k*).)

The functions S 0 2 (R*n* , X ) of Definition 1 are an integral part of the following corollary to Theorem 1; recall that these functions are defined by the necessity for X being a Banach space. We know that a Banach space satisfies all of the conditions on X stated prior to Theorem 1 and also is a complete norm space of type (DF); since a Hilbert space is a Banach space, a Hilbert space also satisfies all of these stated conditions on X . Thus, the abovestated result of Yoshinaga and Theorem 1 of this paper both hold for X being a Banach or Hilbert space.

We obtain a corollary of Theorem 1 now in which more precise information is obtained concerning the spectral function *V* and the boundary value *U* of Theorem 1.

**Corollary 1.** *Let C be an open convex cone and* X *be a Banach space. Let f*(*z*) *be analytic in T C* = R*n* + *iC, have values in* X *, and satisfy* (5)*. There is a continuous function g* ∈ S0 2 (R*n* , X ) *with supp*(*g*) ⊆ *C* ∗ *a*.*e*. *and an n-tuple α of nonnegative integers, such that the spectral function V* ∈ S0 (R*n* , X ) *of Theorem 1 has the form Vt* = *Dα t g*(*t*)*, and there is U* = F[*V*] ∈ S0 (R*n* , X ) *such that*

$$\lim\_{y \to \overline{0}, y \in \mathbb{C}} f(x + iy) = \mathcal{U}$$

*in the weak and strong topologies of* S 0 (R*n* , X )*. Further, for* X = H *being a Hilbert space, we have* F[*g*] ∈ *L*2(R*n* , H)*; and the boundary value U* ∈ S0 (R*n* , H) *has the form*

$$\mathcal{U}\_{\mathbf{x}} = \mathbf{x}^{a} \mathcal{F}[\mathbf{g}]\_{\mathbf{x}} = \mathbf{x}^{a} (1 - \frac{\Delta}{4\pi^{2}})^{m} h(\mathbf{x}) \tag{16}$$

*in* S 0 (R*n* , H) *where h* ∈ *L* 2 (R*n* , X ), *α is an n-tuple of nonnegative integers, and m* ≥ 0 *is a real number that can be taken to be a nonnegative integer.*

**Proof.** We apply the results of Theorem 1 and consider the spectral function *V* ∈ S0 (R*n* , X ) obtained in Theorem 1 where X is a Banach space in this corollary. As per the result of Yoshinaga [22] (Proposition 3) stated above, there is a continuous function **g** on R*n*

with values in X , an n-tuple *α* of nonnegative integers, and an integer *k* ≥ 0, such that *Vt* = *Dα t* **g**(*t*) and { **g**(*t*) (1+|*t*| 2) *kn* ; *t* ∈ R*n*} is a bounded subset of X . (In Yoshinaga's symbolism, *α* is the n-tuple with all components being *k*.) Thus, there is a real constant *R* > 0, such that

$$\mathcal{N}\left(\frac{\mathbf{g}(t)}{(1+|t|^2)^{kn}}\right) = \frac{\mathcal{N}(\mathbf{g}(t))}{(1+|t|^2)^{kn}} \le \mathbb{R}, \ t \in \mathbb{R}^n.$$

For the integer *k* ≥ 0, we have

$$\begin{aligned} &\int\_{\mathbb{R}^n} (\mathcal{N}\left(\frac{\mathbf{g}(t)}{(1+|t|^2)^{(k+2)n}}\right))^2 dt \\ &= \int\_{\mathbb{R}^n} \left(\frac{1}{(1+|t|^2)^{2n}}\right)^2 (\mathcal{N}\left(\frac{\mathbf{g}(t)}{(1+|t|^2)^{kn}}\right))^2 dt \\ &\le R^2 \int\_{\mathbb{R}^n} \frac{1}{(1+|t|^2)^{4n}} dt < \infty \end{aligned}$$

which proves that **g** ∈ S0 2 (R*n* , X ). Further, supp(**g**) ⊆ *C* ∗ a.e. since supp(*V*) ⊆ *C* ∗ . From Theorem 1, the boundary value *U* ∈ S0 (R*n* , X ) in (6) is *U* = F[*V*], the Fourier transform of the spectral function *V* ∈ S0 (R*n* , X ) in S 0 (R*n* , X ). Moreover, from Theorem 1, the boundary value *U* is obtained in both the weak and strong topologies of S 0 (R*n* , X ).

Now, let X = H, a Hilbert space, in this Corollary 1. Since **g** ∈ S0 2 (R*n* , H), then F[**g**] ∈ *L*2(R*n* , H) in S 0 (R*n* , H) by Lemma 1. We know from the above that the boundary value *U* ∈ S0 (R*n* , H) is *U* = F[*V*], and *V* ∈ S0 (R*n* , H) has the form *Vt* = *Dα t* **g**(*t*) in S 0 (R*n* , H). Let *φ* ∈ S(R*n* ). We have

$$
\begin{split}
\langle\mathcal{U},\boldsymbol{\Phi}\rangle &= \langle\mathcal{F}[\boldsymbol{V}],\boldsymbol{\Phi}\rangle = \langle\boldsymbol{V},\boldsymbol{\hat{\Phi}}\rangle = \langle\boldsymbol{D}^{a}\_{t}\mathbf{g}(t),\boldsymbol{\hat{\Phi}}(t)\rangle \\ &= (-1)^{|\boldsymbol{\alpha}|} \int\_{\mathbb{R}^{\boldsymbol{u}}} \mathbf{g}(t) D^{a}\_{t} \int\_{\mathbb{R}^{\boldsymbol{u}}} \boldsymbol{\phi}(\boldsymbol{x}) e^{2\pi i \langle \boldsymbol{x},t\rangle} d\boldsymbol{x} dt \\ &= (-1)^{|\boldsymbol{\alpha}|} \int\_{\mathbb{R}^{\boldsymbol{u}}} \mathbf{g}(t) \int\_{\mathbb{R}^{\boldsymbol{u}}} \boldsymbol{\phi}(\boldsymbol{x}) (-1/2\pi i)^{|\boldsymbol{\alpha}|} (2\pi i)^{|\boldsymbol{\alpha}|} \boldsymbol{x}^{\boldsymbol{\alpha}} e^{2\pi i \langle \boldsymbol{x},t\rangle} d\boldsymbol{x} dt \\ &= \langle\boldsymbol{\Phi}(t),\mathcal{F}[\boldsymbol{x}^{\boldsymbol{a}}\boldsymbol{\phi}(\boldsymbol{x});t]\rangle = \langle\mathcal{F}[\boldsymbol{g}]\_{\boldsymbol{x}},\boldsymbol{x}^{\boldsymbol{a}}\boldsymbol{\phi}(\boldsymbol{x})\rangle = \langle\boldsymbol{x}^{\boldsymbol{a}}\mathcal{F}[\boldsymbol{g}]\_{\boldsymbol{x}},\boldsymbol{\phi}(\boldsymbol{x})\rangle.
\end{split}
$$

Thus, *Ux* = *x α*F[**g**]*x* in S 0 (R*n* , H) with **g** ∈ S0 2 (R*n* , H). Since **g** ∈ S0 2 (R*n* , H), by definition there is a real number *m* ≥ 0, such that **g**(*t*)/(1 + |*t*| 2 ) *m* ∈ *L* 2 (R*n* , H), and *m* can be taken to be a nonnegative integer. We have—by the proof of Lemma 1—that F[**g**]*x* = (1 − (4*π* 2 ) −1∆) *m***h**(*x*) ∈ *L*2(R*n* , H) in S 0 (R*n* , H), where **h** ∈ *L* 2 (R*n* , H) and ∆ is the Laplace operator in the *x* ∈ R*n* variable. Combining equalities, we have

$$\mathcal{U}\_{\mathbf{x}} = \mathfrak{x}^{\mathfrak{a}} \mathcal{F}[\mathbf{g}]\_{\mathfrak{x}} = \mathfrak{x}^{\mathfrak{a}} (1 - \frac{\Delta}{4\pi^2})^m \mathbf{h}(\mathfrak{x})^\mathfrak{x}$$

in S 0 (R*n* , H) with **h** ∈ *L* 2 (R*n* , H), which is (16). The proof is complete.

#### **5. Cauchy Integral**

A Cauchy integral of tempered distributions S 0 (R*n* ) has been defined in one and many dimensions. Of course, the main problem in making such a definition is that the Cauchy kernel is not a tempered function in S(R*n* ); an arbitrary element of S 0 (R*n* ) applied to the Cauchy kernel is not well defined.

Let *C* be a regular cone in R*n* ; that is, *C* is an open convex cone that does not contain an entirely straight line. With *C* ∗ being the dual cone of *C*, the Cauchy kernel function is

$$K(z - t) = \int\_{\mathbb{C}^\*} e^{2\pi i \langle z - t, \mu \rangle} d\mu, \; z \in T^{\mathbb{C}}, \; t \in \mathbb{R}^n.$$

as defined in Section 3. For the tube *T C* being the upper or lower half-planes in C1 or the tube defined by one of the 2 *n* quadrant cones *Cµ* = {*y* ∈ R*n* : (−1) *µjyj* > 0, *j* = 1, . . . , *n*} where *µ* is any of the 2 *n* n-tuples whose components are either 0 or 1, the Cauchy kernel takes the usual form. In order to generate an element of S(R*n* ) from the Cauchy kernel in the half plane setting in C1 and the tube defined by a quadrant cone, one divides the Cauchy kernel by a certain specifically chosen polynomial.

Sebastião e Silva [5] introduced a Cauchy integral for tempered distributions in the half-plane setting. Carmichael [7] defined a Cauchy integral for tempered distributions in the C*n* setting corresponding to analytic functions in the quadrant cone setting *T Cµ* in C*n* and showed that the analytic functions in (C − R) *n* , which have distributional boundary values in S 0 (R*n* ), can be recovered as the Cauchy integral of the boundary value; the results of [7] can be extended to the vector-valued tempered distributions considered in this paper by the same techniques as those in [7]. The Cauchy integrals introduced by both Sebastião e Silva and Carmichael are in fact equivalence classes of analytic functions defined by an integral involving the Cauchy kernel. Vladimirov [8–10] has defined a Cauchy integral for tempered distributions associated with analytic functions in general tubes *T C* = R*n* + *iC* ⊂ C*n* corresponding to regular cones *C* similar to the analytic functions we considered in this paper. Vladimirov showed that the analytic functions that he considered can be recovered by a Cauchy integral involving the tempered distributional boundary values of the analytic functions. The papers mentioned in this paragraph all concern scalar-valued analytic functions and distributions.

In this section, we build on our analyses of Sections 3 and 4 to obtain a Cauchy integral representation of the vector-valued analytic functions, which we considered in Theorem 1 and in Corollary 1. The proof of our results here—and the forms of our results—are different from any of the previous results concerning the Cauchy integral of the tempered distribution representation of the analytic functions. By our technique here, we do not need to divide the Cauchy kernel or the boundary value in (16) by a specified form of the polynomial and do not need to apply other special features of proof previously used by the authors in order to obtain that our Cauchy integral is well defined and that the analytic function considered is represented by a Cauchy integral involving the boundary value.

The Cauchy integral representation of the analytic functions that we considered in this paper follows. Note that cone *C* in the following result is assumed to be a regular cone. In Theorem 1 and Corollary 1, we assumed that cone *C* was an open convex cone. However, an open convex cone could contain an entirely straight line; in this case, the dual cone has measure 0 and *K*(*z* − *t*) = 0, *z* ∈ *T C*, *t* ∈ R*n* . To avoid this triviality, we assume that cone *C* in the following Cauchy integral representation is a regular cone.

**Theorem 2.** *Let C be a regular cone in* R*n and* H *be a Hilbert space. Let f*(*z*) *be analytic in T C* = R*n* + *iC, have values in* H*, and satisfy* (5)*. There is a continuous function g* ∈ S0 2 (R*n* , H) *with supp*(*g*) ⊆ *C* ∗ *a.e. and an n-tuple α of nonnegative integers, such that*

$$f(z) = z^{\mathfrak{a}} \langle \mathcal{F}[\mathfrak{g}]\_{\nu} \mathcal{K}(z - \nu) \rangle, \; z \in T^{\mathbb{C}}, \tag{17}$$

*in* S 0 (R*n* , H)*. Further,*

$$
\langle \mathcal{F}[\mathbf{g}]\_{\nu} \, \mathsf{K}(z - \nu) \rangle = \Theta \, \mathsf{i} \, z \in T^{-\mathsf{C}} \, \mathsf{A} \tag{18}
$$

*in* S 0 (R*n* , H)*.*

**Proof.** From Theorem 1, there is an element *V* ∈ S0 (R*n* , H), the spectral function of **f**(*z*), *z* ∈ *T C*, such that *e* −2*π*h*y*,*t*i*Vt* ∈ S0 (R*n* , H), *y* ∈ *C*; supp(*V*) ⊆ *C* ∗ ; and **f**(*z*) = F[*e* −2*π*h*y*,*t*i*Vt* ]*x*, *y* ∈ *C*, in S 0 (R*n* , H). Further, by Corollary 1, there is a continuous function **g** ∈ S0 2 (R*n* , H) with supp(**g**) ⊆ *C* ∗ a.e. and an n-tuple *α* of nonnegative integers, such that *Vt* = *Dα t* **g**(*t*), *t* ∈ R*n* . Now, let *φ* ∈ S(R*n* ) and *z* = *x* + *iy* ∈ *T C*. Recall that we have defined the differential operator *D* to be *Dt* = (−1/2*πi*)( *∂ ∂t*1 , . . . , *∂ ∂tn* ). We have

h**f**(*x* + *iy*), *φ*(*x*)i = hF[*e* −2*π*h*y*,*t*i*Vt* ]*x*, *φ*(*x*)i = h*e* −2*π*h*y*,*t*i*Vt* , *φ*ˆ(*t*)i = h*Vt* , Z R *φ*(*x*)*e* 2*πi*h*z*,*t*i *dx*i = h*D α t* **g**(*t*), Z R*n φ*(*x*)*e* 2*πi*h*z*,*t*i *dx*i (19) = (−1) |*α*| Z *C*∗ **g**(*t*) Z R*n φ*(*x*)(−1/2*πi*) |*α*| (2*πi*) |*α*| *z α e* 2*πi*h*z*,*t*i *dxdt* = Z *C*∗ **g**(*t*) Z R*n φ*(*x*)*z α e* −2*π*h*y*,*t*i *e* 2*πi*h*x*,*t*i *dxdt* = Z *C*∗ *e* −2*π*h*y*,*t*i**g**(*t*)F[*z αφ*(*x*); *t*]*dt* = h*z α*F[*IC*∗ (*t*)*e* −2*π*h*y*,*t*i**g**(*t*)]*x*, *φ*(*x*)i

where *IC*∗ (*t*) is the characteristic function of *C* ∗ . We have proven in [17] (Lemma 4.2.1, p. 62) that *IC*∗ (*t*)*e* −2*π*h*y*,*t*i ∈ *L p* , *y* ∈ *C*, for all *p*, 1 ≤ *p* ≤ ∞. Since **g** ∈ S0 2 (R*n* , H), then F[**g**]*x* ∈ *L*2(R*n* , H) in S 0 (R*n* , H) by Lemma 1. Recall also from Section 3 that the Cauchy kernel *K*(*z* − ·) ∈ D(∗, *L p* ) ⊂ D*L p* (R*n* ), 1 < *p* ≤ ∞, for *z* ∈ *T C* with *C* being a regular cone and that an element of *L*2(R*n* , H) applied to *K*(*z* − ·), *z* ∈ *T C*, is a well-defined function of *z* ∈ *T C*. Continuing (19) and using convolution, we now have

$$\begin{split} \langle \mathbf{f}(\mathbf{x} + i\mathbf{y}), \boldsymbol{\phi}(\mathbf{x}) \rangle &= \langle z^{\mathbf{a}}(\mathcal{F}[\mathbf{g}] \ast \mathcal{F}[I\_{\mathbb{C}^\*}(t)e^{-2\pi \langle \mathbf{y}, t \rangle}])\_{\mathbf{x} \prime} \boldsymbol{\phi}(\mathbf{x}) \rangle \\ &= \langle z^{\mathbf{a}}(\mathcal{F}[\mathbf{g}]\_{\mathbb{V}^\prime} \mathcal{F}[I\_{\mathbb{C}^\*}(t)e^{-2\pi \langle \mathbf{y}, t \rangle}]\_{\langle \mathbf{x} - \mathbf{v} \rangle} \boldsymbol{\phi}(\mathbf{x}) \rangle \\ &= \langle z^{\mathbf{a}}(\mathcal{F}[\mathbf{g}]\_{\mathbb{V}^\prime} \int\_{\mathbb{C}^\*} e^{2\pi i \langle z - \mathbf{v}, t \rangle} dt \rangle, \boldsymbol{\phi}(\mathbf{x}) \rangle \\ &= \langle z^{\mathbf{a}}(\mathcal{F}[\mathbf{g}]\_{\mathbb{V}^\prime} \mathcal{K}(z - \mathbf{v})), \boldsymbol{\phi}(\mathbf{x}) \rangle \end{split} \tag{20}$$

where *IC*∗ (*t*) is the characteristic function of *C* ∗ . Since **g** ∈ S0 2 (R*n* , H), then F[**g**] ∈ *L*2(R*n* , H) by Lemma 1; and as previously noted, F[**g**] applied to the Cauchy kernel is a well-defined function of *z* ∈ *T C* and is an analytic function of *z* ∈ *T C* with values in H. Thus, from (20) we have obtained

$$\mathbf{f}(z) = z^{\kappa} \langle \mathcal{F}[\mathbf{g}]\_{\nu} \, K(z - \nu) \rangle\_{\nu} \, z \in T^{\mathbb{C}}\_{\nu}$$

in S 0 (R*n* , H), and (17) is obtained.

To prove (18), first note that for a regular cone, *C*, −*C* is also a regular cone; and (−*C*) ∗ = −*C* ∗ . Thus, for *z* ∈ *T* −*C* and *φ* ∈ S(R*n* ),

$$\begin{split} \langle \langle \mathcal{F}[\mathbf{g}]\_{\nu}, K(z-\nu) \rangle, \boldsymbol{\phi}(\mathbf{x}) \rangle &= \langle \langle \mathcal{F}[\mathbf{g}]\_{\nu}, \int\_{-\mathbb{C}^{\*}} e^{-2\pi \langle \boldsymbol{y}, t \rangle} e^{2\pi i \langle \mathbf{x} - \boldsymbol{y}, t \rangle} d\boldsymbol{t} \rangle, \boldsymbol{\phi}(\mathbf{x}) \rangle \\ &= \langle \langle \mathcal{F}[\mathbf{g}]\_{\nu}, \mathcal{F}[I\_{-\mathbb{C}^{\*}}(t) e^{-2\pi \langle \boldsymbol{y}, t \rangle}]\_{(\mathbf{x}-\boldsymbol{y})}, \boldsymbol{\phi}(\mathbf{x}) \rangle \\ &= \langle \langle (\mathcal{F}[\mathbf{g}] \* \mathcal{F}[I\_{-\mathbb{C}^{\*}}(t) e^{-2\pi \langle \boldsymbol{y}, t \rangle}])\_{\mathbf{x}} \rangle, \boldsymbol{\phi}(\mathbf{x}) \rangle \\ &= \langle \mathcal{F}[I\_{-\mathbb{C}^{\*}}(t) e^{-2\pi \langle \boldsymbol{y}, t \rangle} \mathbf{g}(t)]\_{\mathbf{x}}, \boldsymbol{\phi}(\mathbf{x}) \rangle. \end{split} \tag{21}$$

Now *I*−*C*∗ (*t*) = 0 if *t* ∈ − / *C* ∗ and, hence, if *t* ∈ *C* ∗ . This fact coupled with the fact that supp(**g**) ⊆ *C* ∗ a.e. yields *I*−*C*∗ (*t*)*e* −2*π*h*y*,*t*i**g**(*t*) = Θ a.e. for *t* ∈ R*n* and *y* ∈ −*C*. Hence F[*I*−*C*∗ (*t*)*e* −2*π*h*y*,*t*i**g**(*t*)]*x* = Θ, *x* ∈ R*n* , *y* ∈ −*C*, in (21). Thus, from (21), we have hF[**g**]*ν*, *K*(*z* − *ν*)i = Θ, *z* ∈ *T* −*C*, in S 0 (R*n* , H); and (18) is obtained.

#### **6. Conclusions**

Tillmann [1] obtained the original analysis concerning the scalar-valued tempered distributions S 0 (R*n* ) as boundary values of analytic functions. We proved a boundary value result concerning vector-valued tempered distributions S 0 (R*n* , X ) as boundary values of

vector-valued analytic functions in [4] (Theorem 8) but used a norm growth condition on the analytic functions, which was a special case for the growth of Tillmann. We desired to obtain a result, such as [4] (Theorem 8), but under the general norm growth on the analytic function, which was equivalent to the growth of Tillmann. We achieved this first goal of this paper in Theorem 1 for vector-valued analytic functions **f**(*z*) on tubes *T C* = R*n* + *iC* with *C* being an open convex cone. The values of the analytic functions and the tempered distributions were in a very general type of topological vector space. We achieved additional information in Theorem 1 concerning the spectral function of the analytic function.

We asked if additional information concerning the spectral function and the boundary value could be obtained if the topological vector space X was restricted somewhat. We obtained the desired information in Corollary 1 by restricting X to be a Banach space and then a Hilbert space; we showed the structure of the spectral function and the boundary value in these cases for X . Integral to this analysis was the Lemma 1 result, which proved the relation under the Fourier transform between two important subsets of S 0 (R*n* , H) for our results in Corollary 1. It is important to note that the reason to restrict to Hilbert space H (which we do in our results) is that the Plancherel theory for the Fourier transform of the functions holds if and only if the functions have value in the Hilbert space.

The second principal goal of this paper was to obtain a Cauchy integral representation of the analytic functions considered in Theorem 1 and Corollary 1. Sebastião e Silva, Carmichael, and Vladimirov have obtained and studied the Cauchy integral of tempered distributions S 0 (R*n* ) in the scalar-valued case and in one and several dimensions; see the papers of these authors in the references. Their analyses basically concerned dividing the Cauchy kernel or the boundary value by a suitable polynomial whose order was large enough to make the quotient when evaluated by the tempered distribution to be well defined, or used other special features of proof that we do not use here.

In Section 5 of this paper, we constructed our Cauchy integral used in the representation of the assumed analytic function in a different manner by using the general known structure of the spectral function and our proven structure of the tempered distributional boundary value in S 0 (R*n* , H) for H being a Hilbert space. The analytic function obtaining the boundary value in S 0 (R*n* , H) was shown to be equated to the product of a polynomial and the constructed Cauchy integral.

This paper concerns theoretical mathematics, yet the topics considered find applications in mathematical physics and in mathematics that are applied to physical problems. We survey historically some areas of application in the scalar-valued case. We recall the work of Streater and Wightman [23] in studying quantum field theory. In a field theory, the "vacuum expectation values" are tempered distributions, which are boundary values in the tempered distribution topology of analytic functions with the analytic functions being Fourier–Laplace transforms. In addition, a field theory can be recovered from its "vacuum expectation values"; see [23] (Chapter 3). A similar field theory analysis using boundary values of analytic functions is contained in the work by Simon [24]. We also reference Raina [25] concerning "form factor bounds" in particle physics in which tempered distributional boundary values, which are of a special form, imply that the analytic functions that obtain these boundary values are Hardy *Hp* functions; this fact is then used in the analysis of the "form factor bounds". See also the associated papers listed in the references of [25].

As noted in Vladimirov [8], scalar-valued analytic functions of the type that we considered in this paper can arise in applying the Fourier–Laplace transform to convolution equations, which describe linear homogeneous processes with causality that find application in the quantum field theory, theory of electrical circuits, scattering of electromagnetic waves, and linear thermodynamic systems; refer to the list of references in [8]. We also note paper [26] by Vladimirov, concerning the linear conjugacy of scalar-valued analytic functions of several complex variables, which are again of the type that we considered in this paper with respect to growth. The linear conjugacy analysis involves scalar-valued tempered distributional boundary values of analytic functions represented as Fourier– Laplace integrals. Vladimirov [26] (p. 207) states that many problems arising in mathematical physics reduce to the problem of linear conjugacy involving tempered distributions; Vladimirov [26] provides examples of such problems.

The survey of applications above (concerning the type of analysis used in this paper) involve scalar-valued functions and distributions. Yet, a close consideration of the linear conjugacy problem of [26], together with the vector-valued analysis of this paper, leads one to believe that the linear conjugacy problem can be extended to the vector-valued case. Further, in an analysis of the stated applications above, one must sometimes obtain a distributional solution of a partial differential equation; such calculations can be extended to the vector-valued case. We suggest that the considerable related analyses to the results of this paper and the results of related references in this paper can be achieved in the vector-valued case and will work toward this end in the future.

**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 author declares no conflict of interest.

#### **References**

## *Article* **On Special Fuzzy Differential Subordinations Obtained for Riemann–Liouville Fractional Integral of Ruscheweyh and S˘al˘agean Operators**

**Alina Alb Lupa¸s**

Department of Mathematics and Computer Science, University of Oradea, 1 Universitatii Street, 410087 Oradea, Romania<EMAIL_ADDRESS>
**Abstract:** New results concerning fuzzy differential subordination theory are obtained in this paper using the operator denoted by *D*−*λ z L n α* , previously introduced by applying the Riemann–Liouville fractional integral to the convex combination of well-known Ruscheweyh and S˘al˘agean differential operators. A new fuzzy subclass *DL*F *n* (*δ*, *α*, *λ*) is defined and studied involving the operator *D*−*λ z L n α* . Fuzzy differential subordinations are obtained considering functions from class *DL*F *n* (*δ*, *α*, *λ*) and the fuzzy best dominants are also given. Using particular functions interesting corollaries are obtained and an example shows how the obtained results can be applied.

**Keywords:** differential operator; fuzzy differential subordination; fuzzy best dominant; fractional integral

**MSC:** 30C45; 30A10; 33C05

#### **1. Introduction**

The concept of fuzzy set, introduced by Lotfi A. Zadeh in 1965 [1], has opened the way for a new theory called fuzzy set theory. It has developed intensely, nowadays having applications in many branches of science and technology.

The fuzzy set concept was applied for developing new directions of study in many mathematical theories. In geometric function theory, it was used for introducing the new concepts of fuzzy subordination [2] and fuzzy differential subordinations [3] as generalizations of the classical notion of differential subordination due to Miller and Mocanu [4,5]. The main aspects regarding the theory of differential subordination can be found in [6]. Steps in the evolution of the theory of fuzzy differential subordination can be followed in [7].

The general context of the study presented in this paper contains notions familiar to geometric function theory merged with fuzzy set theory. We first present the main classes of analytic functions involved and the definitions regarding fuzzy differential subordination theory.

*U* = {*z* ∈ C : |*z*| < 1} represents the unit disc of the complex plane and H(*U*) the space of holomorphic functions in *U*.

Consider A = { *f* ∈ H(*U*) : *f*(*z*) = *z* + *a*2*z* 2 + . . . , *z* ∈ *U*}, and H[*a*, *m*] = { *f* ∈ H(*U*) : *f*(*z*) = *a* + *amz m* + *am*+1*z m*+1 + . . . , *z* ∈ *U*}, for *a* ∈ C and *m* ∈ N.

We remember the usual definitions needed for fuzzy differential subordination:

**Definition 1** ([8])**.** *A fuzzy subset of X is a pair* (*M*, *FA*)*, with M* = {*x* ∈ *X* : 0 < *FM*(*x*) ≤ 1} *the support of the fuzzy set and FM* : *X* → [0, 1] *the membership function of the fuzzy set. It is denoted M* = supp(*M*, *FM*)*.*

$$\textbf{Remark 1.}\text{ When }M\subset X\text{, we have }\mathsf{F}\_M(\mathsf{x})=\left\{\begin{array}{c}1,\mathsf{if}\ x\in M,\\0,\mathsf{if}\ x\notin M.\end{array}\right.$$

**Citation:** Alb Lupa¸s, A. On Special Fuzzy Differential Subordinations Obtained for Riemann–Liouville Fractional Integral of Ruscheweyh and S˘al˘agean Operators. *Axioms* **2022**, *11*, 428. https://doi.org/10.3390/ axioms11090428

Academic Editor: Radko Mesiar

Received: 18 July 2022 Accepted: 23 August 2022 Published: 25 August 2022

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

**Copyright:** © 2022 by the author. 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/).

*Evidently F*∅(*x*) = 0*, x* ∈ *X, and FX*(*x*) = 1*, x* ∈ *X.*

**Definition 2** ([2])**.** *Let D* ⊂ C *and let z*0 ∈ *D be a fixed point. We take the functions f* , *g* ∈ H(*D*)*. The function f is said to be fuzzy subordinate to g and we write f* ≺F *g, if there exists a function F* : C → [0, 1] *such that f*(*z*0) = *g*(*z*0) *and Ff*(*D*) *f*(*z*) ≤ *Fg*(*D*) *g*(*z*)*, z* ∈ *D*.

**Remark 2.** *(1) If g is univalent, then f* ≺F *g if and only if f*(*z*0) = *g*(*z*0) *and f*(*D*) ⊂ *g*(*D*). |*z*| .

*(2) Such a function F* : C → [0, 1] *can be consider F*(*z*) = 1+|*z*| *, F*(*z*) = 1 1+|*z*|

*(3) If D* = *U the conditions become f*(0) = *g*(0) *and f*(*U*) ⊂ *g*(*U*)*, which is equivalent to the classical definition of subordination.*

**Definition 3** ([3])**.** *Consider h an univalent function in U and ψ* : C3 × *U* → C*, such that h*(0) = *ψ*(*a*, 0; 0) = *a. When the fuzzy differential subordination*

$$F\_{\boldsymbol{\upPsi}(\mathbb{C}^{3}\times\mathcal{U})}\psi(p(\boldsymbol{z}),zp'(\boldsymbol{z}),\boldsymbol{z}^{2}p''(\boldsymbol{z});\boldsymbol{z}) \leq F\_{\boldsymbol{\upPsi}(\mathcal{U})}h(\boldsymbol{z}), \quad \boldsymbol{z}\in\mathcal{U}.\tag{1}$$

*is satisfied for an analytic function p in U, such that p*(0) = *a, then p is called a fuzzy solution of the fuzzy differential subordination. A fuzzy dominant of the fuzzy solutions of the fuzzy differential subordination is an univalent function q for which Fp*(*U*) *p*(*z*) ≤ *Fq*(*U*) *q*(*z*)*, z* ∈ *U, for all p satisfying (1). The fuzzy best dominant of (1) is a fuzzy dominant q*e*, such that Fq*e(*U*) *q*˜(*z*) ≤ *Fq*(*U*) *q*(*z*)*, z* ∈ *U, for all fuzzy dominants q of (1).*

**Lemma 1** ([6])**.** *Consider h* ∈ A*. If Re zh*00(*z*) *h* 0(*z*) + 1 > −1 2 *, z* ∈ *U*, *then* 1 *z* R *z* 0 *h*(*t*)*dt is a convex function, z* ∈ *U.*

**Lemma 2** ([9])**.** *Consider a convex function h with h*(0) = *a, and γ* ∈ C∗ *such that Re γ* ≥ 0*. When p* ∈ H[*a*, *m*]*, ψ* : C2 × *U* → C*, ψ*(*p*(*z*), *zp*0 (*z*); *z*) = 1 *γ zp*0 (*z*) + *p*(*z*) *is an analytic function in U and*

$$F\_{\Psi(\mathbb{C}^2 \times \mathcal{U})} \left( \frac{1}{\gamma} z p'(z) + p(z) \right) \le F\_{h(\mathcal{U})} h(z), \ z \in \mathcal{U}\_{\nu}$$

*then*

$$F\_{p(\mathcal{U})}p(z) \le F\_{\mathcal{G}(\mathcal{U})}g(z) \le F\_{h(\mathcal{U})}h(z), \quad z \in \mathcal{U}\_{\mathcal{A}}$$

*with the convex function g*(*z*) = *γ mz γ m* R *z* 0 *h*(*t*)*t γ m* −1 *dt*, *z* ∈ *U as the fuzzy best dominant.*

**Lemma 3** ([9])**.** *Consider a convex function g in U and define h*(*z*) = *mαzg*0 (*z*) + *g*(*z*), *z* ∈ *U*, *with m* ∈ N *and α* > 0*.*

*If p*(*z*) = *g*(0) + *pmz m* + *pm*+1*z m*+1 + . . . , *z* ∈ *U*, *is a holomorphic function in U and*

$$F\_{p(\mathcal{U})}\left(p(z) + \alpha z p'(z)\right) \le F\_{h(\mathcal{U})}h(z), \quad z \in \mathcal{U}\_{\mathcal{H}}$$

*then we obtain the sharp result*

$$F\_{p(\mathcal{U})}p(z) \le F\_{\mathcal{g}(\mathcal{U})}\mathcal{g}(z), z \in \mathcal{U}.$$

The original results exposed in this paper are obtained using the well-known Ruscheweyh and S˘al˘agean differential operators combined with Riemann–Liouville fractional integral. The resulting operator was introduced in [10], where it was used for obtaining results involving classical differential subordination theory. The necessary definitions are reminded:

**Definition 4** (Ruscheweyh [11])**.** *The Ruscheweyh operator Rn is introduced by Rn* : A → A*,*

$$\begin{array}{rcl} \mathcal{R}^0 f(z) &=& f(z) \\ \mathcal{R}^1 f(z) &=& zf'(z) \\ &\cdots \\ (n+1)\mathcal{R}^{n+1} f(z) &=& n\mathcal{R}^n f(z) + z(\mathcal{R}^n f(z))' \end{array}$$

*for f* ∈ A, *n* ∈ N, *z* ∈ *U*.

**Remark 3.** *For a function f*(*z*) = *z* + ∑ ∞ *j*=2 *ajz j* ∈ A, *the Ruscheweyh operator can be written using the following form R n f*(*z*) = *z* + ∑ ∞ *j*=2 Γ(*n*+*j*) Γ(*n*+1)Γ(*j*) *ajz j , z* ∈ *U, where* Γ *denotes the gamma function.*

**Definition 5** (S˘al˘agean [12])**.** *The S˘al˘agean operator Sn is introduced by Sn* : A → A*,*

$$\begin{array}{rcl} S^0 f(z) &=& f(z), \\ S^1 f(z) &=& zf'(z), \\ & & \cdots \\ S^{n+1} f(z) &=& z(S^n f(z))', \end{array}$$

*for f* ∈ A*, n* ∈ N*, z* ∈ *U*.

**Remark 4.** *For a function f*(*z*) = *z* + ∑ ∞ *j*=2 *ajz j* ∈ A*, the S˘al˘agean operator can be written using the following form Sn f*(*z*) = *z* + ∑ ∞ *j*=2 *j n ajz j , z* ∈ *U.*

**Definition 6** ([13])**.** *Define the linear operator Ln α* : A → A, *given by*

$$L\_{\mathfrak{a}}^{\mathfrak{n}}f(z) = \mathfrak{a}\mathcal{S}^{\mathfrak{n}}f(z) + (1-\mathfrak{a})\mathcal{R}^{\mathfrak{n}}f(z), \ z \in \mathcal{U}\_{\mathfrak{n}}$$

*where α* ≥ 0*, n* ∈ N*.*

**Remark 5.** *For a function f*(*z*) = *z* + ∑ ∞ *j*=2 *ajz j* ∈ A, *the defined operator can be written using the following form Ln α f*(*z*) = *z* + ∑ ∞ *j*=2 h *αj n* + (1 − *α*) Γ(*n*+*j*) Γ(*n*+1)Γ(*j*) i *ajz j* , *z* ∈ *U*.

We also remind the definition of Riemann–Liouville fractional integral:

**Definition 7** ([14])**.** *The Riemann–Liouville fractional integral of order λ applied to an analytic function f is defined by*

$$D\_z^{-\lambda}f(z) = \frac{1}{\Gamma(\lambda)} \int\_0^z \frac{f(t)}{\left(z - t\right)^{1 - \lambda}} dt\_\lambda$$

*with λ* > 0.

In [10] we defined the Riemann–Liouville fractional integral applied to the operator *L n α* as follows:

**Definition 8** ([10])**.** *The Riemann–Liouville fractional integral applied to the differential operator L n α f is introduced by*

$$D\_z^{-\lambda}L\_a^n f(z) = \frac{1}{\Gamma(\lambda)} \int\_0^z \frac{L\_a^n f(t)}{(z-t)^{1-\lambda}} dt = $$

$$\frac{1}{\Gamma(\lambda)} \int\_0^z \frac{t}{(z-t)^{1-\lambda}} dt + \sum\_{j=2}^\infty \left( aj^n + (1-a)\frac{\Gamma(n+j)}{\Gamma(n+1)\Gamma(j)} \right) a\_j \int\_0^z \frac{t^j}{(z-t)^{1-\lambda}} dt,$$

*where α* ≥ 0*, λ* > 0 *and n* ∈ N*.*

**Remark 6.** *For a function f*(*z*) = *z* + ∑ ∞ *j*=2 *ajz j* ∈ A, *the Riemann–Liouville fractional integral of Ln α f has the following form*

$$D\_z^{-\lambda}L\_a^{\mathfrak{n}}f(z) = \frac{1}{\Gamma(2+\lambda)}z^{1+\lambda} + \sum\_{j=2}^{\infty} \left[ \frac{aj^m\Gamma(j+1)}{\Gamma(j+\lambda+1)} + \frac{(1-a)j\Gamma(m+j)}{\Gamma(m+1)\Gamma(j+\lambda+1)} \right] a\_j z^{j+\lambda}/\lambda$$

*and D*−*λ z L n α f*(*z*) ∈ H[0, *λ* + 1].

The results exposed in this paper follow a line of research concerned with fuzzy differential subordinations which is popular nowadays, namely introducing new operators and using them for defining and studying new fuzzy classes of functions.

Fuzzy differential subordinations involving Ruscheweyh and S˘al˘agean differential operators were obtained in many studies, such as [15]. New operators introduced using fractional integral and applied in fuzzy differential subordination theory were studied in [16] where Riemann–Liouville fractional integral is applied for Gaussian hypergeometric function and in [17] where Riemann–Liouville fractional integral is combined with confluent hypergeometric function.

Motivated by the nice results obtained in fuzzy differential subordination theory using Ruscheweyh and S˘al˘agean differential operators and fractional integral applied to different known operators, the study presented in this paper uses the previously defined operator *D*−*λ z L n α* given in Definition 8 applied for obtaining new fuzzy differential subordinations. In the next section, a new fuzzy class will be defined and studied in order to obtain fuzzy differential subordinations inspired by recently published studies concerned with the same topic seen in [18–20].

The main results contained in Section 2 of the paper, begin with the definition of a new fuzzy class *DL*F *n* (*δ*, *α*, *λ*) for which the operator *D*−*λ z L n α* given in Definition 8 is used. The property of this class to be convex is proved and certain fuzzy differential subordinations involving functions from the class and the operator *D*−*λ z L n α* are obtained. The fuzzy best dominants are given for the considered fuzzy differential subordinations in theorems which generate interesting corollaries when specific functions with remarkable geometric properties are used as fuzzy best dominants. An example is also shown in order to prove the applicability of the new results.

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

The usage of the operator *D*−*λ z L n α* seen in Definition 8 defines a new fuzzy subclass of analytic functions as follows:

**Definition 9.** *The class DL*F *n* (*δ*, *α*, *λ*) *is composed of all functions f* ∈ A *with the property*

$$F\_{\left(D\_z^{-\lambda}L\_a^nf\right)'(\mathcal{U})} ^{\prime} \left(D\_z^{-\lambda}L\_a^nf(z)\right)^{\prime} > \delta, \quad z \in \mathcal{U}\_{\prime}$$

*where n* ∈ N, *δ* ∈ [0, 1)*, α* ≥ 0*, λ* > 0*.*

We begin studying this subclass of functions:

**Theorem 1.** *DL*F *n* (*δ*, *α*, *λ*) *is a convex set.*

**Proof.** Taking the functions

$$f\_k(z) = z + \sum\_{j=2}^{\infty} a\_{jk} z^j, \quad k = 1, 2, \quad z \in \mathcal{U}\_2$$

belonging to the class *DL*F *n* (*δ*, *α*, *λ*), we have to prove that the function

$$h(z) = \gamma\_1 f\_1(z) + \gamma\_2 f\_2(z)$$

belongs to the class *DL*F *n* (*δ*, *α*, *λ*) with *γ*1, *γ*2 ≥ 0, *γ*1 + *γ*2 = 1.

We have *h* 0 (*z*) = (*γ*1 *f*1 + *γ*2 *f*2) 0 (*z*) = *γ*1 *f* 0 1 (*z*) + *γ*2 *f* 0 2 (*z*), *z* ∈ *U*, and *D*−*λ z L n αh*(*z*) 0 = *D*−*λ z L n α* (*γ*1 *f*1 + *γ*2 *f*2)(*z*) 0 = *γ*1 *D*−*λ z L n α f*1(*z*) 0 + *γ*2 *D*−*λ z L n α f*2(*z*) 0 . and we can write *F* (*D* −*λ z L n α h*) 0 (*U*) *D*−*λ z L n αh*(*z*) 0 = *F* (*D* −*λ z L n α* (*γ*1 *f*1+*γ*2 *f*2)) 0 (*U*) *D*−*λ z L n α* (*γ*1 *f*1 + *γ*2 *f*2)(*z*) 0 = *F* (*D* −*λ z L n α* (*γ*1 *f*1+*γ*2 *f*2)) 0 (*U*) *γ*1 *D*−*λ z L n α f*1(*z*) 0 + *γ*2 *D*−*λ z L n α f*2(*z*) 0 = *F* (*γ*1*D* −*λ z L n α f*1) 0 (*U*) *γ*1(*D*−*λ z L n α f*1(*z*)) 0 +*F* (*γ*2*D* −*λ z L n α f*2) 0 (*U*) *γ*2(*D*−*λ z L n α f*2(*z*)) 0 2 = *F* (*D* −*λ z L n α f*1) 0 (*U*) (*D*−*λ z L n α f*1(*z*)) 0 +*F* (*D* −*λ z L n α f*2) 0 (*U*) (*D*−*λ z L n α f*2(*z*)) 0 2 .

Having *f*1, *f*2 ∈ *DL*F *n* (*δ*, *α*, *λ*) we get *δ* < *F* (*D* −*λ z L n α f*1) 0 (*U*) *D*−*λ z L n α f*1(*z*) 0 ≤ 1 and *δ* < *F* (*D* −*λ z L n α f*2) 0 (*U*) *D*−*λ z L n α f*2(*z*) 0 ≤ 1, *z* ∈ *U*.

In these conditions *δ* < *F* (*D* −*λ z L n α f*1) 0 (*U*) (*D*−*λ z L n α f*1(*z*)) 0 +*F* (*D* −*λ z L n α f*2) 0 (*U*) (*D*−*λ z L n α f*2(*z*)) 0 2 ≤ 1 and we get *δ* < *F* (*D* −*λ z L n α h*) 0 (*U*) *D*−*λ z L n αh*(*z*) 0 ≤ 1, equivalently with *h* ∈ *DL*F *n* (*δ*, *α*, *λ*) and *DL*F *n* (*δ*, *α*, *λ*) is a convex set.

We give fuzzy differential subordinations obtained for the operator *D*−*λ z L n a* .

**Theorem 2.** *Considering a convex function g in U and defining h*(*z*) = *g*(*z*) + 1 *c*+2 *zg*0 (*z*), *with c* > 0, *z* ∈ *U*, *when f* ∈ *DL*F *n* (*δ*, *α*, *λ*) *and G*(*z*) = *c*+2 *z c*+1 R *z* 0 *t c f*(*t*)*dt, z* ∈ *U*, *then*

$$\left(\mathrm{F}\_{\left(D\_z^{-\lambda}L\_u^{\mathrm{u}}f\right)'(\mathsf{U})}^{\prime}(\mathsf{U})\left(\mathrm{D}\_z^{-\lambda}L\_u^{\mathrm{u}}f(z)\right)' \leq \mathrm{F}\_{\mathbb{H}(\mathsf{U})}h(z), \quad z \in \mathsf{U},\tag{2}$$

*implies the sharp result*

$$\left(F\_{\left(D\_z^{-\lambda}L\_a^{\eta}G\right)'(\mathcal{U})}\right)'(\mathcal{U})\left(D\_z^{-\lambda}L\_a^{\eta}G(z)\right)' \le F\_{\mathcal{G}(\mathcal{U})}g(z), \ z \in \mathcal{U}.\,\,\,\xi$$

**Proof.** Differentiating relation

$$z^{\mathfrak{c}+1}G(z) = (\mathfrak{c}+\mathfrak{2})\int\_0^z t^{\mathfrak{c}}f(t)dt,$$

considering *z* as variable, we get (*c* + 1)*G*(*z*) + *zG*0 (*z*) = (*c* + 2)*f*(*z*) and

$$\left( (\mathfrak{c} + 1) D\_z^{-\lambda} L\_\mathfrak{a}^\mathfrak{n} G(z) + z \left( D\_z^{-\lambda} L\_\mathfrak{a}^\mathfrak{n} G(z) \right)' = (\mathfrak{c} + 2) D\_z^{-\lambda} L\_\mathfrak{a}^\mathfrak{n} f(z), \quad z \in \mathsf{U}\_\sigma$$

and differentiating it again with respect to *z*, we obtain

$$\left(D\_z^{-\lambda}L\_a^{\underline{n}}G(z)\right)' + \frac{1}{c+2}z\left(D\_z^{-\lambda}L\_a^{\underline{n}}G(z)\right)'' = \left(D\_z^{-\lambda}L\_a^{\underline{n}}f(z)\right)', \ z \in \mathcal{U}.\,\,\,\xi$$

and the inequality (2) representing the fuzzy differential subordination can be written

$$\left(F\_{D\_z^{-\lambda}L\_a^nG(\mathcal{U})}\left(\frac{1}{c+2}z\left(D\_z^{-\lambda}L\_a^nG(z)\right)'' + \left(D\_z^{-\lambda}L\_a^nG(z)\right)'\right) \le F\_{\mathcal{S}}(\mathcal{U})\left(\frac{1}{c+2}zg'(z) + g(z)\right).$$

Denoted

$$p(z) = \left(D\_z^{-\lambda} L\_a^n G(z)\right)', \ z \in \mathcal{U}\_\nu$$

where *p* ∈ H[1, *n*], we obtain

$$F\_{p(\mathcal{U})}\left(\frac{1}{\varepsilon+2}zp'(z)+p(z)\right) \le F\_{\mathcal{S}(\mathcal{U})}\left(\frac{1}{\varepsilon+2}zg'(z)+g(z)\right), \ z \in \mathcal{U}.$$

Applying Lemma 3, we get

$$F\_{\left(D\_z^{-\lambda}L\_{\mathfrak{a}}^{\mathfrak{n}}G\right)'(\mathcal{U})} ^{\prime} \left(D\_z^{-\lambda}L\_{\mathfrak{a}}^{\mathfrak{n}}G(z)\right)^{\prime} \leq F\_{\mathcal{G}(\mathcal{U})}g(z), \ z \in \mathcal{U}\_{\prime}$$

and *g* is the best dominant.

We give an inclusion result for the class *DL*F *n* (*δ*, *α*, *λ*):

**Theorem 3.** *Taking h*(*z*) = 1+(2*δ*−1)*z* 1+*z and G*(*z*) = *c*+2 *z c*+1 R *z* 0 *t c f*(*t*)*dt, z* ∈ *U, with δ* ∈ [0, 1), *c* > 0*, n* ∈ N, *α* ≥ 0*, λ* > 0*, then*

$$\mathbb{G}\left[DL\_{\mathfrak{n}}^{\mathcal{F}}(\delta,\mathfrak{a},\lambda)\right] \subset DL\_{\mathfrak{n}}^{\mathcal{F}}(\delta^\*,\mathfrak{a},\lambda),\tag{3}$$

*where δ* ∗ = 2*δ* − 1 + 2(2 + *c*)(1 − *δ*) R 1 0 *t c*+1 *t*+1 *dt*.

**Proof.** Making the same steps such as in the proof of Theorem 2, taking account the hypothesis of Theorem 3 and that *h*(*z*) = 1+(2*δ*−1)*z* 1+*z* is a convex function, we obtain

$$F\_{p(\mathcal{U})}\left(\frac{1}{c+2}zp'(z)+p(z)\right) \le f\_{h(\mathcal{U})}h(z).$$

with *p*(*z*) = *D*−*λ z L n αG*(*z*) 0 , *z* ∈ *U*. Applying Lemma 2, we get

$$F\_{\left(D\_z^{-\lambda}L\_{\mathfrak{a}}^{\eta}G\right)'(\mathcal{U})} \left(D\_z^{-\lambda}L\_{\mathfrak{a}}^{\eta}G(z)\right)' \le F\_{\mathcal{G}(\mathcal{U})}\mathcal{g}(z) \le F\_{h(\mathcal{U})}h(z).$$

where

$$g(z) = \frac{2+c}{nz^{\frac{2+\zeta}{n}}} \int\_0^z t^{\frac{2+\zeta}{n}-1} \frac{1+(2\delta-1)t}{1+t} dt = (2\delta-1) + \frac{2(c+2)(1-\delta)}{nz^{\frac{\zeta+2}{n}}} \int\_0^z t^{\frac{2+\zeta}{n}-1} dt.$$

Since the function *g* is convex and *g*(*U*) is symmetric with respect to the real axis, we can write

$$F\_{D\_z^{-\lambda}L\_R^nG(\mathcal{U})}\left(D\_z^{-\lambda}L\_R^nG(z)\right)' \geq \min\_{|z|=1} F\_{\mathcal{S}}(\mathcal{U})\mathcal{g}(z) = F\_{\mathcal{S}}(\mathcal{U})\mathcal{g}(1)\tag{4}$$

$$\text{and } \delta^\* = g(1) = 2\delta - 1 + \frac{2(2+c)(1-\delta)}{n} \int\_0^1 \frac{t^{\frac{2+c}{n}-1}}{t+1}dt, \text{ that give the inclusion (3).} \quad \square$$

**Theorem 4.** *Taking a convex function g with the property g*(0) = 0, *define h*(*z*) = *g*(*z*) + *zg*0 (*z*), *z* ∈ *U*. *When f* ∈ A, *n* ∈ N, *α* ≥ 0, *λ* > 0*, and the fuzzy differential subordination holds*

$$\left(\mathrm{F}\_{\left(D\_{z}^{-\lambda}L\_{u}^{n}f\right)'(\mathsf{U})}^{\prime}\left(\mathrm{D}\_{z}^{-\lambda}L\_{u}^{n}f(z)\right)' \leq \mathrm{F}\_{h(\mathsf{U})}h(z), \quad z \in \mathsf{U},\tag{5}$$

*then we get the sharp result*

$$F\_{D\_z^{-\lambda}L\_a^n f(\mathcal{U})} \frac{D\_z^{-\lambda}L\_a^n f(z)}{z} \le F\_{\mathcal{S}(\mathcal{U})} \mathcal{g}(z), \quad z \in \mathcal{U}.$$

**Proof.** Considering *p*(*z*) = *D*−*λ z L n α f*(*z*) *z* ∈ H[0, *λ*], we can write *zp*(*z*) = *D*−*λ z L n α f*(*z*), *z* ∈ *U*, and differentiating it we get *zp*0 (*z*) + *p*(*z*) = *D*−*λ z L n α f*(*z*) 0 , *z* ∈ *U*.

The inequality (5) can be written as following

$$F\_{p(\mathcal{U})}\left(zp'(z) + p(z)\right) \le F\_{\mathbb{H}(\mathcal{U})}h(z) = F\_{\mathcal{g}(\mathcal{U})}\left(zg'(z) + \mathcal{g}(z)\right), \quad z \in \mathcal{U}\_{\mathcal{H}}$$

and applying Lemma 3, we get the sharp result

$$F\_{\left(D\_z^{-\lambda}L\_{\alpha}^nf\right)'(U)} ^{\prime} \frac{D\_z^{-\lambda}L\_{\alpha}^nf(z)}{z} \le F\_{\mathcal{S}(U)}g(z), \quad z \in \mathcal{U}.$$

**Example 1.** *Consider*

$$g(z) = \frac{-2z}{1+z}$$

*a convex function in U and we obtain that g*(0) = 0*, g*0 (*z*) = −2 (1+*z*) 2 *. Define*

$$h(z) = g(z) + zg'(z) = \frac{-2z}{1+z} - \frac{2z}{\left(1+z\right)^2} = \frac{-2z^2 - 4z}{\left(1+z\right)^2}.$$

*Take α* = 2*, n* = 1*, f*(*z*) = *z* + *z* 2 *, z* ∈ *U, and after a short computation we obtain*

$$L\_2^1 f(z) = z + 2z^2$$

*and*

$$D\_z^{-\lambda}L\_2^1f(z) = \frac{1}{\Gamma(\lambda)}\int\_0^z \frac{L\_2^1f(t)}{\left(z-t\right)^{1-\lambda}}dt = \frac{1}{\Gamma(\lambda)}\int\_0^z \frac{t+2t^2}{\left(z-t\right)^{1-\lambda}}dt$$

$$= \frac{1}{\Gamma(\lambda+2)}z^{1+\lambda} + \frac{4}{\Gamma(\lambda+3)}z^{2+\lambda}$$

*and differentiating it*

$$\left(D\_z^{-\lambda}L\_2^1f(z)\right)' = \frac{1}{\Gamma(\lambda+1)}z^{\lambda} + \frac{4}{\Gamma(\lambda+2)}z^{\lambda+1}.$$

*Applying Theorem 4 we get the following fuzzy differential subordination*

$$\frac{1}{\Gamma(1+\lambda)}z^{\lambda} + \frac{4}{\Gamma(2+\lambda)}z^{1+\lambda} \preccurlyeq \frac{-2z^2 - 4z}{\left(1+z\right)^2}, z \in \mathsf{U}\_{\lambda}$$

*induce the following fuzzy differential subordination*

$$\frac{1}{\Gamma(2+\lambda)}z^{\lambda} + \frac{4}{\Gamma(3+\lambda)}z^{1+\lambda} \prec\_{\mathcal{F}} \frac{-2z}{1+z}, \; z \in \mathcal{U}.$$

**Theorem 5.** *Taking a holomorphic function h, such that h*(0) = 0 *and* Re 1 + *zh*00(*z*) *h* 0(*z*) > −1 2 , *z* ∈ *U*, *when f* ∈ A, *n* ∈ N, *α* ≥ 0, *λ* > 0*, and the fuzzy differential subordination holds*

$$\left(F\_{\left(D\_z^{-\lambda}L\_u^n f\right)'(\mathcal{U})} \left(D\_z^{-\lambda}L\_u^n f(z)\right)' \le F\_{h(\mathcal{U})}h(z), \quad z \in \mathcal{U},\tag{6}$$

*then*

$$F\_{D\_z^{-\lambda}L\_a^n f(\mathcal{U})} \frac{D\_z^{-\lambda}L\_a^n f(z)}{z} \le F\_{q(\mathcal{U})}q(z), \quad z \in \mathcal{U}\_\lambda$$

*where the fuzzy best dominant q*(*z*) = 1 *z* R *z* 0 *h*(*t*)*dt is convex.* **Proof.** Considering Re 1 + *zh*00(*z*) *h* 0(*z*) > −1 2 , *z* ∈ *U*, and using Lemma 1, we deduce that *q*(*z*) = 1 *z* R *z* 0 *h*(*t*)*dt* is a convex function and it is a solution of the differential equation defining the fuzzy differential subordination (6) *zq*0 (*z*) + *q*(*z*) = *h*(*z*), therefore it is the fuzzy best dominant.

Differentiating *zp*(*z*) = *D*−*λ z L n α f*(*z*), we get *D*−*λ z L n α f*(*z*) 0 = *zp*0 (*z*) + *p*(*z*), *z* ∈ *U*, and (6) can be written

$$F\_{p(\mathcal{U})} \left( zp'(z) + p(z) \right) \le F\_{h(\mathcal{U})} h(z), \quad z \in \mathcal{U}.$$

Applying Lemma 3, we get

$$F\_{D\_z^{-\lambda}L\_a^nf(\mathcal{U})} \frac{D\_z^{-\lambda}L\_a^nf(z)}{z} \le F\_{q(\mathcal{U})}q(z), \ z \in \mathcal{U}.$$

**Corollary 1.** *Taking the convex function in U*, *h*(*z*) = 1+(2*δ*−1)*z* 1+*z* , *with δ* ∈ [0, 1)*, when f* ∈ A *and the fuzzy differential subordination holds*

$$\left(\mathrm{F}\_{\left(D\_z^{-\lambda}L\_a^n f\right)'(\mathcal{U})}^{'}\left(\mathcal{U}\right)\left(\mathrm{D}\_z^{-\lambda}L\_a^n f(z)\right)' \leq \mathrm{F}\_{\hbar(\mathcal{U})}\hbar(z), z \in \mathcal{U},\tag{7}$$

*then*

$$F\_{D\_z^{-\lambda}L\_\alpha^n f(\mathcal{U})} \frac{D\_z^{-\lambda}L\_\alpha^n f(z)}{z} \le F\_{q(\mathcal{U})}q(z), z \in \mathcal{U}\_\alpha$$

*where the fuzzy best dominant q*(*z*) = 2*δ* − 1 + 2(1 − *δ*) ln(*z*+1) *z* , *z* ∈ *U*, *is convex.*

**Proof.** Taking *h*(*z*) = 1+(2*δ*−1)*z* 1+*z* , we obtain *h*(0) = 1, *h* 0 (*z*) = −2(1−*δ*) (1+*z*) 2 and *h* 00(*z*) = 4(1−*δ*) (1+*z*) 3 , therefore *Re zh*00(*z*) *h* 0(*z*) + 1 = *Re* 1−*z* 1+*z* = *Re* 1−*ρ* cos *θ*−*iρ* sin *θ* 1+*ρ* cos *θ*+*iρ* sin *θ* = 1−*ρ* 2 1+2*ρ* cos *θ*+*ρ* 2 > 0 > −1 2 . *n*

Following the same steps like in the proof of Theorem 5 with *p*(*z*) = *D*−*λ z L α f*(*z*) *z* , the fuzzy differential subordination (7) can be written

$$F\_{D\_z^{-\lambda}L\_{\mathfrak{a}}^{\mathfrak{n}}f(\mathsf{U})}\left(zp'(z)+p(z)\right)\leq F\_{\mathsf{h}(\mathsf{U})}\mathsf{h}(z),\quad z\in\mathsf{U}.$$

Applying Lemma 2 for *m* = 1 and *γ* = 1, we obtain

$$F\_{D\_z^{-\lambda}L\_\alpha^n f(\mathcal{U})} \frac{D\_z^{-\lambda}L\_\alpha^n f(z)}{z} \le F\_{q(\mathcal{U})}q(z),$$

where

$$q(z) = \frac{1}{z} \int\_0^z h(t)dt = \frac{1}{z} \int\_0^z \frac{1 + (2\delta - 1)t}{t + 1} dt = $$

$$2\delta - 1 + \frac{2(1 - \delta)}{z} \int\_0^z \frac{1}{t + 1} dt = 2\delta - 1 + 2(1 - \delta)\frac{\ln(z + 1)}{z}, \ z \in \mathcal{U}\_z$$

−2*z*

**Example 2.** *Consider*

*h*(*z*) = 1 + *z and we obtain that h*(0) = 0*, h*0 (*z*) = −2 (1+*z*) 2 *and h*00(*z*) = 4 (1+*z*) 3 *. Taking account that*

$$\operatorname{Re}\left(1+\frac{zh''(z)}{h'(z)}\right) = \operatorname{Re}\left(\frac{1-z}{1+z}\right) = \operatorname{Re}\left(\frac{1-\rho\cos\theta-i\rho\sin\theta}{1+\rho\cos\theta+i\rho\sin\theta}\right).$$

$$=\frac{1-\rho^2}{1+2\rho\cos\theta+\rho^2} > 0 > -\frac{1}{2}\rho$$

*h is a convex function in U.*

*Taking α* = 2*, n* = 1*, f*(*z*) = *z* + *z* 2 *, z* ∈ *U, as in Example 1, we have*

$$L\_2^1 f(z) = z + 2z^2$$

*and*

$$D\_z^{-\lambda}L\_2^1f(z) = \frac{1}{\Gamma(\lambda+2)}z^{1+\lambda} + \frac{4}{\Gamma(\lambda+3)}z^{2+\lambda}$$

*and differentiating it*

$$\left(D\_z^{-\lambda}L\_2^1f(z)\right)' = \frac{1}{\Gamma(\lambda+1)}z^{\lambda} + \frac{4}{\Gamma(\lambda+2)}z^{\lambda+1}.$$

*Additionally, we get*

$$q(z) = \frac{1}{z} \int\_0^z \frac{-2t}{1+t} dt = \frac{2\ln(1+z)}{z} - 2.$$

*Applying Theorem 5 we get the following fuzzy differential subordination*

$$\frac{1}{\Gamma(1+\lambda)}z^{\lambda} + \frac{4}{\Gamma(2+\lambda)}z^{1+\lambda} \preccurlyeq\_{\mathcal{F}} \frac{2z}{1+z'}, z \in \mathcal{U}\_{\lambda}$$

*induce the following fuzzy differential subordination*

$$\frac{1}{\Gamma(2+\lambda)}z^{1+\lambda} + \frac{4}{\Gamma(3+\lambda)}z^{2+\lambda} \preccurlyeq\_{\mathcal{F}} \frac{2\ln(1+z)}{z} - 2, \; z \in \mathcal{U}.$$

**Theorem 6.** *Taking a convex function g with the property g*(0) = 0 *and defining h*(*z*) = *zg*0 (*z*) + *g*(*z*)*, z* ∈ *U, when f* ∈ A, *n* ∈ N, *α* ≥ 0, *λ* > 0*, and the fuzzy differential subordination*

$$F\_{D\_z^{-\lambda}L\_a^n f(\mathsf{II})} \left( \frac{zD\_z^{-\lambda}L\_a^{n+1}f(z)}{D\_z^{-\lambda}L\_a^n f(z)} \right)' \le F\_{h(\mathsf{II})}h(z), \quad z \in \mathsf{II},\tag{8}$$

*holds, then we obtain the sharp result*

$$\mathbb{P}\_{D\_z^{-\lambda}L\_a^n f(\mathcal{U})} \frac{z D\_z^{-\lambda} L\_a^{n+1} f(z)}{D\_z^{-\lambda} L\_a^n f(z)} \le \mathbb{P}\_{\mathcal{S}(\mathcal{U})} g(z), \quad z \in \mathcal{U}.$$

**Proof.** Considering *p*(*z*) = *D*−*λ z L n*+1 *α f*(*z*) *D* −*λ z L n α f*(*z*) and differentiating it we get *zp*0 (*z*) + *p*(*z*) = *zLn*+1 *α f*(*z*) *L n α f*(*z*) 0 . With this notation, inequality (8) can be written as

$$F\_{p(\mathcal{U})} \left( zp'(z) + p(z) \right) \le F\_{h(\mathcal{U})} h(z) = F\_{\mathcal{S}(\mathcal{U})} \left( zg'(z) + g(z) \right), \ z \in \mathcal{U}.$$

Applying Lemma 3, we get

$$F\_{D\_z^{-\lambda}L\_a^n f(\mathcal{U})} \frac{D\_z^{-\lambda}L\_a^{n+1}f(z)}{D\_z^{-\lambda}L\_a^n f(z)} \le F\_{\mathcal{S}(\mathcal{U})}g(z), \quad z \in \mathcal{U}.$$

**Example 3.** *Consider*

$$g(z) = \frac{-2z}{1+z}$$

*and*

$$h(z) = g(z) + zg'(z) = \frac{-2z^2 - 4z}{\left(1 + z\right)^2}$$

*as given in Example 1.*

*Taking α* = 2*, n* = 1*, f*(*z*) = *z* + *z* 2 *, z* ∈ *U, as in Example 1, we get*

$$L\_2^1 f(z) = z + 2z^2$$

*and*

$$L\_2^2 f(z) = z + 2z^2$$

*and applying Riemann–Liouville fractional integral of order λ we have*

$$D\_z^{-\lambda}L\_2^1f(z) = \frac{1}{\Gamma(\lambda+2)}z^{1+\lambda} + \frac{4}{\Gamma(\lambda+3)}z^{2+\lambda} = D\_z^{-\lambda}L\_2^2f(z).$$

*Applying Theorem 6 we get the following fuzzy differential subordination*

$$1 \preccurlyeq\_{\mathcal{F}} \frac{-2z^2 - 4z}{\left(1 + z\right)^2}, \; z \in \mathcal{U}\_{\star}$$

*induce the following fuzzy differential subordination*

$$z \preccurlyeq \frac{-2z}{1+z}, \; z \in \mathcal{U}.$$

**Theorem 7.** *Taking a convex function g with the property g*(0) = 0 *and defining h*(*z*) = *λzg*0 (*z*) + *g*(*z*), *z* ∈ *U*, *α* ≥ 0, *λ*, *δ* > 0*, when f* ∈ A *and the fuzzy differential subordination*

$$\left(F\_{\mathcal{D}\_z^{-\lambda}L\_\mathcal{H}^q(\mathcal{U})}\left(\left(\frac{\mathcal{D}\_z^{-\lambda}L\_\mathcal{u}^\eta f(z)}{z}\right)^{\delta-1}\left(D\_z^{-\lambda}L\_\mathcal{u}^\eta f(z)\right)'\right) \le F\_{\mathcal{H}(\mathcal{U})}h(z) \quad z \in \mathcal{U},\tag{9}$$

*holds, then we obtain the sharp result*

$$F\_{D\_z^{-\lambda}L\_a^n f(\mathcal{U})} \left( \frac{D\_z^{-\lambda}L\_a^n f(z)}{z} \right)^\delta \le F\_{\mathcal{S}(\mathcal{U})} g(z), \quad z \in \mathcal{U}.$$

**Proof.** Considering *p*(*z*) = *D*−*λ z L n α f*(*z*) *z δ* ∈ H[0, *λδ*], and differentiating it we obtain

$$z p'(z) = \delta \left(\frac{D\_z^{-\lambda} L\_a^n f(z)}{z}\right)^{\delta - 1} \left(D\_z^{-\lambda} L\_a^n f(z)\right)' - \delta \left(\frac{D\_z^{-\lambda} L\_a^n f(z)}{z}\right)^{\delta}.$$

$$= \delta \left(\frac{D\_z^{-\lambda} L\_a^n f(z)}{z}\right)^{\delta - 1} \left(D\_z^{-\lambda} L\_a^n f(z)\right)' - \delta p(z),$$

$$p(z) + p(z) = \left(\frac{D\_z^{-\lambda} L\_a^n f(z)}{z}\right)^{\delta - 1} \left(D\_z^{-\lambda} L\_a^n f(z)\right)', z \in \mathsf{U}.$$

*δ z* Inequality (9) can be written

and 1

*zp*0

$$\mathbb{P}\_{p(\mathcal{U})}\left(\frac{1}{\delta}zp'(z)+p(z)\right) \le \mathbb{P}\_{h(\mathcal{U})}h(z) = \mathbb{P}\_{\mathcal{S}(\mathcal{U})}\{\lambda z \mathbf{g}'(z) + \mathbf{g}(z)\}, \ z \in \mathcal{U}.$$

*α*

Applying Lemma 3 for *α* = 1 *δ* and *m* = *λδ*, we get

$$F\_{D\_z^{-\lambda}L\_\alpha^n f(\mathcal{U})} \left( \frac{D\_z^{-\lambda}L\_\alpha^n f(z)}{z} \right)^\delta \le F\_{\mathcal{G}(\mathcal{U})} \mathcal{g}(z), \quad z \in \mathcal{U}.$$

**Example 4.** *Consider*

*and*

$$h(z) = g(z) + zg'(z) = \frac{-2z^2 - 4z}{(1+z)^2}$$

−2*z* 1 + *z*

*g*(*z*) =

*as given in Example 1.*

*Taking α* = 2*, n* = 1*, f*(*z*) = *z* + *z* 2 *, z* ∈ *U, as in Example 1, we obtain*

> *L* 1 2 *f*(*z*) = *z* + 2*z* 2

*and*

$$D\_z^{-\lambda}L\_2^1f(z) = \frac{1}{\Gamma(\lambda+2)}z^{1+\lambda} + \frac{4}{\Gamma(\lambda+3)}z^{2+\lambda}$$

*and differentiating it*

$$\left(D\_z^{-\lambda}L\_2^1f(z)\right)' = \frac{1}{\Gamma(\lambda+1)}z^{\lambda} + \frac{4}{\Gamma(\lambda+2)}z^{\lambda+1}.$$

*Applying Theorem 7 we get the following fuzzy differential subordination*

$$\left(\frac{1}{\Gamma(\lambda+2)}z^{\lambda}+\frac{4}{\Gamma(\lambda+3)}z^{1+\lambda}\right)^{\delta-1}\left(\frac{1}{\Gamma(1+\lambda)}z^{\lambda}+\frac{4}{\Gamma(2+\lambda)}z^{1+\lambda}\right)\prec\_{\mathcal{F}}\frac{-2z^2-4z}{(1+z)^2},\;z\in\mathsf{U},\;\lambda\geqslant 1$$

*induce the following fuzzy differential subordination*

$$\left(\frac{1}{\Gamma(\lambda+2)}z^{\lambda} + \frac{4}{\Gamma(\lambda+3)}z^{1+\lambda}\right)^{\delta} \prec\_{\mathcal{F}} \frac{-2z}{1+z}, \; z \in \mathcal{U}.$$

**Theorem 8.** *Considering a holomorphic function h, such that h*(0) = 0 *and* Re 1 + *zh*00(*z*) *h* 0(*z*) > −1 2 , *z* ∈ *U*, *when f* ∈ A, *α* ≥ 0, *λ*, *δ* > 0*, and the fuzzy differential subordination*

$$\left(\mathcal{F}\_{\mathcal{D}\_z^{-\lambda}L\_{\mathfrak{a}}^nf(\mathsf{U})}\left(\left(\frac{\mathcal{D}\_z^{-\lambda}L\_{\mathfrak{a}}^nf(z)}{z}\right)^{\delta-1}\left(\mathcal{D}\_z^{-\lambda}L\_{\mathfrak{a}}^nf(z)\right)'\right) \le \mathcal{F}\_{\mathcal{H}(\mathsf{U})}h(z), \quad z \in \mathsf{U},\tag{10}$$

*holds, then*

$$F\_{D\_z^{-\lambda}L\_a^n f(\mathcal{U})} \left(\frac{D\_z^{-\lambda}L\_a^n f(z)}{z}\right)^\delta \le F\_{q(\mathcal{U})}q(z), \quad z \in \mathcal{U}\_\nu$$

*where the fuzzy best dominant q*(*z*) = 1 *z* R *z* 0 *h*(*t*)*dt is convex.*

**Proof.** Considering *p*(*z*) = *D*−*λ z L n α f*(*z*) *z δ* ∈ H[0, *λδ*], after differentiating it and making an easy computation, we get

$$\frac{1}{\delta}zp'(z) + p(z) = \left(\frac{D\_z^{-\lambda}L\_a^n(z)}{z}\right)^{\delta-1} \left(D\_z^{-\lambda}L\_a^n f(z)\right)' , \ z \in \mathcal{U}\_{\lambda}$$

and inequality (10) can be written

$$\left(F\_{p(\mathcal{U})}\left(\frac{1}{\delta}zp'(z)+p(z)\right)\right)\le F\_{\hbar(\mathcal{U})}h(z),\ z\in\mathcal{U}.$$

Applying Lemma 2, we obtain

$$F\_{D\_z^{-\lambda}L\_a^n f(\mathcal{U})} \left( \frac{D\_z^{-\lambda}L\_a^n f(z)}{z} \right)^\delta \le F\_{q(\mathcal{U})} q(z), \ z \in \mathcal{U}.$$

Taking into account that Re 1 + *zh*00(*z*) *h* 0(*z*) > −1 2 , *z* ∈ *U*, applying Lemma 1 we obtain that *q*(*z*) = 1 *z* R *z* 0 *h*(*t*)*dt* is a convex function and it is a solution of the differential equation of the fuzzy differential subordination (10) *zq*0 (*z*) + *q*(*z*) = *h*(*z*), thus it is the fuzzy best dominant.

**Example 5.** *Considering*

$$h(z) = \frac{-2z}{1+z},$$

*as in Example 2, a convex function which satisfy conditions from Theorem 8, and taking α* = 2*, n* = 1*, f*(*z*) = *z* + *z* 2 *, z* ∈ *U, we obtain*

$$L\_2^1 f(z) = z + 2z^2$$

*and*

$$D\_z^{-\lambda}L\_2^1f(z) = \frac{1}{\Gamma(\lambda+2)}z^{1+\lambda} + \frac{4}{\Gamma(\lambda+3)}z^{2+\lambda}$$

*and differentiating it*

$$\left(D\_z^{-\lambda}L\_2^1f(z)\right)' = \frac{1}{\Gamma(\lambda+1)}z^{\lambda} + \frac{4}{\Gamma(\lambda+2)}z^{\lambda+1}.$$

*Additionally, we get*

$$q(z) = \frac{1}{z} \int\_0^z \frac{-2t}{1+t} dt = \frac{2\ln(1+z)}{z} - 2.$$

*Applying Theorem 8 we get the following fuzzy differential subordination*

$$\left(\frac{1}{\Gamma(\lambda+2)}z^{\lambda}+\frac{4}{\Gamma(\lambda+3)}z^{1+\lambda}\right)^{\delta-1}\left(\frac{1}{\Gamma(1+\lambda)}z^{\lambda}+\frac{4}{\Gamma(2+\lambda)}z^{1+\lambda}\right)\prec\_{\mathbb{F}}\frac{2z}{1+z},\ z\in\mathsf{U}\_{\lambda}$$

*induce the following fuzzy differential subordination*

$$\left(\frac{1}{\Gamma(\lambda+2)}z^{\lambda} + \frac{4}{\Gamma(\lambda+3)}z^{1+\lambda}\right)^{\delta} \precneq \frac{2\ln(1+z)}{z} - 2, \; z \in \mathcal{U}.$$

**Theorem 9.** *Considering a convex function g with the property g*(0) = 1 *λ*+1 *and defining h*(*z*) = *zg*0 (*z*) + *g*(*z*), *z* ∈ *U, λ* > 0*, α* ≥ 0*, n* ∈ N, *when f* ∈ A *and the fuzzy differential subordination*

$$\left(\mathcal{F}\_{D\_z^{-\lambda}L\_a^nf(\mathcal{U})}\left(1-\frac{D\_z^{-\lambda}L\_a^nf(z)\left(D\_z^{-\lambda}L\_a^nf(z)\right)'}{\left[\left(D\_z^{-\lambda}L\_a^nf(z)\right)'\right]^2}\right) \le \mathcal{F}\_{h(\mathcal{U})}h(z), \quad z \in \mathcal{U}\_{\mathcal{U}}$$

*holds, then we obtain the sharp result*

$$F\_{D\_z^{-\lambda}L\_a^nf(\mathcal{U})}\left(\frac{D\_z^{-\lambda}L\_a^nf(z)}{z\left(D\_z^{-\lambda}L\_a^nf(z)\right)'}\right) \le F\_{\mathcal{S}(\mathcal{U})}\mathcal{g}(z), \ z \in \mathcal{U}.\mathcal{Z}$$

**Proof.** Differentiating *p*(*z*) = *D*−*λ z L n α f*(*z*) *z*(*D* −*λ z L n α f*(*z*)) 0 we obtain *zp*0 (*z*) + *p*(*z*) = 1− *D*−*λ z L n α f*(*z*)(*D*−*λ z L n α f*(*z*)) 00 h (*D* −*λ z L n α f*(*z*)) 0 i2 , *z* ∈ *U*.

Using this notation, the fuzzy differential subordination can be written

$$F\_{p(\mathcal{U})} \left( zp'(z) + p(z) \right) \le F\_{h(\mathcal{U})}h(z) = F\_{\mathcal{g}(\mathcal{U})} \left( zg'(z) + g(z) \right), \ z \in \mathcal{U},$$

and applying Lemma 3, we obtain the sharp result

$$\left(F\_{D\_z^{-\lambda}L\_n^nf(\mathcal{U})}\left(\frac{D\_z^{-\lambda}L\_n^nf(z)}{z\left(D\_z^{-\lambda}L\_n^nf(z)\right)'}\right) \le F\_{\mathcal{S}(\mathcal{U})}g(z), \quad z \in \mathcal{U}.\right)$$

#### **3. Conclusions**