& + \left(1+ \frac{2m}{\gamma_k \lambda_k \mu_h}\right) \frac{C_h^2}{\mu_h^2}\left  |1-\frac{\lambda_{k-1}}{\lambda_k} \right|^2.
    \end{aligned}$$ Plugging this inequality into we obtain $$\begin{aligned}
    &\left \|x_{k+1} - x_{\lambda_k}^* \right \|^2  \leq \left( 1- \frac{\gamma_k \lambda_k \mu_h}{m} \right) \left( 1+ \frac{\gamma_k \lambda_k \mu_h}{2m} \right) \left \|x_{k} - x_{\lambda_{k-1}}^* \right \|^2 \\& +\left( 1- \frac{\gamma_k \lambda_k \mu_h}{m} \right) \left(1+ \frac{2m}{\gamma_k \lambda_k \mu_h}\right) \frac{C_h^2}{\mu_h^2}\left    |1-\frac{\lambda_{k-1}}{\lambda_k} \right|^2 \\& +  6 m^2 \gamma_k^2 (C_f^2 + \lambda_k^2 C_h^2).     \end{aligned}$$ The desired result is obtained [[from]{}]{} $0<\mu_k\lambda_k \mu_h \leq 2m$.

We will make use of the following result in Proposition \[conv x\_k INC\].

\[lemma conv lemma\] Let $\{\nu_k\}$ be a sequence of nonnegative scalars and let $\{\alpha_k\}$ and $\{\beta_k\}$ be scalar sequences such that: $$\begin{aligned}
    &\nu_{k+1} \leq (1-\alpha_k)\nu_k + \beta_k \qquad \hbox{for all } k \geq 0,\\
    &0 \leq \alpha_k \leq 1, \ \beta_k \geq0, \ \sum_{k=0}^{\infty}\alpha_k=\infty, \ \sum_{k=0}^{\infty}\beta_k <\infty,  \ \lim_{k \to \infty} \frac{\beta_k}{\alpha_k}=0.
    \end{aligned}$$ Then, $\lim_{k \to \infty} \nu_k = 0$.

\[assumptions for conv INC\] Assume that for all $k \geq 0$ we have $(a)\{\gamma_k\}$ and $\{\lambda_k\}$ are non-increasing positive sequences with $\gamma_0\lambda_0 \leq \frac{2m}{\mu_h}.$\
$(b) \sum_{k=0}^{\infty}\gamma_k\lambda_k= \infty.$ $ \qquad (c) \sum_{k=0}^{\infty}\frac{1}{\gamma_k\lambda_k}\left(\frac{\lambda_{k-1}}{\lambda_k}-1\right)^2 <\infty.$\
$(d) \sum_{k=0}^{\infty} \gamma_k^2 < \infty.$\
$(e) \lim_{k\to \infty} \frac{1}{\gamma_k^2\lambda_k^2}\left(\frac{\lambda_{k-1}}{\lambda_k}-1\right)^2=0.$ $\qquad (f) \lim_{k\to \infty} \frac{\gamma_k}{\lambda_k}=0.$

\[conv x\_k INC\] Consider problem 