Stochastic proximal splitting algorithm for composite minimization

Abstract

Supported by the recent contributions in multiple domains, the first-order splitting became algorithms of choice for structured nonsmooth optimization. The large-scale noisy contexts make available stochastic information on the objective function and thus, the extension of proximal gradient schemes to stochastic oracles is heavily based on the tractability of the proximal operator corresponding to nonsmooth component, which has been highly exploited in the literature. However, some questions remained about the complexity of the composite models with proximal untractable terms. In this paper we tackle composite optimization problems, assuming only the access to stochastic information on both smooth and nonsmooth components, with a stochastic proximal first-order scheme with stochastic proximal updates. We provide sublinear $\mathcal {O}\left( \frac{1}{k} \right)$ convergence rates (in expectation of squared distance to the optimal set) under the strong convexity assumption on the objective function. Also, linear convergence is achieved for convex feasibility problems. The empirical behavior is illustrated by numerical tests on parametric sparse representation models.

Appendix

Proof (of Corollary 3)

For simplicity denote $\theta _k = (1 - \mu _k\sigma _{f})$ , then Theorem 2 implies that:

$$\begin{aligned} \mathbb {E}\left[ \Vert x^{k+1}-x^*\Vert ^2 \right]&\le \left( \prod _{i=0}^k \theta _i\right) \Vert x^0-x^*\Vert ^2 + \varSigma \sum \limits _{i=0}^k \left( \prod \limits _{j=i+1}^{k} \theta _j\right) \mu _i^2. \end{aligned}$$

By using the Bernoulli inequality $1- tx \le \frac{1}{1 + tx} \le (1 + x)^{-t}$ for $t \in [0,1], x \ge 0$, then we have:

$$\begin{aligned} \prod \limits _{i=l}^u \theta _i&= \prod \limits _{i=l}^u \left( 1 - \frac{\mu _0}{i^{\gamma }} \sigma _{f}\right) \le \prod \limits _{i=l}^u (1 + \mu _0 \sigma _f)^{-1/i^{\gamma }} = (1 + \mu _0 \sigma _{f})^{- \sum \limits _{i=l}^u \frac{1}{i^{\gamma }}}. \end{aligned}$$

(18)

On the other hand, if we use the lower bound

$$\begin{aligned} \sum \limits _{i=l}^u \frac{1}{i^{\gamma }} \ge \int \limits _{l}^{u + 1} \frac{1}{\tau ^{\gamma }} d\tau = \varphi _{1-\gamma }(u+1) - \varphi _{1-\gamma }(l). \end{aligned}$$

(19)

then we can finally derive:

$$\begin{aligned}&\sum \limits _{i=0}^k \left( \prod \limits _{j=i+1}^{k} \theta _j\right) \mu _i^2 = \sum \limits _{i=0}^m \left( \prod \limits _{j=i+1}^{k} \theta _j\right) \mu _i^2 + \sum \limits _{i=m+1}^k \left( \prod \limits _{j=i+1}^{k} \theta _j\right) \mu _i^2\\&\quad \overset{(18) + (19)}{\le } \sum \limits _{i=0}^m (1 + \mu _0 \sigma _f)^{ \varphi _{1-\gamma }(i+1) - \varphi _{1-\gamma }(k) } \mu _i^2 + \mu _{m+1} \sum \limits _{i=m+1}^k \left[ \prod \limits _{j=i+1}^{k} (1 - \mu _j\sigma _f) \right] \mu _i \\&\quad \le (1 + \mu _0 \sigma _f)^{ \varphi _{1-\gamma }(m) - \varphi _{1-\gamma }(k) } \sum \limits _{i=0}^m \mu _i^2\\&\qquad + \frac{\mu _{m+1}}{\sigma _f} \sum \limits _{i=m+1}^k \left[ \prod \limits _{j=i+1}^{k} (1 - \mu _j\sigma _f) \right] (1 - (1- \sigma _f\mu _i)) \\&\quad = (1 + \mu _0 \sigma _f)^{ \varphi _{1-\gamma }(m) - \varphi _{1-\gamma }(k) } \mu _0^2 \sum \limits _{i=0}^m \frac{1}{i^{2\gamma }} \\&\qquad +\frac{\mu _{m+1}}{\sigma _f} \sum \limits _{i=m+1}^k \left[ \prod \limits _{j=i+1}^{k} (1 - \mu _j\sigma _f) - \prod \limits _{j=i}^{k} (1 - \mu _j\sigma _f) \right] \\&\quad \le (1 + \mu _0 \sigma _f)^{ \varphi _{1-\gamma }(m) - \varphi _{1-\gamma }(k) } \frac{m^{1- 2\gamma } - 1}{1 - 2\gamma } + \frac{\mu _{m+1}}{\sigma _f} \left[ 1 - \prod \limits _{j=m+1}^{k} (1 - \mu _j\sigma _f) \right] \\&\quad \le (1 + \mu _0 \sigma _f)^{ \varphi _{1-\gamma }(m) - \varphi _{1-\gamma }(k) } \varphi _{1 - 2\gamma }(m) + \frac{\mu _{m+1}}{\sigma _f}. \end{aligned}$$

By denoting the second constant $\tilde{\theta }_0 = \frac{1}{1+\mu _0 \sigma _f}$, then the last relation implies the following bound:

$$\begin{aligned} \mathbb {E}\left[ \Vert x^{k+1}-x^*\Vert ^2\right] \le \tilde{\theta }_0^{\varphi _{1-\gamma }(k)} \Vert x^{0}-x^*\Vert ^2 + \tilde{\theta }_0^{ \varphi _{1-\gamma }(k) - \varphi _{1-\gamma }(m) } \varphi _{1 - 2\gamma }(m)\varSigma + \frac{\mu _{m+1}}{\sigma _f} \varSigma . \end{aligned}$$

Denote $r_k^2 = \mathbb {E}[\Vert x^k-x^*\Vert ^2]$. To derive an explicit convergence rate order we analyze upper bounds on function $\phi$.

(i) First assume that $\gamma \in (0, \frac{1}{2})$. This implies that $1 - 2\gamma > 0$ and that:

$$\begin{aligned} \varphi _{1-2\gamma }\left( \left\lfloor \frac{k}{2} \right\rfloor \right) \le \varphi _{1-2\gamma }\left( \frac{k}{2}\right) = \frac{\left( \frac{k}{2} \right) ^{1-2\gamma } - 1}{1-2\gamma }\le \frac{\left( \frac{k}{2} \right) ^{1-2\gamma }}{1-2\gamma }. \end{aligned}$$

(20)

On the other hand, by using the inequality $e^{-x} \le \frac{1}{1 + x}$ for all $x \ge 0$, we obtain:

$$\begin{aligned}&\tilde{\theta }_0^{\varphi _{1-\gamma }(k) - \varphi _{1-\gamma }(\frac{k-2}{2})} \varphi _{1-2\gamma }\left( \frac{k}{2}\right) = e^{(\varphi _{1-\gamma }(k) - \varphi _{1-\gamma }(\frac{k-2}{2}))\ln {\tilde{\theta }_0}} \varphi _{1-2\gamma }\left( \frac{k}{2} \right) \\&\quad \le \frac{\varphi _{1-2\gamma }\left( \frac{k}{2} \right) }{1 + [\varphi _{1-\gamma }(k) - \varphi _{1-\gamma }(\frac{k}{2}-1)]\ln {\frac{1}{\tilde{\theta }_0}}} \overset{(20)}{\le } \frac{\frac{k^{1-2\gamma }}{2^{1-2\gamma } (1-2\gamma )} }{\frac{1}{1-\gamma }[k^{1-\gamma } - (\frac{k}{2}-1)^{1-\gamma }]\ln {\frac{1}{\tilde{\theta }_0}}} \\&\quad = \frac{\frac{k^{1-2\gamma }}{2^{1-2\gamma } (1-2\gamma )}}{\frac{k^{1-\gamma }}{1-\gamma }[1 - (\frac{1}{6})^{1-\gamma }]\ln {\frac{1}{\tilde{\theta }_0}}} = \frac{1-\gamma }{1-2\gamma }\frac{2^{\gamma }k^{-\gamma }}{2^{1-2\gamma }[1 - (\frac{1}{6})^{1-\gamma }]\ln {\frac{1}{\theta _0}}} = \mathcal {O}\left( \frac{1}{k^{\gamma }}\right) . \end{aligned}$$

Therefore, in this case, the overall rate will be given by:

$$\begin{aligned} r_{k+1}^2 \le \theta _0^{\mathcal {O}(k^{1-\gamma })}r_0^2 + \mathcal {O}\left( \frac{1}{k^{\gamma }}\right) \approx \mathcal {O}\left( \frac{1}{k^{\gamma }}\right) . \end{aligned}$$

If $\gamma = \frac{1}{2}$, then the definition of $\varphi _{1-2\gamma }(\frac{k}{2})$ provides that:

$$\begin{aligned} r_{k+1}^2 \le \tilde{\theta }_0^{\mathcal {O}(\sqrt{k})}r_0^2 + \tilde{\theta }_0^{\mathcal {O}(\sqrt{k})}\mathcal {O}(\ln {k}) + \mathcal {O}\left( \frac{1}{\sqrt{k}}\right) \approx \mathcal {O}\left( \frac{1}{\sqrt{k}}\right) . \end{aligned}$$

When $\gamma \in (\frac{1}{2}, 1)$, it is obvious that $\varphi _{1-2\gamma }\left( \frac{k}{2}\right) \le \frac{1}{2\gamma - 1}$ and therefore the order of the convergence rate changes into:

$$\begin{aligned} r_{k+1}^2 \le \tilde{\theta }_0^{\mathcal {O}(k^{1-\gamma })}[r_0^2 + \mathcal {O}(1)] + \mathcal {O}\left( \frac{1}{k^{\gamma }}\right) \approx \mathcal {O}\left( \frac{1}{k^{\gamma }}\right) . \end{aligned}$$

(ii) Lastly, if $\gamma = 1$, by using $\tilde{\theta }_0^{\ln {k+1}} \le \left( \frac{1}{k}\right) ^{\ln {\frac{1}{\tilde{\theta }_0}}}$ we obtain the second part of our result. $\square$