acceleratedzerothordermethodfornonsmoothstochasticconvexoptimizationproblemwithinfinitevariance/full.md

Accelerated Zeroth-order Method for Non-Smooth Stochastic Convex Optimization Problem with Infinite Variance

Nikita Kornilov

MIPT, SkolTech

kornilov.nm@phystech.edu

Ohad Shamir

Weizmann Institute of Science

ohad.shamir@weizmann.ac.il

Aleksandr Lobanov

MIPT, ISP RAS

lobanov.av@mipt.ru

Darina Dvinskikh

HSE University, ISP RAS

dmdvinskikh@hse.ru

Alexander Gasnikov

MIPT,

ISP RAS, SkolTech

gasnikov@yandex.ru

Innokentiy Shibaev

MIPT,

IITP RAS

innokenti.shibayev@phystech.edu

Eduard Gorbunov

MBZUAI

eduard.gorbunov@mbzuai.ac.ae

Samuel Horváth

MBZUAI

samuel.horvath@mbzuai.ac.ae

Abstract

In this paper, we consider non-smooth stochastic convex optimization with two function evaluations per round under infinite noise variance. In the classical setting when noise has finite variance, an optimal algorithm, built upon the batched accelerated gradient method, was proposed in [17]. This optimality is defined in terms of iteration and oracle complexity, as well as the maximal admissible level of adversarial noise. However, the assumption of finite variance is burdensome and it might not hold in many practical scenarios. To address this, we demonstrate how to adapt a refined clipped version of the accelerated gradient (Stochastic Similar Triangles) method from [35] for a two-point zero-order oracle. This adaptation entails extending the batching technique to accommodate infinite variance — a non-trivial task that stands as a distinct contribution of this paper.

1 Introduction

In this paper, we consider stochastic non-smooth convex optimization problem

$$\begin{array}{cccc}& \underset{x\in {\mathbb{R}}^d}{\min }\{f(x){\mathbb{E}}_{\xi \sim \mathcal{D}}[f(x,\xi )]\},& & \text{(1)}\end{array}\backslash min\; \_\; \{x\; \backslash in\; \backslash mathbb\; \{R\}\; ^\; \{d\}\}\; \backslash left\backslash \{f\; (x)\; \backslash stackrel\; \{\backslash text\; \{d\; e\; f\}\}\; \{=\}\; \backslash mathbb\; \{E\}\; \_\; \{\backslash xi\; \backslash sim\; \backslash mathcal\; \{D\}\}\; [\; f\; (x,\; \backslash xi)\; ]\; \backslash right\backslash \},\; \backslash tag\; \{1\}$$

where $f(x,\xi)$ is $M_2(\xi)$ -Lipschitz continuous in $x$ w.r.t. the Euclidean norm, and the expectation $\mathbb{E}_{\xi \sim \mathcal{D}}[f(x,\xi)]$ is with respect to a random variable $\xi$ with unknown distribution $\mathcal{D}$ . The optimization is performed only by accessing two function evaluations per round rather than sub-gradients, i.e., for any pair of points $x,y\in \mathbb{R}^d$ , an oracle returns $f(x,\xi)$ and $f(y,\xi)$ with the same $\xi$ . The primary motivation for this gradient-free oracle arises from different applications where calculating gradients is computationally infeasible or even impossible. For instance, in medicine, biology, and physics, the objective function can only be computed through numerical simulation or as the result of a real experiment, i.e., automatic differentiation cannot be employed to calculate function derivatives. Usually, a black-box function we are optimizing is affected by stochastic or computational noise. This noise can arise naturally from modeling randomness within a simulation or by computer discretization.

In classical setting, this noise has light tails. However, usually in black-box optimization, we know nothing about the function, only its values at the requested points are available/computable, so light tails assumption may be violated. In this case, gradient-free algorithms may diverge. We aim to develop an algorithm that is robust even to heavy-tailed noise, i.e., noise with infinite variance. In theory, one can consider heavy-tailed noise to simulate situations where noticeable outliers may occur (even if the nature of these outliers is non-stochastic). Therefore we relax classical finite variance assumption and consider less burdensome assumption of finite $\alpha$ -th moment, where $\alpha \in (1,2]$ .

In machine learning, interest in gradient-free methods is mainly driven by the bandit optimization problem [14, 2, 5], where a learner engages in a game with an adversary: the learner selects a point $x$ , and the adversary chooses a point $\xi$ . The learner's goal is to minimize the average regret based solely on observations of function values (losses) $f(x, \xi)$ . As feedback, at each round, the learner receives losses at two points. This corresponds to a zero-order oracle in stochastic convex optimization with two function evaluations per round. The vast majority of researches assume sub-Gaussian distribution of rewards. However, in some practical cases (e.g., in finance [33]) reward distribution has heavy-tails or can be adversarial. For the heavy-tailed bandit optimization, we refer to [9].

Two-point gradient-free optimization for non-smooth (strongly) convex objective is a well-researched area. Numerous algorithms have been proposed which are optimal with respect to two criteria: oracle and iteration complexity. For a detailed overview, see the recent survey [15] and the references therein. Optimal algorithms, in terms of oracle call complexity, are presented in [10, 36, 3]. The distinction between the number of successive iterations (which cannot be executed in parallel) and the number of oracle calls was initiated with the lower bound obtained in [6]. It culminated with the optimal results from [17], which provides an algorithm which is optimal in both criteria. Specifically, the algorithm produces $\hat{x}$ , an $\varepsilon$ -solution of (1), such that we can guarantee $\mathbb{E}[f(\hat{x})] - \min_{x \in \mathbb{R}^d} f(x)$ after

$$\begin{array}{c}\sim d^{\frac{1}{4}}{\epsilon }^{-1}\text{s u c c e s s i v e i t e r a t i o n s},\\ \sim d{\epsilon }^{-2}\text{o r a c l e c a l l s (n u m b e r o f}f(x,\xi )\text{c a l c u l a t i o n s)}\mathrm{.}\end{array}\backslash begin\{array\}\{l\}\; \backslash sim\; d\; ^\; \{\backslash frac\; \{1\}\{4\}\}\; \backslash varepsilon^\; \{-\; 1\}\; \backslash quad\; \backslash text\; \{s\; u\; c\; c\; e\; s\; s\; i\; v\; e\; i\; t\; e\; r\; a\; t\; i\; o\; n\; s\},\; \backslash \backslash \; \backslash sim\; d\; \backslash varepsilon^\; \{-\; 2\}\; \backslash quad\; \backslash text\; \{o\; r\; a\; c\; l\; e\; c\; a\; l\; l\; s\; (n\; u\; m\; b\; e\; r\; o\; f\}\; f\; (x,\; \backslash xi)\; \backslash text\; \{c\; a\; l\; c\; u\; l\; a\; t\; i\; o\; n\; s)\}.\; \backslash \backslash \; \backslash end\{array\}$$

The convergence guarantee for this optimal algorithm from [17] was established in the classical setting of light-tailed noise, i.e., when noise has finite variance: $\mathbb{E}{\xi}[M_2(\xi)^2] < \infty$ . However, in many modern learning problems the variance might not be finite, leading the aforementioned algorithms to potentially diverge. Indeed, heavy-tailed noise is prevalent in contemporary applications of statistics and deep learning. For example, heavy-tailed behavior can be observed in training attention models [41] and convolutional neural networks [37, 20]. Consequently, our goal is to develop an optimal algorithm whose convergence is not hampered by this restrictive assumption. To the best of our knowledge, no existing literature on gradient-free optimization allows for $\mathbb{E}{\xi}[M_2(\xi)^2]$ to be infinite. Furthermore, convergence results for all these gradient-free methods were provided in expectation, that is, without (non-trivial) high-probability bounds. Although the authors of [17] mentioned (without proof) that their results can be formulated in high probability using [19], this aspect notably affects the oracle calls complexity bound and complicates the analysis.

A common technique to relax finite variance assumption is gradient clipping [31]. Starting from the work of [26] (see also [8, 19]), there has been increased interest in algorithms employing gradient clipping to achieve high-probability convergence guarantees for stochastic optimization problems with heavy-tailed noise. Particularly, in just the last two years there have been proposed

• an optimal algorithm with a general proximal setup for non-smooth stochastic convex optimization problems with infinite variance [39] that converges in expectation (also referenced in [27]),
• an optimal adaptive algorithm with a general proximal setup for non-smooth online stochastic convex optimization problems with infinite variance [42] that converges with high probability,
• optimal algorithms using the Euclidean proximal setup for both smooth and non-smooth stochastic convex optimization problems and variational inequalities with infinite variance [35, 30, 29] that converge with high probability,
• an optimal variance-adaptive algorithm with the Euclidean proximal setup for non-smooth stochastic (strongly) convex optimization problems with infinite variance [24] that converges with high probability.

None of these papers discuss a gradient-free oracle. Moreover, they do not incorporate acceleration (given the non-smooth nature of the problems) with the exception of [35]. Acceleration is a crucial

component to achieving optimal bounds on the number of successive iterations. However, the approach in [35] presumes smoothness and does not utilize batching. To apply the convergence results from [35] to our problem, we need to adjust our problem formulation to be smooth. This is achieved by using the Smoothing technique [27, 36, 17]. In the work [22] authors proposed an algorithm based on Smoothing technique and non-accelerated Stochastic Mirror Descent with clipping. However, this work also does not support acceleration, minimization on the whole space and batching. Adapting the technique from [35] to incorporate batching necessitates a substantial generalization. We regard this aspect of our work as being of primary interest.

Heavy-tailed noise can also be handled without explicit gradient clipping, e.g., by using Stochastic Mirror Descent algorithm with a particular class of uniformly convex mirror maps [39]. However, the convergence guarantee for this algorithm is given in expectation. Moreover, applying batching and acceleration is a non-trivial task. Without this, we are not able to get the optimal method in terms of the number of iterations and not only in terms of oracle complexity. There are also some studies on the alternatives to gradient clipping [21] but the results for these alternatives are given in expectation and are weaker than the state-of-the-art results for the methods with clipping. This is another reason why we have chosen gradient clipping to handle the heavy-tailed noise.

1.1 Contributions

We generalize the optimal result from [17] to accommodate a weaker assumption that allows the noise to exhibit heavy tails. Instead of the classical assumption of finite variance, we require the boundedness of the $\alpha$ -moment: there exists $\alpha \in (1,2]$ such that $\mathbb{E}_{\xi}[M_2(\xi)^{\alpha}] < \infty$ . Notably, when $\alpha < 2$ , this assumption is less restrictive than the assumption of a finite variance and thus it has garnered considerable interest recently, see [41, 35, 29] and the references therein. Under this assumption we prove that for convex $f$ , an $\varepsilon$ -solution can be found with high probability after

$$\begin{array}{c}\sim d^{\frac{1}{4}}{\epsilon }^{-1}\text{s u c c e s s i v e i t e r a t i o n s},\\ \sim {\left(\sqrt{d}\mathrm{/}\epsilon \right)}^{\frac{\alpha }{\alpha -1}}\text{o r a c l e}\end{array}\backslash begin\{array\}\{l\}\; \backslash sim\; d\; ^\; \{\backslash frac\; \{1\}\{4\}\}\; \backslash varepsilon^\; \{-\; 1\}\; \backslash quad\; \backslash text\; \{s\; u\; c\; c\; e\; s\; s\; i\; v\; e\; i\; t\; e\; r\; a\; t\; i\; o\; n\; s\},\; \backslash \backslash \; \backslash sim\; \backslash left(\backslash sqrt\; \{d\}\; /\; \backslash varepsilon\backslash right)\; ^\; \{\backslash frac\; \{\backslash alpha\}\{\backslash alpha\; -\; 1\}\}\; \backslash quad\; \backslash text\; \{o\; r\; a\; c\; l\; e\}\; \backslash \backslash \; \backslash end\{array\}$$

and for $\mu$ -strongly convex $f$ , the $\varepsilon$ -solution can be found with high probability after

$$\begin{array}{c}\sim d^{\frac{1}{4}}(\mu \epsilon {)}^{-\frac{1}{2}}\text{s u c c e s s i v e i t e r a t i o n s},\\ \sim {(d\mathrm{/}(\mu \epsilon ))}^{\frac{\alpha }{2(\alpha -1)}}\text{o r a c l e}\end{array}\backslash begin\{array\}\{l\}\; \backslash sim\; d\; ^\; \{\backslash frac\; \{1\}\{4\}\}\; (\backslash mu\; \backslash varepsilon)\; ^\; \{-\; \backslash frac\; \{1\}\{2\}\}\; \backslash quad\; \backslash text\; \{s\; u\; c\; c\; e\; s\; s\; i\; v\; e\; i\; t\; e\; r\; a\; t\; i\; o\; n\; s\},\; \backslash \backslash \; \backslash sim\; \backslash left(d\; /\; (\backslash mu\; \backslash varepsilon)\backslash right)\; ^\; \{\backslash frac\; \{\backslash alpha\}\{2\; (\backslash alpha\; -\; 1)\}\}\; \backslash quad\; \backslash text\; \{o\; r\; a\; c\; l\; e\}\; \backslash \backslash \; \backslash end\{array\}$$

In both instances, the number of oracle calls is optimal in terms of $\varepsilon$ -dependency within the non-smooth setting [27, 39]. For first-order optimization under heavy-tailed noise, the optimal $\varepsilon$ -dependency remains consistent, as shown in [35, Table 1].

In what follows, we highlight the following several important aspects of our results

• High-probability guarantees. We provide upper-bounds on the number of iterations/oracle calls needed to find a point $\hat{x}$ such that $f(\hat{x}) - \min_{x\in \mathbb{R}^d}f(x)\leq \varepsilon$ with probability at least $1 - \beta$ . The derived bounds have a poly-logarithmic dependence on $1 / \beta$ . To the best of our knowledge, there are no analogous high-probability results, even for noise with bounded variance.
• Generality of the setup. Our results are derived under the assumption that gradient-free oracle returns values of stochastic realizations $f(x, \xi)$ subject to (potentially adversarial) bounded noise. We further provide upper bounds for the magnitude of this noise, contingent upon the target accuracy $\varepsilon$ and confidence level $\beta$ . Notably, our assumptions about the objective and noise are confined to a compact subset of $\mathbb{R}^d$ . This approach, which differs from standard ones in derivative-free optimization, allows us to encompass a wide range of problems. This approach differs from standard ones in derivative-free optimization
• Batching without bounded variance. To establish the aforementioned upper bounds, we obtain the following: given $X_{1},\ldots ,X_{B}$ as independent random vectors in $\mathbb{R}^d$ where $\mathbb{E}[X_i] = x\in \mathbb{R}^d$ and $\mathbb{E}| X_i - x| _2^\alpha \leq \sigma^\alpha$ for some $\sigma \geq 0$ and $\alpha \in (1,2]$ , then

$$\begin{array}{cccc}& \mathbb{E}\left[{\parallel \frac{1}{B}\sum _{i=1}^B{X}_i-x\parallel }_2^{\alpha }\right]\le \frac{2{\sigma }^{\alpha }}{B^{\alpha -1}}\mathrm{.}& & \text{(2)}\end{array}\backslash mathbb\; \{E\}\; \backslash left[\; \backslash left\backslash |\; \backslash frac\; \{1\}\{B\}\; \backslash sum\_\; \{i\; =\; 1\}\; ^\; \{B\}\; X\; \_\; \{i\}\; -\; x\; \backslash right\backslash |\; \_\; \{2\}\; ^\; \{\backslash alpha\}\; \backslash right]\; \backslash leq\; \backslash frac\; \{2\; \backslash sigma^\; \{\backslash alpha\}\}\{B\; ^\; \{\backslash alpha\; -\; 1\}\}.\; \backslash tag\; \{2\}$$

When $\alpha = 2$ , this result aligns with the conventional case of bounded variance (accounting for a numerical factor of 2). Unlike existing findings, such as [40, Lemma 7] where $\alpha < 2$ , the relation (2) does not exhibit a dependency on the dimension $d$ . Moreover, (2) offers a theoretical basis to highlight the benefits of mini-batching, applicable to methods highlighted in this paper as well as first-order methods presented in [35, 29].

• Dependency on $d$ . As far as we are aware, an open question remains: is the bound $\left(\sqrt{d} / \varepsilon\right)^{\frac{\alpha}{\alpha - 1}}$ optimal regarding its dependence on $d$ ? For smooth stochastic convex optimization problems using a $(d + 1)$ -points stochastic zeroth-order oracle, the answer is negative. The optimal bound is proportional to $d\varepsilon^{-\frac{\alpha}{\alpha - 1}}$ . Consequently, for $\alpha \in (1,2)$ , our results are intriguing because the dependence on $d$ in our bound differs from known results in the classical case where $\alpha = 2$ .

1.2 Paper organization

The paper is organized as follows. In Section 2, we give some preliminaries, such as smoothing technique and gradient estimation, that are workhorse of our algorithms. Section 3 is the main section presenting two novel gradient-free algorithms along with their convergence results in high probability. These algorithms solve non-smooth stochastic optimization under heavy-tailed noise, and they will be referred to as ZO-clipped-SSTM and R-ZO-clipped-SSTM (see Algorithms 1 and 2 respectively). In Section 4, we extend our results to gradient-free oracle corrupted by additive deterministic adversarial noise. In Section 5, we describe the main ideas behind the proof and emphasize key lemmas. In Section 6, we provide numerical experiments on the synthetic task that demonstrate the robustness of the proposed algorithms towards heavy-tailed noise.

2 Preliminaries

Assumptions on a subset. Although we consider an unconstrained optimization problem, our analysis does not require any assumptions to hold on the entire space. For our purposes, it is sufficient to introduce all assumptions only on some convex set $Q \in \mathbb{R}^d$ since we prove that the considered methods do not leave some ball around the solution or some level-set of the objective function with high probability. This allows us to consider fairly large classes of problems.

Assumption 1 (Strong convexity) There exist a convex set $Q \subset \mathbb{R}^d$ and $\mu \geq 0$ such that function $f(x, \xi)$ is $\mu$ -strongly convex on $Q$ for any fixed $\xi$ , i.e.

$$f(\lambda x_1+(1-\lambda )x_2)\le \lambda f(x_1)+(1-\lambda )f(x_2)-\frac{1}{2}\mu \lambda (1-\lambda )\mathrm{\parallel }{x}_1-x_2{\mathrm{\parallel }}_2^2,f\; (\backslash lambda\; x\; \_\; \{1\}\; +\; (1\; -\; \backslash lambda)\; x\; \_\; \{2\})\; \backslash leq\; \backslash lambda\; f\; (x\; \_\; \{1\})\; +\; (1\; -\; \backslash lambda)\; f\; (x\; \_\; \{2\})\; -\; \backslash frac\; \{1\}\{2\}\; \backslash mu\; \backslash lambda\; (1\; -\; \backslash lambda)\; \backslash |\; x\; \_\; \{1\}\; -\; x\; \_\; \{2\}\; \backslash |\; \_\; \{2\}\; ^\; \{2\},$$

for all $x_{1},x_{2}\in Q,\lambda \in [0,1]$

This assumption implies that $f(x)$ is $\mu$ -strongly convex as well.

For a small constant $\tau > 0$ , let us define an expansion of set $Q$ namely $Q_{\tau} = Q + B_2^d$ , where $+$ stands for Minkowski addition. Using this expansion we make the following assumption.

Assumption 2 (Lipschitz continuity and boundedness of $\alpha$ -moment) There exist a convex set $Q \subset \mathbb{R}^d$ and $\tau > 0$ such that function $f(x, \xi)$ is $M_2(\xi)$ -Lipschitz continuous w.r.t. the Euclidean norm on $Q_{\tau}$ , i.e., for all $x_1, x_2 \in Q_{\tau}$

$$\mid f(x_1,\xi )-f(x_2,\xi )\mid \le M_2(\xi )\mathrm{\parallel }{x}_1-x_2{\mathrm{\parallel }}_2\mathrm{.}\backslash left|\; f\; \backslash left(x\; \_\; \{1\},\; \backslash xi\backslash right)\; -\; f\; \backslash left(x\; \_\; \{2\},\; \backslash xi\backslash right)\; \backslash right|\; \backslash leq\; M\; \_\; \{2\}\; (\backslash xi)\; \backslash |\; x\; \_\; \{1\}\; -\; x\; \_\; \{2\}\; \backslash |\; \_\; \{2\}.$$

Moreover, there exist $\alpha \in (1,2]$ and $M_2 > 0$ such that $\mathbb{E}_{\xi}[M_2(\xi)^{\alpha}]\leq M_2^{\alpha}$ .

If $\alpha < 2$ , we say that noise is heavy-tailed. When $\alpha = 2$ , the above assumption recovers the standard uniformly bounded variance assumption.

Lemma 1 Assumption 2 implies that $f(x)$ is $M_2$ -Lipschitz on $Q$ .

Randomized smoothing. The main scheme that allows us to develop batch-parallel gradient-free methods for non-smooth convex problems is randomized smoothing [13, 17, 27, 28, 38] of a non-smooth function $f(x,\xi)$ . The smooth approximation to a non-smooth function $f(x,\xi)$ is defined as

$$\begin{array}{cccc}& {\widehat{f}}_{\tau }(x){\mathbb{E}}_{\mathbf{u},\xi }[f(x+\tau \mathbf{u},\xi )],& & \text{(3)}\end{array}\backslash hat\; \{f\}\; \_\; \{\backslash tau\}\; (x)\; \backslash stackrel\; \{\backslash text\; \{d\; e\; f\}\}\; \{=\}\; \backslash mathbb\; \{E\}\; \_\; \{\backslash mathbf\; \{u\},\; \backslash xi\}\; [\; f\; (x\; +\; \backslash tau\; \backslash mathbf\; \{u\},\; \backslash xi)\; ],\; \backslash tag\; \{3\}$$

where $\mathbf{u} \sim U(B_2^d)$ is a random vector uniformly distributed on the Euclidean unit ball $B_2^d \stackrel{\mathrm{def}}{=} {x \in \mathbb{R}^d : |x|_2 \leq 1}$ . In this approximation, a new type of randomization appears in addition to stochastic variable $\xi$ .

The next lemma gives estimates for the quality of this approximation. In contrast to $f(x)$ , function $\hat{f}_{\tau}(x)$ is smooth and has several useful properties.

Lemma 2 [17, Theorem 2.1.] Let there exist a subset $Q \subset \mathbb{R}^d$ and $\tau > 0$ such that Assumptions 1 and 2 hold on $Q_{\tau}$ . Then,

1. Function $\hat{f}_{\tau}(x)$ is convex, Lipschitz with constant $M_2$ on $Q$ , and satisfies

$$\underset{x\in Q}{\sup }\mathrm{\mid }{\widehat{f}}_{\tau }(x)-f(x)\mathrm{\mid }\le \tau M_2\mathrm{.}\backslash sup\; \_\; \{x\; \backslash in\; Q\}\; |\; \backslash hat\; \{f\}\; \_\; \{\backslash tau\}\; (x)\; -\; f\; (x)\; |\; \backslash leq\; \backslash tau\; M\; \_\; \{2\}.$$

2. Function $\hat{f}_{\tau}(x)$ is differentiable on $Q$ with the following gradient

$$\mathrm{\nabla }{\widehat{f}}_{\tau }(x)={\mathbb{E}}_{\mathbf{e}}[\frac{d}{\tau }f(x+\tau \mathbf{e})\mathbf{e}],\backslash nabla\; \backslash hat\; \{f\}\; \_\; \{\backslash tau\}\; (x)\; =\; \backslash mathbb\; \{E\}\; \_\; \{\backslash mathbf\; \{e\}\}\; \backslash left[\; \backslash frac\; \{d\}\{\backslash tau\}\; f\; (x\; +\; \backslash tau\; \backslash mathbf\; \{e\})\; \backslash mathbf\; \{e\}\; \backslash right],$$

where $\mathbf{e} \sim U(S_2^d)$ is a random vector uniformly distributed on unit Euclidean sphere $S_2^d \stackrel{\text{def}}{=} {x \in \mathbb{R}^d : |x|_2 = 1}$ .

3. Function $\hat{f}_{\tau}(x)$ is $L$ -smooth with $L = \sqrt{d} M_2 / \tau$ on $Q$ .

Our algorithms will aim at minimizing the smooth approximation $\hat{f}_{\tau}(x)$ . Given Lemma 2, the output of the algorithm will also be a good approximate minimizer of $f(x)$ when $\tau$ is sufficiently small.

Gradient estimation. Our algorithms will be based on randomized gradient estimates, which will then be used in a first order algorithm. Following [36], the gradient can be estimated by the following vector:

$$\begin{array}{cccc}& g(x,\xi ,\mathbf{e})=\frac{d}{2\tau }(f(x+\tau \mathbf{e},\xi )-f(x-\tau \mathbf{e},\xi ))\mathbf{e},& & \text{(4)}\end{array}g\; (x,\; \backslash xi\; ,\; \backslash mathbf\; \{e\})\; =\; \backslash frac\; \{d\}\{2\; \backslash tau\}\; \backslash left(f\; (x\; +\; \backslash tau\; \backslash mathbf\; \{e\},\; \backslash xi)\; -\; f\; (x\; -\; \backslash tau\; \backslash mathbf\; \{e\},\; \backslash xi)\backslash right)\; \backslash mathbf\; \{e\},\; \backslash tag\; \{4\}$$

where $\tau > 0$ and $\mathbf{e} \sim U(S_2^d)$ . We also use batching technique in order to allow parallel calculation of gradient estimation and acceleration. Let $B$ be a batch size, we sample ${\mathbf{e}i}{i=1}^B$ and ${\xi_i}_{i=1}^B$ independently, then

$$\begin{array}{cccc}& g^B(x,\{\xi \},\{\mathbf{e}\})=\frac{d}{2B\tau }\sum _{i=1}^B(f(x+\tau {\mathbf{e}}_i,{\xi }_i)-f(x-\tau {\mathbf{e}}_i,{\xi }_i)){\mathbf{e}}_i\mathrm{.}& & \text{(5)}\end{array}g\; ^\; \{B\}\; (x,\; \backslash \{\backslash xi\; \backslash \},\; \backslash \{\backslash mathbf\; \{e\}\; \backslash \})\; =\; \backslash frac\; \{d\}\{2\; B\; \backslash tau\}\; \backslash sum\_\; \{i\; =\; 1\}\; ^\; \{B\}\; \backslash left(f\; \backslash left(x\; +\; \backslash tau\; \backslash mathbf\; \{e\}\; \_\; \{i\},\; \backslash xi\_\; \{i\}\backslash right)\; -\; f\; \backslash left(x\; -\; \backslash tau\; \backslash mathbf\; \{e\}\; \_\; \{i\},\; \backslash xi\_\; \{i\}\backslash right)\backslash right)\; \backslash mathbf\; \{e\}\; \_\; \{i\}.\; \backslash tag\; \{5\}$$

The next lemma states that $g^{B}(x, {\xi}, {\mathbf{e}})$ from (5) is an unbiased estimate of the gradient of $\hat{f}_{\tau}(x)$ (3). Moreover, under heavy-tailed noise (Assumption 2) with bounded $\alpha$ -moment $g^{B}(x, {\xi}, {\mathbf{e}})$ has also bounded $\alpha$ -moment.

Lemma 3 Under Assumptions 1 and 2, it holds

$${\mathbb{E}}_{\xi ,\mathbf{e}}[g(x,\xi ,\mathbf{e})]={\mathbb{E}}_{\{\xi \},\{\mathbf{e}\}}[g^B(x,\{\xi \},\{\mathbf{e}\})]=\mathrm{\nabla }{\widehat{f}}_{\tau }(x)\mathrm{.}\backslash mathbb\; \{E\}\; \_\; \{\backslash xi\; ,\; \backslash mathbf\; \{e\}\}\; [\; g\; (x,\; \backslash xi\; ,\; \backslash mathbf\; \{e\})\; ]\; =\; \backslash mathbb\; \{E\}\; \_\; \{\backslash \{\backslash xi\; \backslash \},\; \backslash \{\backslash mathbf\; \{e\}\; \backslash \}\}\; [\; g\; ^\; \{B\}\; (x,\; \backslash \{\backslash xi\; \backslash \},\; \backslash \{\backslash mathbf\; \{e\}\; \backslash \})\; ]\; =\; \backslash nabla\; \backslash hat\; \{f\}\; \_\; \{\backslash tau\}\; (x).$$

and

$${\mathbb{E}}_{\xi ,\mathbf{e}}[\mathrm{\parallel }g(x,\xi ,\mathbf{e})-{\mathbb{E}}_{\xi ,\mathbf{e}}[g(x,\xi ,\mathbf{e})]{\mathrm{\parallel }}_2^{\alpha }]\le {\sigma }^{\alpha }{\left(\frac{\sqrt{d}{M}_2}{2^{\frac{1}{4}}}\right)}^{\alpha },\backslash mathbb\; \{E\}\; \_\; \{\backslash xi\; ,\; \backslash mathbf\; \{e\}\}\; [\; \backslash |\; g\; (x,\; \backslash xi\; ,\; \backslash mathbf\; \{e\})\; -\; \backslash mathbb\; \{E\}\; \_\; \{\backslash xi\; ,\; \backslash mathbf\; \{e\}\}\; [\; g\; (x,\; \backslash xi\; ,\; \backslash mathbf\; \{e\})\; ]\; \backslash |\; \_\; \{2\}\; ^\; \{\backslash alpha\}\; ]\; \backslash leq\; \backslash sigma^\; \{\backslash alpha\}\; \backslash stackrel\; \{\{d\; e\; f\}\}\; \{\{=\}\}\; \backslash left(\backslash frac\; \{\backslash sqrt\; \{d\}\; M\; \_\; \{2\}\}\{2\; ^\; \{\backslash frac\; \{1\}\{4\}\}\}\backslash right)\; ^\; \{\backslash alpha\},$$

$$\begin{array}{cccc}& {\mathbb{E}}_{\{\xi \},\{\mathbf{e}\}}[\mathrm{\parallel }{g}^B(x,\{\xi \},\{\mathbf{e}\})-{\mathbb{E}}_{\{\xi \},\{\mathbf{e}\}}[g^B(x,\{\xi \},\{\mathbf{e}\})]{\mathrm{\parallel }}_2^{\alpha }]\le \frac{2{\sigma }^{\alpha }}{B^{\alpha -1}}\mathrm{.}& & \text{(6)}\end{array}\backslash mathbb\; \{E\}\; \_\; \{\backslash \{\backslash xi\; \backslash \},\; \backslash \{\backslash mathbf\; \{e\}\; \backslash \}\}\; [\; \backslash |\; g\; ^\; \{B\}\; (x,\; \backslash \{\backslash xi\; \backslash \},\; \backslash \{\backslash mathbf\; \{e\}\; \backslash \})\; -\; \backslash mathbb\; \{E\}\; \_\; \{\backslash \{\backslash xi\; \backslash \},\; \backslash \{\backslash mathbf\; \{e\}\; \backslash \}\}\; [\; g\; ^\; \{B\}\; (x,\; \backslash \{\backslash xi\; \backslash \},\; \backslash \{\backslash mathbf\; \{e\}\; \backslash \})\; ]\; \backslash |\; \_\; \{2\}\; ^\; \{\backslash alpha\}\; ]\; \backslash leq\; \backslash frac\; \{2\; \backslash sigma^\; \{\backslash alpha\}\}\{B\; ^\; \{\backslash alpha\; -\; 1\}\}.\; \backslash tag\; \{6\}$$

3 Main Results

In this section, we present two our new zero-order algorithms, which we will refer to as ZO-clipped-SSTM and R-ZO-clipped-SSTM, to solve problem (1) under heavy-tailed noise assumption. To deal with heavy-tailed noise, we use clipping technique which clips heavy tails. Let $\lambda > 0$ be clipping constant and $g \in \mathbb{R}^d$ , then clipping operator clip is defined as

We apply clipping operator for batched gradient estimate $g^{B}(x,{\xi } ,{\mathbf{e}})$ from (5) and then feed it into first-order Clipped Stochastic Similar Triangles Method (clipped-SSTM) from [18]. We will refer to our zero-order versions of clipped-SSTM as ZO-clipped-SSTM and R-ZO-clipped-SSTM for the convex case and strongly convex case respectively.

3.1 Convex case

Let us suppose that Assumption 1 is satisfied with $\mu = 0$

Algorithm 1 ZO-clipped-SSTM $\left(x^{0},K,B,a,\tau ,{\lambda_{k}}{k = 0}^{K - 1}\right)$
Input: starting point $x^0$ , number of iterations $K$ , batch size $B$ , stepsize $a > 0$ , smoothing parameter $\tau$ , clipping levels ${\lambda_k}{k = 0}^{K - 1}$
1: Set $L = \sqrt{d} M_2 / \tau$ $A_0 = \alpha_0 = 0$ $y^{0} = z^{0} = x^{0}$
2: for $k = 0,\dots ,K - 1$ do
3: Set $\alpha_{k + 1} = k + 2 / 2aL$ $A_{k + 1} = A_{k} + \alpha_{k + 1}$
4: $x^{k + 1} = \frac{A_ky^k + \alpha_{k + 1}z^k}{A_{k + 1}}.$
5: Sample ${\xi_i^k}{i = 1}^B\sim \mathcal{D}$ and ${\mathbf{e}i^k}{i = 1}^B\sim S_2^d$ independently.
6: Compute gradient estimation $g^{B}(x^{k + 1},{\xi^{k}} ,{\mathbf{e}^{k}})$ as defined in (5).
7: Compute clipped $\tilde{g}{k + 1} = \mathrm{clip}\left(g^{B}(x^{k + 1},{\xi^{k}} ,{\mathbf{e}^{k}}),\lambda_{k}\right)$ as defined in (7).
8: $z^{k + 1} = z^{k} - \alpha_{k + 1}\tilde{g}{k + 1}$
9: $y^{k + 1} = \frac{A_ky^k + \alpha{k + 1}z^{k + 1}}{A_{k + 1}}.$
10: end for
Output: $y^{K}$

Theorem 1 (Convergence of ZO-clipped-SSTM) Let Assumptions 1 and 2 hold with $\mu = 0$ and $Q = \mathbb{R}^d$ . Let $| x^0 - x^|^2 \leq R^2$ , where $x^0$ is a starting point and $x^$ is an optimal solution to (1). Then for the output $y^K$ of ZO-clipped-SSTM, run with batchsize $B$ , $A = \ln^{4K} / \beta \geq 1$ , $a = \Theta(\max{A^2, \sqrt{d}M_2K^{\frac{(\alpha + 1)}{\alpha}} A^{\frac{(\alpha - 1)}{\alpha}} /{LRB}\frac{(\alpha - 1)}{\alpha}})$ , $\tau = \varepsilon / 4M_2$ and $\lambda_k = \Theta(R / (\alpha{k+1}A))$ , we can guarantee $f(y^K) - f(x^*) \leq \varepsilon$ with probability at least $1 - \beta$ (for any $\beta \in (0,1]$ ) after

$$\begin{array}{cccc}& K=\tilde{\mathcal{O}}(\max \{\frac{M_2\sqrt[4]{d}R}{\epsilon },\frac{1}{B}{\left(\frac{\sqrt{d}{M}_{2}R}{\epsilon }\right)}^{\frac{\alpha }{\alpha -1}}\})& & \text{(8)}\end{array}K\; =\; \backslash widetilde\; \{\backslash mathcal\; \{O\}\}\; \backslash left(\backslash max\; \backslash left\backslash \{\backslash frac\; \{M\; \_\; \{2\}\; \backslash sqrt\; [\; 4\; ]\{d\}\; R\}\{\backslash varepsilon\},\; \backslash frac\; \{1\}\{B\}\; \backslash left(\backslash frac\; \{\backslash sqrt\; \{d\}\; M\; \_\; \{2\}\; R\}\{\backslash varepsilon\}\backslash right)\; ^\; \{\backslash frac\; \{\backslash alpha\}\{\backslash alpha\; -\; 1\}\}\; \backslash right\backslash \}\backslash right)\; \backslash tag\; \{8\}$$

successive iterations and $K \cdot B$ oracle calls. Moreover, with probability at least $1 - \beta$ the iterates of ZO-clipped-SSTM remain in the ball $B_{2R}(x^*)$ , i.e., ${x^k}{k=0}^{K+1}, {y^k}{k=0}^{K}, {z^k}{k=0}^{K} \subseteq B{2R}(x^*)$ .

Here and below notation $\tilde{\mathcal{O}}$ means an upper bound on the growth rate hiding logarithms. The first term in bound (8) is optimal for the deterministic case for non-smooth problem (see [6]) and the second term in bound (8) is optimal in $\varepsilon$ for $\alpha \in (1,2]$ and zero-point oracle (see [27]).

We notice that increasing the batch size $B$ to reduce the number of successive iterations makes sense only as long as the first term of (8) lower than the second one, i.e. there exists optimal value of batchsize

$$B\le {\left(\frac{\sqrt{d}{M}_{2}R}{\epsilon }\right)}^{\frac{1}{\alpha -1}}\mathrm{.}B\; \backslash leq\; \backslash left(\backslash frac\; \{\backslash sqrt\; \{d\}\; M\; \_\; \{2\}\; R\}\{\backslash varepsilon\}\backslash right)\; ^\; \{\backslash frac\; \{1\}\{\backslash alpha\; -\; 1\}\}.$$

3.2 Strongly-convex case

Now we suppose that Assumption 1 is satisfied with $\mu >0$ . For this case we employ ZO-clippedSSTM with restarts technique. We will call this algorithm as R-ZO-clipped-SSTM (see Algorithm 2). At each round R-ZO-clipped-SSTM call ZO-clipped-SSTM with starting point $\hat{x}^t$ , which is the output from the previous round, and for $K_{t}$ iterations.

Algorithm 2 R-ZO-clipped-SSTM

Input: starting point $x^0$ , number of restarts $N$ , number of steps ${K_t}{t=1}^N$ , batchesizes ${B_t}{t=1}^N$ , stepsizes ${a_t}{t=1}^N$ , smoothing parameters ${\tau_t}{t=1}^N$ , clipping levels ${\lambda_k^1}{k=0}^{K_1-1}, \dots, {\lambda_k^N}{k=0}^{K_N-1}$

1: $\hat{x}^0 = x^0$
2: for $t = 1, \dots, N$ do
3: $\hat{x}^t = \mathsf{ZO}$ -clipped-SSTM $\left(\hat{x}^{t - 1},K_t,B_t,a_t,\tau_t,{\lambda_k^t}_{k = 0}^{K_t - 1}\right)$ .
4: end for

Output: $\hat{x}^N$

Theorem 2 (Convergence of R-ZO-clipped-SSTM) Let Assumptions 1, 2 hold with $\mu >0$ and $Q = \mathbb{R}^d$ . Let $| x^0 -x^| ^2\leq R^2$ , where $x^0$ is a starting point and $x^{}$ is the optimal solution to (1). Let also $N = \lceil \log_2(\mu R^2 /2\varepsilon)\rceil$ be the number of restarts. Let at each stage $t = 1,\dots,N$ of R-ZOclipped-SSTM, ZO-clipped-SSTM is run with batchsize $B_{t}$ , $\tau_t = \varepsilon_t / 4M_2$ , $L_{t} = M_{2}\sqrt{d} /\tau_{t}$ , $K_{t} = \widetilde{\Theta} (\max {\sqrt{L_{t}R_{t - 1}^{2} / \varepsilon_{t}},(\sigma R_{t - 1} / \varepsilon_{t})^{\frac{\alpha}{(\alpha - 1)}} / B_{t}})$ , $a_{t} = \widetilde{\Theta} (\max {1,\sigma K_{t}^{\frac{\alpha + 1}{\alpha}} / L_{t}R_{t}})$ and $\lambda_k^t = \widetilde{\Theta} (R / \alpha_{k + 1}^t)$ where $R_{t - 1} = 2^{-\frac{(t - 1)}{2}}R$ , $\varepsilon_t = \mu R_{t - 1}^2 /4$ , $\ln^{4NK_t / \beta}\geq 1$ , $\beta \in (0,1]$ . Then to guarantee $f(\hat{x}^N) - f(x^*)\leq \varepsilon$ with probability at least $1 - \beta$ , R-ZO-clipped-SSTM requires

$$\begin{array}{cccc}& \tilde{\mathcal{O}}(\max \{\sqrt{\frac{M_2^2\sqrt{d}}{\mu \epsilon }},{\left(\frac{d{M}_2^2}{\mu \epsilon }\right)}^{\frac{\alpha }{2(\alpha -1)}}\})& & \text{(9)}\end{array}\backslash widetilde\; \{\backslash mathcal\; \{O\}\}\; \backslash left(\backslash max\; \backslash left\backslash \{\backslash sqrt\; \{\backslash frac\; \{M\; \_\; \{2\}\; ^\; \{2\}\; \backslash sqrt\; \{d\}\}\{\backslash mu\; \backslash varepsilon\}\},\; \backslash left(\backslash frac\; \{d\; M\; \_\; \{2\}\; ^\; \{2\}\}\{\backslash mu\; \backslash varepsilon\}\backslash right)\; ^\; \{\backslash frac\; \{\backslash alpha\}\{2\; (\backslash alpha\; -\; 1)\}\}\; \backslash right\backslash \}\backslash right)\; \backslash tag\; \{9\}$$

oracle calls. Moreover, with probability at least $1 - \beta$ the iterates of R-ZO-clipped-SSTM at stage $t = 1,\ldots ,N$ stay in the ball $B_{2R_{t - 1}}(x^{*})$ .

The obtained complexity bound (see the proof in Appendix C.2) is the first optimal (up to logarithms) high-probability complexity bound under Assumption 2 for the smooth strongly convex problems. Indeed, the first term cannot be improved in view of the deterministic lower bound [27], and the second term is optimal [41].

4 Setting with Adversarial Noise

Often, black-box access of $f(x,\xi)$ are affected by some deterministic noise $\delta (x)$ . Thus, now we suppose that a zero-order oracle instead of objective values $f(x,\xi)$ returns its noisy approximation

$$\begin{array}{cccc}& f_{\delta }(x,\xi )f(x,\xi )+\delta (x)\mathrm{.}& & \text{(10)}\end{array}f\; \_\; \{\backslash delta\}\; (x,\; \backslash xi)\; \backslash stackrel\; \{\backslash text\; \{d\; e\; f\}\}\; \{=\}\; f\; (x,\; \backslash xi)\; +\; \backslash delta\; (x).\; \backslash tag\; \{10\}$$

This noise $\delta (x)$ can be interpreted, e.g., as a computer discretization when calculating $f(x,\xi)$ . For our analysis, we need this noise to be uniformly bounded. Recently, noisy «black-box» optimization with bounded noise has been actively studied [11, 25]. The authors of [11] consider deterministic adversarial noise, while in [25] stochastic adversarial noise was explored.

Assumption 3 (Boundedness of noise) There exists a constant $\Delta >0$ such that $|\delta (x)|\leq \Delta$ for all $x\in Q$

This is a standard assumption often used in the literature (e.g., [17]). Moreover, in some applications [4] the bigger the noise the cheaper the zero-order oracle. Thus, it is important to understand the maximum allowable level of adversarial noise at which the convergence of the gradient-free algorithm is unaffected.

4.1 Non-smooth setting

In noisy setup, gradient estimate from (4) is replaced by

$$\begin{array}{cccc}& g(x,\xi ,\mathbf{e})=\frac{d}{2\tau }(f_{\delta }(x+\tau \mathbf{e},\xi )-f_{\delta }(x-\tau \mathbf{e},\xi ))\mathbf{e}\mathrm{.}& & \text{(11)}\end{array}g\; (x,\; \backslash xi\; ,\; \backslash mathbf\; \{e\})\; =\; \backslash frac\; \{d\}\{2\; \backslash tau\}\; \backslash left(f\; \_\; \{\backslash delta\}\; (x\; +\; \backslash tau\; \backslash mathbf\; \{e\},\; \backslash xi)\; -\; f\; \_\; \{\backslash delta\}\; (x\; -\; \backslash tau\; \backslash mathbf\; \{e\},\; \backslash xi)\backslash right)\; \backslash mathbf\; \{e\}.\; \backslash tag\; \{11\}$$

Then (6) from Lemma 3 will have an extra factor driven by noise (see Lemma 2.3 from [22])

$${\mathbb{E}}_{\{\xi \},\{\mathbf{e}\}}\left[{\parallel g^B(x,\{\xi \},\{\mathbf{e}\})-{\mathbb{E}}_{\{\xi \},\{\mathbf{e}\}}[g^B(x,\{\xi \},\{\mathbf{e}\})]\parallel }_2^{\alpha }\right]\le \frac{2}{B^{\alpha -1}}{(\frac{\sqrt{d}{M}_2}{2^{\frac{1}{4}}}+\frac{d\mathrm{\Delta }}{\tau })}^{\alpha }\mathrm{.}\backslash mathbb\; \{E\}\; \_\; \{\backslash \{\backslash xi\; \backslash \},\; \backslash \{\backslash mathbf\; \{e\}\; \backslash \}\}\; \backslash left[\; \backslash left\backslash |\; g\; ^\; \{B\}\; (x,\; \backslash \{\backslash xi\; \backslash \},\; \backslash \{\backslash mathbf\; \{e\}\; \backslash \})\; -\; \backslash mathbb\; \{E\}\; \_\; \{\backslash \{\backslash xi\; \backslash \},\; \backslash \{\backslash mathbf\; \{e\}\; \backslash \}\}\; [\; g\; ^\; \{B\}\; (x,\; \backslash \{\backslash xi\; \backslash \},\; \backslash \{\backslash mathbf\; \{e\}\; \backslash \})\; ]\; \backslash right\backslash |\; \_\; \{2\}\; ^\; \{\backslash alpha\}\; \backslash right]\; \backslash leq\; \backslash frac\; \{2\}\{B\; ^\; \{\backslash alpha\; -\; 1\}\}\; \backslash left(\backslash frac\; \{\backslash sqrt\; \{d\}\; M\; \_\; \{2\}\}\{2\; ^\; \{\backslash frac\; \{1\}\{4\}\}\}\; +\; \backslash frac\; \{d\; \backslash Delta\}\{\backslash tau\}\backslash right)\; ^\; \{\backslash alpha\}.$$

To guarantee the same convergence of the algorithm as in Theorem 1 (see (8)) for the adversarial deterministic noise case, the variance term must dominate the noise term, i.e. $d\Delta \tau^{-1} \lesssim \sqrt{d} M_2$ . Note that if the term with noise dominates the term with variance, it does not mean that the gradient-free algorithm will not converge. In contrast, algorithm will still converge, only slower (oracle complexity will be $\sim \varepsilon^{-2}$ times higher). Thus, if we were considering the zero-order oracle concept with adversarial stochastic noise, it would be enough to express the noise level $\Delta$ , and substitute the value of the smoothing parameter $\tau$ to obtain the maximum allowable noise level. However, since we are considering the concept of adversarial noise in a deterministic setting, following previous work [11, 1] we can say that adversarial noise accumulates not only in the variance, but also in the bias:

$${\mathbb{E}}_{\xi ,\mathbf{e}}⟨[g(x,\xi ,\mathbf{e})]-\mathrm{\nabla }{\widehat{f}}_{\tau }(x),r⟩\lesssim \sqrt{d}\mathrm{\Delta }\mathrm{\parallel }r{\mathrm{\parallel }}_2{\tau }^{-1},\text{f o r a l l}r\in {\mathbb{R}}^d\mathrm{.}\backslash mathbb\; \{E\}\; \_\; \{\backslash xi\; ,\; \backslash mathbf\; \{e\}\}\; \backslash langle\; [\; g\; (x,\; \backslash xi\; ,\; \backslash mathbf\; \{e\})\; ]\; -\; \backslash nabla\; \backslash hat\; \{f\}\; \_\; \{\backslash tau\}\; (x),\; r\; \backslash rangle\; \backslash lesssim\; \backslash sqrt\; \{d\}\; \backslash Delta\; \backslash |\; r\; \backslash |\; \_\; \{2\}\; \backslash tau^\; \{-\; 1\},\; \backslash quad\; \backslash text\; \{f\; o\; r\; a\; l\; l\}\; r\; \backslash in\; \backslash mathbb\; \{R\}\; ^\; \{d\}.$$

This bias can be controlled by the noise level $\Delta$ , i.e., in order to achieve the $\varepsilon$ -accuracy algorithm considered in this paper, the noise condition must be satisfied:

$$\begin{array}{cccc}& \mathrm{\Delta }\lesssim \frac{\tau \epsilon }{R\sqrt{d}}\mathrm{.}& & \text{(12)}\end{array}\backslash Delta\; \backslash lesssim\; \backslash frac\; \{\backslash tau\; \backslash varepsilon\}\{R\; \backslash sqrt\; \{d\}\}.\; \backslash tag\; \{12\}$$

As we can see, we have a more restrictive condition on the noise level in the bias (12) than in the variance $(\Delta \lesssim \gamma^{M_2} / \sqrt{d})$ . Thus, the maximum allowable level of adversarial deterministic noise, which guarantees the same convergence of ZO-clipped-SSTM, as in Theorem 1 (see (8)) is as follows

$$\mathrm{\Delta }\lesssim \frac{{\epsilon }^2}{R{M}_2\sqrt{d}},\backslash Delta\; \backslash lesssim\; \backslash frac\; \{\backslash varepsilon^\; \{2\}\}\{R\; M\; \_\; \{2\}\; \backslash sqrt\; \{d\}\},$$

where $\tau = \varepsilon /2M_2$ the smoothing parameter from Lemma 2.

Remark 1 ( $\mu$ -strongly convex case) Let us assume that $f(x)$ is also $\mu$ -strongly convex (see Assumption 1). Then, following the works [11, 22], we can conclude that the R-ZO-clipped-SSTM has the same oracle and iteration complexity as in Theorem 2 at the following maximum allowable level of adversarial noise: $\Delta \lesssim \mu^{1/2} \varepsilon^{3/2} / \sqrt{d} M_2$ .

4.2 Smooth setting

Now we examine the maximum allowable level of noise at which we can solve optimization problem (1) with $\varepsilon$ -precision under the following additional assumption

Assumption 4 (Smoothness) The function $f$ is $L$ -smooth, i.e., it is differentiable on $Q$ and for all $x, y \in Q$ with $L > 0$ :

$$\mathrm{\parallel }\mathrm{\nabla }f(y)-\mathrm{\nabla }f(x){\mathrm{\parallel }}_2\le L\mathrm{\parallel }y-x{\mathrm{\parallel }}_2\mathrm{.}\backslash |\; \backslash nabla\; f\; (y)\; -\; \backslash nabla\; f\; (x)\; \backslash |\; \_\; \{2\}\; \backslash leq\; L\; \backslash |\; y\; -\; x\; \backslash |\; \_\; \{2\}.$$

If Assumption 4 holds, then Lemma 2 can be rewritten as

$$\underset{x\in Q}{\sup }\mathrm{\mid }{\widehat{f}}_{\tau }(x)-f(x)\mathrm{\mid }\le \frac{L{\tau }^2}{2}\mathrm{.}\backslash sup\; \_\; \{x\; \backslash in\; Q\}\; |\; \backslash hat\; \{f\}\; \_\; \{\backslash tau\}\; (x)\; -\; f\; (x)\; |\; \backslash leq\; \backslash frac\; \{L\; \backslash tau^\; \{2\}\}\{2\}.$$

Thus, we can now present the convergence results of the gradient-free algorithm in the smooth setting. Specifically, if the Assumptions 2-4 are satisfied, then $\widetilde{ZO}$ -clipped-SSTM converges to $\varepsilon$ -accuracy after $K = \widetilde{\mathcal{O}}\left(\sqrt{LR^2\varepsilon^{-1}}\right)$ iterations with probability at least $1 - \beta$ . It is easy to see that the iteration complexity improves in the smooth setting (since the Lipschitz gradient constant $L$ already exists,

i.e., no smoothing is needed), but oracle complexity remained unchanged (since we are still using the gradient approximation via $l_{2}$ randomization (11) instead of the true gradient $\nabla f(x)$ ), consistent with the already optimal estimate on oracle calls: $\widetilde{\mathcal{O}}\left(\left(\sqrt{d} M_2R\varepsilon^{-1}\right)^{\frac{\alpha}{\alpha - 1}}\right)$ . And to obtain the maximum allowable level of adversarial noise $\Delta$ in the smooth setting, which guarantees such convergence, it is sufficient to substitute the smoothing parameter $\tau = \sqrt{\varepsilon / L}$ in the inequality (12):

$$\mathrm{\Delta }\lesssim \frac{{\epsilon }^{3\mathrm{/}2}}{R\sqrt{dL}}\mathrm{.}\backslash Delta\; \backslash lesssim\; \backslash frac\; \{\backslash varepsilon^\; \{3\; /\; 2\}\}\{R\; \backslash sqrt\; \{d\; L\}\}.$$

Thus, we can conclude that smooth setting improves iteration complexity and the maximum allowable noise level for the gradient-free algorithm, but the oracle complexity remains unchanged.

Remark 2 ( $\mu$ -strongly convex case) Suppose that $f(x)$ is also $\mu$ -strongly convex (see Assumption 1). Then we can conclude that R-ZO-clipped-SSTM has the oracle and iteration complexity just mentioned above at the following maximum allowable level of adversarial noise: $\Delta \lesssim \mu^{1/2} \varepsilon / \sqrt{dL}$ .

Remark 3 (Upper bounds optimality) The upper bounds on maximum allowable level of adversarial noise obtained in this section in both non-smooth and smooth settings are optimal in terms of dependencies on $\varepsilon$ and $d$ according to the works [32, 34].

Remark 4 (Better oracle complexity) In the aforementioned approach in the case when $f(x, \xi)$ has Lipschitz gradient in $x$ (for all $\xi$ ) one can improve oracle complexity from $\widetilde{\mathcal{O}}\left(\left(\sqrt{d}M_2R\varepsilon^{-1}\right)^{\frac{\alpha}{\alpha - 1}}\right)$ to $\widetilde{\mathcal{O}}\left(d(M_2R\varepsilon^{-1})^{\frac{\alpha}{\alpha - 1}}\right)$ . This can be done by using component-wise finite-difference stochastic gradient approximation [15]. Iteration complexity remains $\widetilde{\mathcal{O}}\left(\sqrt{LR^2\varepsilon^{-1}}\right)$ . The same can be done for $\mu$ -strongly convex case: from $\widetilde{\mathcal{O}}\left((dM_2^2(\mu\varepsilon)^{-1})^{\frac{\alpha}{2(\alpha - 1)}}\right)$ to $\widetilde{\mathcal{O}}\left(d(M_2^2(\mu\varepsilon)^{-1})^{\frac{\alpha}{2(\alpha - 1)}}\right)$ .

5 Details of the proof

The proof is built upon a combination of two techniques. The first one is the Smoothing technique from [17] that is used to develop a gradient-free method for convex non-smooth problems based on full-gradient methods. The second technique is the Accelerated Clipping technique that has been recently developed for smooth problems with the noise having infinite variance and first-order oracle [35]. The authors of [35] propose clipped-SSTM method that we develop in our paper. We modify clipped-SSTM by introducing batching into it. Note that due to the infinite variance, such a modification is interesting in itself. Then we run batched clipped-SSTM with gradient estimations of function $f$ obtained via Smoothing technique and two-point zeroth-order oracle. To do this, we need to estimate the variance of the clipped version of the batched gradient estimation.

In more detail, we replace the initial problem of minimizing $f$ by minimizing its smoothed approximation $\hat{f}{\tau}$ , see Lemma 2 In order to use estimated gradient of $\hat{f}{\tau}$ defined in (4) or (5), we make sure that it has bounded $\alpha$ -th moment. For these purposes we proof Lemma 3. First part shows boundness of unbatched estimated gradient $g$ defined in (4). It follows from measure concentration phenomenon for the Euclidean sphere for $\frac{M_2}{\tau}$ -Lipschitz function $f(x + \mathbf{e}\tau)$ . According to this phenomenon probability of the functions deviation from its math expectation becomes exponentially small and $\alpha$ -th moment of this deviation becomes bounded. Furthermore, the second part of Lemma 3 shows that batching helps to bound $\alpha$ -th moment of batched gradient $g^B$ defined in (5) even more. Also batching allows parallel calculations reducing number of necessary iteration with the same number of oracle calls. All this is possible thanks to the result, interesting in itself, presented in the following Lemma.

Lemma 4 Let $X_{1},\ldots ,X_{B}$ be $d$ -dimensional martingale difference sequence (i.e. $\mathbb{E}[X_i|X_{i - 1},\dots ,X_1] = 0$ for $1 < i\leq B)$ satisfying for $1\leq \alpha \leq 2$

$$\mathbb{E}[\mathrm{\parallel }{X}_i{\mathrm{\parallel }}^{\alpha }\mid X_{i-1},\dots ,X_1]\le {\sigma }^{\alpha }\mathrm{.}\backslash mathbb\; \{E\}\; \backslash left[\; \backslash |\; X\; \_\; \{i\}\; \backslash |\; ^\; \{\backslash alpha\}\; \backslash mid\; X\; \_\; \{i\; -\; 1\},\; \backslash dots\; ,\; X\; \_\; \{1\}\; \backslash right]\; \backslash leq\; \backslash sigma^\; \{\backslash alpha\}.$$

Then we have

$$\mathbb{E}\left[{\parallel \frac{1}{B}\sum _{i=1}^B{X}_i\parallel }_2^{\alpha }\right]\le \frac{2{\sigma }^{\alpha }}{B^{\alpha -1}}\mathrm{.}\backslash mathbb\; \{E\}\; \backslash left[\; \backslash left\backslash |\; \backslash frac\; \{1\}\{B\}\; \backslash sum\_\; \{i\; =\; 1\}\; ^\; \{B\}\; X\; \_\; \{i\}\; \backslash right\backslash |\; \_\; \{2\}\; ^\; \{\backslash alpha\}\; \backslash right]\; \backslash leq\; \backslash frac\; \{2\; \backslash sigma^\; \{\backslash alpha\}\}\{B\; ^\; \{\backslash alpha\; -\; 1\}\}.$$

Next, we use the clipped-SSTM for function $\hat{f}_{\tau}$ with heavy-tailed gradient estimates. This algorithm was initially proposed for smooth functions in the work [35]. The scheme for proving convergence with high probability is also taken from it, the only difference is additional randomization caused by smoothing scheme.

6 Numerical experiments

We tested ZO-clipped-SSTM on the following problem

$$\underset{x\in {\mathbb{R}}^d}{\min }\mathrm{\parallel }Ax-b{\mathrm{\parallel }}_2+⟨\xi ,x⟩,\backslash min\; \_\; \{x\; \backslash in\; \backslash mathbb\; \{R\}\; ^\; \{d\}\}\; \backslash |\; A\; x\; -\; b\; \backslash |\; \_\; \{2\}\; +\; \backslash langle\; \backslash xi\; ,\; x\; \backslash rangle\; ,$$

where $\xi$ is a random vector with independent components sampled from the symmetric Levy $\alpha$ -stable distribution with $\alpha = 3/2$ , $A \in \mathbb{R}^{m \times d}$ , $b \in \mathbb{R}^m$ (we used $d = 16$ and $m = 500$ ). For this problem, Assumption 1 holds with $\mu = 0$ and Assumption 2 holds with $\alpha = 3/2$ and $M_2(\xi) = | A |_2 + | \xi |_2$ .

We compared ZO-clipped-SSTM, proposed in this paper, with ZOSGD and ZO-SSTM. For these algorithms, we gridsearched batchsize $B:5,10,50,100,500$ and stepsize $\gamma :1e - 3,1e - 4,1e - 5,1e - 6$ . The best convergence was achieved with the following parameters:

ZO-clipped-SSTM: $\gamma = 1e - 3,$ $B = 10,\lambda = 0.01$

• ZO-SSTM: $\gamma = 1e - 5, B = 500$
• ZO-SGD: $\gamma = 1e - 4, B = 100$ , $\omega = 0.9$ , where $\omega$ is a heavy-ball momentum parameter.

Figure 1: Convergence of ZO-clipped-SSTM, ZO-SGD and ZO-clipped-SSTM in terms of a gap function w.r.t. the number of consumed samples.

The code is written on Pythone and is publicly available at https://github.com/ClippedStochasticMethods/Z0-clipped-SSTM. Figure 1 presents the comparison of convergences averaged over 15 launches with different noise. In contrast to ZO-clipped-SSTM, the last two methods are unclipped and therefore failed to converge under haivy-tailed noise.

7 Conclusion and future directions

In this paper, we propose a first gradient-free algorithm ZO-clipped-SSTM and to solve problem (1) under heavy-tailed noise assumption. By using the restart technique we extend this algorithm for strongly convex objective, we refer to this algorithm as R-ZO-clipped-SSTM. The proposed algorithms are optimal with respect to oracle complexity (in terms of the dependence on $\varepsilon$ ), iteration complexity and the maximal level of noise (possibly adversarial). The algorithms can be adapted to composite and distributed minimization problems, saddle-point problems, and variational inequalities. Despite the fact that the algorithms utilize the two-point feedback, they can be modified to the one-point feedback. We leave it for future work.

Moreover, we provide theoretical basis to demonstrate benefits of batching technique in case of heavy-tailed stochastic noise and apply it to methods from this paper. Thanks to this basis, it is possible to use batching in other methods with heavy-tiled noise, e.g. first-order methods presented in [35, 29].

8 Acknowledgment

The work of Alexander Gasnikov, Aleksandr Lobanov, and Darina Dvinskikh was supported by a grant for research centers in the field of artificial intelligence, provided by the Analytical Center for the Government of the Russian Federation in accordance with the subsidy agreement (agreement identifier 000000D730321P5Q0002) and the agreement with the Ivannikov Institute for System Programming of dated November 2, 2021 No. 70-2021-00142.

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A Batching with unbounded variance

To prove batching Lemma 4 we generalize Lemma 4.2 from [7] not only for i.i.d. random variables with zero mean but for martingale difference sequence.

Lemma 5 Let $g(x) = \text{sign}(x)|x|^{\alpha - 1} = \nabla \left(\frac{|x|^{\alpha}}{\alpha}\right)$ for $1 < \alpha \leq 2$ . Then for any $h \geq 0$

$$\underset{h}{\max }g(x+h)-g(x)=2^{2-\alpha }{h}^{\alpha -1}=2^{2-\alpha }g(h)\mathrm{.}\backslash max\; \_\; \{h\}\; g\; (x\; +\; h)\; -\; g\; (x)\; =\; 2\; ^\; \{2\; -\; \backslash alpha\}\; h\; ^\; \{\backslash alpha\; -\; 1\}\; =\; 2\; ^\; \{2\; -\; \backslash alpha\}\; g\; (h).$$

Proof. Consider $l(x) = g(x + h) - g(x)$ . We see that $l$ is differentiable everywhere except $x = 0$ and $x = -h$ . As long as $x \neq 0, -h$ , we have

$$l^{\mathrm{\prime }}(x)=g^{\mathrm{\prime }}(x+h)-g^{\mathrm{\prime }}(x)=(\alpha -1)(\mathrm{\mid }x+h{\mathrm{\mid }}^{\alpha -2}-\mathrm{\mid }x{\mathrm{\mid }}^{\alpha -2})\mathrm{.}l\; ^\; \{\backslash prime\}\; (x)\; =\; g\; ^\; \{\backslash prime\}\; (x\; +\; h)\; -\; g\; ^\; \{\backslash prime\}\; (x)\; =\; (\backslash alpha\; -\; 1)\; \backslash left(|\; x\; +\; h\; |\; ^\; \{\backslash alpha\; -\; 2\}\; -\; |\; x\; |\; ^\; \{\backslash alpha\; -\; 2\}\backslash right).$$

Since, we have $\alpha > 1$ , $x = -\frac{h}{2}$ is local maxima for $l(x)$ . Furthermore, note that $l'(x) \geq 0$ for $x \in (-\infty, -\frac{h}{2}) \setminus {-h}$ and $l'(x) \leq 0$ for $x \in \left(-\frac{h}{2}, \infty\right) \setminus {0}$ . Therefore, $\frac{-h}{2}$ is global maxima.

Lemma 6 Let $x_{1}, \ldots, x_{B}$ be one-dimensional martingale difference sequence, i.e. $\mathbb{E}[x_i | x_{i-1}, \ldots, x_1] = 0$ for $1 < i \leq B$ , satisfying for $1 \leq \alpha \leq 2$

$$\mathbb{E}[\mathrm{\mid }{x}_i{\mathrm{\mid }}^{\alpha }\mathrm{\mid }{x}_{i-1},\dots ,x_1]\le {\sigma }^{\alpha }\mathrm{.}\backslash mathbb\; \{E\}\; [\; |\; x\; \_\; \{i\}\; |\; ^\; \{\backslash alpha\}\; |\; x\; \_\; \{i\; -\; 1\},\; \backslash dots\; ,\; x\; \_\; \{1\}\; ]\; \backslash leq\; \backslash sigma^\; \{\backslash alpha\}.$$

We have:

$$\mathbb{E}\left[{\mid \frac{1}{B}\sum _{i=1}^B{x}_i\mid }^{\alpha }\right]\le \frac{2{\sigma }^{\alpha }}{B^{\alpha -1}}\mathrm{.}\backslash mathbb\; \{E\}\; \backslash left[\; \backslash left|\; \backslash frac\; \{1\}\{B\}\; \backslash sum\_\; \{i\; =\; 1\}\; ^\; \{B\}\; x\; \_\; \{i\}\; \backslash right|\; ^\; \{\backslash alpha\}\; \backslash right]\; \backslash leq\; \backslash frac\; \{2\; \backslash sigma^\; \{\backslash alpha\}\}\{B\; ^\; \{\backslash alpha\; -\; 1\}\}.$$

Proof. Everywhere below we will use the following notations.

For $\alpha = 1$ proof follows from triangle inequality for $|\cdot|$ . When $\alpha > 1$ , we start by defining

$$S_i=\sum _{j=1}^i{x}_j,S_0=0,f(x)=\mathrm{\mid }x{\mathrm{\mid }}^{\alpha }\mathrm{.}S\; \_\; \{i\}\; =\; \backslash sum\_\; \{j\; =\; 1\}\; ^\; \{i\}\; x\; \_\; \{j\},\; \backslash qquad\; S\; \_\; \{0\}\; =\; 0,\; \backslash qquad\; f\; (x)\; =\; |\; x\; |\; ^\; \{\backslash alpha\}.$$

Then we can calculate desired expectation as

$$\begin{array}{cccc}& \begin{array}{c}\mathbb{E}[f(S_B)]=\mathbb{E}[{\sum }_{i=1}^{B}f(S_i)-f(S_{i-1})]={\sum }_{i=1}^B\mathbb{E}[f(S_i)-f(S_{i-1})]\\ ={\sum }_{i=1}^B\mathbb{E}[{\int }_{S_{i-1}}^{S_i}{f}^{\mathrm{\prime }}(x)dx]\\ ={\sum }_{i=1}^B\mathbb{E}[x_i{f}^{\mathrm{\prime }}\left(S_{i-1}\right)+{\int }_{S_{i-1}}^{S_i}{f}^{\mathrm{\prime }}(x)-f^{\mathrm{\prime }}\left(S_{i-1}\right)dx]\mathrm{.}\end{array}& & \text{(13)}\end{array}\backslash begin\{array\}\{l\}\; \backslash mathbb\; \{E\}\; [\; f\; (S\; \_\; \{B\})\; ]\; =\; \backslash mathbb\; \{E\}\; \backslash left[\; \backslash sum\_\; \{i\; =\; 1\}\; ^\; \{B\}\; f\; (S\; \_\; \{i\})\; -\; f\; (S\; \_\; \{i\; -\; 1\})\; \backslash right]\; =\; \backslash sum\_\; \{i\; =\; 1\}\; ^\; \{B\}\; \backslash mathbb\; \{E\}\; [\; f\; (S\; \_\; \{i\})\; -\; f\; (S\; \_\; \{i\; -\; 1\})\; ]\; \backslash \backslash \; =\; \backslash sum\_\; \{i\; =\; 1\}\; ^\; \{B\}\; \backslash mathbb\; \{E\}\; \backslash left[\; \backslash int\_\; \{S\; \_\; \{i\; -\; 1\}\}\; ^\; \{S\; \_\; \{i\}\}\; f\; ^\; \{\backslash prime\}\; (x)\; d\; x\; \backslash right]\; \backslash \backslash \; =\; \backslash sum\_\; \{i\; =\; 1\}\; ^\; \{B\}\; \backslash mathbb\; \{E\}\; \backslash left[\; x\; \_\; \{i\}\; f\; ^\; \{\backslash prime\}\; \backslash left(S\; \_\; \{i\; -\; 1\}\backslash right)\; +\; \backslash int\_\; \{S\; \_\; \{i\; -\; 1\}\}\; ^\; \{S\; \_\; \{i\}\}\; f\; ^\; \{\backslash prime\}\; (x)\; -\; f\; ^\; \{\backslash prime\}\; \backslash left(S\; \_\; \{i\; -\; 1\}\backslash right)\; d\; x\; \backslash right].\; \backslash tag\; \{13\}\; \backslash \backslash \; \backslash end\{array\}$$

While ${x_i}$ is martingale difference sequence then $\mathbb{E}[x_if'(S_{i - 1})] = \mathbb{E}{< i}[\mathbb{E}{| < i}[x_if'(S_{i - 1})]] = 0$ From (13) and Lemma 5 $(g(x) = f^{\prime}(x) / \alpha)$ we obtain

Now we are ready to prove batching lemma for random variables with infinite variance.

Lemma 7 Let $X_{1},\ldots ,X_{B}$ be $d$ -dimensional martingale difference sequence, i.e., $\mathbb{E}[X_i|X_{i - 1},\dots ,X_1] = 0$ for $1 < i\leq B$ satisfying for $1\leq \alpha \leq 2$

$$\mathbb{E}[\mathrm{\parallel }{X}_i{\mathrm{\parallel }}_2^{\alpha }\mathrm{\mid }{X}_{i-1},\dots ,X_1]\le {\sigma }^{\alpha }\mathrm{.}\backslash mathbb\; \{E\}\; \backslash left[\; \backslash |\; X\; \_\; \{i\}\; \backslash |\; \_\; \{2\}\; ^\; \{\backslash alpha\}\; |\; X\; \_\; \{i\; -\; 1\},\; \backslash dots\; ,\; X\; \_\; \{1\}\; \backslash right]\; \backslash leq\; \backslash sigma^\; \{\backslash alpha\}.$$

We have

$$\mathbb{E}\left[{\parallel \frac{1}{B}\sum _{i=1}^B{X}_i\parallel }_2^{\alpha }\right]\le \frac{2{\sigma }^{\alpha }}{B^{\alpha -1}}\mathrm{.}\backslash mathbb\; \{E\}\; \backslash left[\; \backslash left\backslash |\; \backslash frac\; \{1\}\{B\}\; \backslash sum\_\; \{i\; =\; 1\}\; ^\; \{B\}\; X\; \_\; \{i\}\; \backslash right\backslash |\; \_\; \{2\}\; ^\; \{\backslash alpha\}\; \backslash right]\; \backslash leq\; \backslash frac\; \{2\; \backslash sigma^\; \{\backslash alpha\}\}\{B\; ^\; \{\backslash alpha\; -\; 1\}\}.$$

Proof.

Let $g \sim \mathcal{N}(0, I)$ and $y_i \stackrel{\text{def}}{=} X_i^\top g$ . Firstly, we prove that $\mathbb{E}[y_i^\alpha] \leq \mathbb{E}[|X_i|^{\alpha}]$ . Indeed, using conditional expectation we get

$$\begin{array}{c}\mathbb{E}\left[{\mid y_i\mid }^{\alpha }\right]=\mathbb{E}\left[{\left(\mid X_i^{\mathrm{\top }}g\mid \right)}^{\alpha }\right]={\mathbb{E}}_{X_i}\left[{\mathbb{E}}_{g\mathrm{\mid }{X}_i}\left[{\left(\mid X_i^{\mathrm{\top }}g)\right)}^{\alpha }\right]\right]\\ ={\mathbb{E}}_{X_i}\left[{\mathbb{E}}_{g\mathrm{\mid }{X}_i}\left[{\left({\left(\mathrm{\mid }{X}_i^{\mathrm{\top }}g\mathrm{\mid }\right)}^2\right)}^{\alpha \mathrm{/}2}\right]\right]\end{array}\backslash begin\{array\}\{l\}\; \backslash mathbb\; \{E\}\; \backslash left[\; \backslash left|\; y\; \_\; \{i\}\; \backslash right|\; ^\; \{\backslash alpha\}\; \backslash right]\; =\; \backslash mathbb\; \{E\}\; \backslash left[\; \backslash left(\backslash left|\; X\; \_\; \{i\}\; ^\; \{\backslash top\}\; g\; \backslash right|\backslash right)\; ^\; \{\backslash alpha\}\; \backslash right]\; =\; \backslash mathbb\; \{E\}\; \_\; \{X\; \_\; \{i\}\}\; \backslash left[\; \backslash mathbb\; \{E\}\; \_\; \{g\; |\; X\; \_\; \{i\}\}\; \backslash left[\; \backslash left(\backslash left|\; X\; \_\; \{i\}\; ^\; \{\backslash top\}\; g\backslash right)\backslash right)\; ^\; \{\backslash alpha\}\; \backslash right]\; \backslash right]\; \backslash \backslash \; =\; \backslash mathbb\; \{E\}\; \_\; \{X\; \_\; \{i\}\}\; \backslash left[\; \backslash mathbb\; \{E\}\; \_\; \{g\; |\; X\; \_\; \{i\}\}\; \backslash left[\; \backslash left(\backslash left(|\; X\; \_\; \{i\}\; ^\; \{\backslash top\}\; g\; |\backslash right)\; ^\; \{2\}\backslash right)\; ^\; \{\backslash alpha\; /\; 2\}\; \backslash right]\; \backslash right]\; \backslash \backslash \; \backslash end\{array\}$$

$$\begin{array}{cccc}& \begin{array}{c}\begin{array}{c}\text{J e s e n i n q}\\ \le \end{array}{\mathbb{E}}_{X_i}\left[{\left({\mathbb{E}}_{g\mathrm{\mid }{X}_i}[(X_i^{\mathrm{\top }}g{)}^2]\right)}^{\alpha \mathrm{/}2}\right]\\ ={\mathbb{E}}_{X_i}\left[{\left(\mathrm{\parallel }{X}_i{\mathrm{\parallel }}_2^2\right)}^{\alpha \mathrm{/}2}\right]=\mathbb{E}[\mathrm{\parallel }{X}_i{\mathrm{\parallel }}_2^{\alpha }]\mathrm{.}\end{array}& & \text{(15)}\end{array}\backslash begin\{array\}\{l\}\; \backslash begin\{array\}\{l\}\; \backslash text\; \{J\; e\; s\; e\; n\; i\; n\; q\}\; \backslash \backslash \; \backslash leq\; \backslash end\{array\}\; \backslash quad\; \backslash mathbb\; \{E\}\; \_\; \{X\; \_\; \{i\}\}\; \backslash left[\; \backslash left(\backslash mathbb\; \{E\}\; \_\; \{g\; |\; X\; \_\; \{i\}\}\; \backslash left[\; (X\; \_\; \{i\}\; ^\; \{\backslash top\}\; g)\; ^\; \{2\}\; \backslash right]\backslash right)\; ^\; \{\backslash alpha\; /\; 2\}\; \backslash right]\; \backslash \backslash \; =\; \backslash mathbb\; \{E\}\; \_\; \{X\; \_\; \{i\}\}\; \backslash left[\; \backslash left(\backslash |\; X\; \_\; \{i\}\; \backslash |\; \_\; \{2\}\; ^\; \{2\}\backslash right)\; ^\; \{\backslash alpha\; /\; 2\}\; \backslash right]\; =\; \backslash mathbb\; \{E\}\; [\; \backslash |\; X\; \_\; \{i\}\; \backslash |\; \_\; \{2\}\; ^\; \{\backslash alpha\}\; ].\; \backslash tag\; \{15\}\; \backslash \backslash \; \backslash end\{array\}$$

Next, considering $X_{i}^{\top}g\sim \mathcal{N}(0,| X|^{2})$ and, thus, $\mathbb{E}_g|X_i^\top g| = | X|$ , we bound desired expectation as

$$\begin{array}{cccc}& {\mathbb{E}}_X\left[{\parallel \sum _{i=1}^B{X}_i\parallel }_2^{\alpha }\right]={\mathbb{E}}_X\left[{\left({\mathbb{E}}_g\mid \sum _{i=1}^B{X}_i^{\mathrm{\top }}g\mid \right)}^{\alpha }\right]& & \text{(16)}\end{array}\backslash mathbb\; \{E\}\; \_\; \{X\}\; \backslash left[\; \backslash left\backslash |\; \backslash sum\_\; \{i\; =\; 1\}\; ^\; \{B\}\; X\; \_\; \{i\}\; \backslash right\backslash |\; \_\; \{2\}\; ^\; \{\backslash alpha\}\; \backslash right]\; \backslash quad\; =\; \backslash quad\; \backslash mathbb\; \{E\}\; \_\; \{X\}\; \backslash left[\; \backslash left(\backslash mathbb\; \{E\}\; \_\; \{g\}\; \backslash left|\; \backslash sum\_\; \{i\; =\; 1\}\; ^\; \{B\}\; X\; \_\; \{i\}\; ^\; \{\backslash top\}\; g\; \backslash right|\backslash right)\; ^\; \{\backslash alpha\}\; \backslash right]\; \backslash tag\; \{16\}$$

$$\begin{array}{cccc}& {\le }^{\text{J e s s e n ’ s i n q}}{\mathbb{E}}_{X,g}\left[{\mid \sum _{i=1}^B{X}_i^{\mathrm{\top }}g\mid }^{\alpha }\right]={\mathbb{E}}_{X,g}\left[{\mid \sum _{i=1}^B{y}_i\mid }^{\alpha }\right]\mathrm{.}& & \text{(17)}\end{array}\backslash leq^\; \{\backslash text\; \{J\; e\; s\; s\; e\; n\; \text{'}\; s\; i\; n\; q\}\}\; \backslash mathbb\; \{E\}\; \_\; \{X,\; g\}\; \backslash left[\; \backslash left|\; \backslash sum\_\; \{i\; =\; 1\}\; ^\; \{B\}\; X\; \_\; \{i\}\; ^\; \{\backslash top\}\; g\; \backslash right|\; ^\; \{\backslash alpha\}\; \backslash right]\; =\; \backslash mathbb\; \{E\}\; \_\; \{X,\; g\}\; \backslash left[\; \backslash left|\; \backslash sum\_\; \{i\; =\; 1\}\; ^\; \{B\}\; y\; \_\; \{i\}\; \backslash right|\; ^\; \{\backslash alpha\}\; \backslash right].\; \backslash tag\; \{17\}$$

Finally, we apply Lemma 6 for $y_{i}$ sequence with bounded $\alpha$ -th moment from (15) and get

$${\mathbb{E}}_X{\parallel \mid \sum _{i=1}^B{X}_i\mid \parallel }_2^{\alpha }]\le {\mathbb{E}}_{X,g}\left[{\mid \sum _{i=1}^B{y}_i\mid }^{\alpha }\right]\le 2{\sigma }^{\alpha }B\mathrm{.}\backslash mathbb\; \{E\}\; \_\; \{X\}\; \backslash left\backslash |\; \backslash left|\; \backslash sum\_\; \{i\; =\; 1\}\; ^\; \{B\}\; X\; \_\; \{i\}\; \backslash right|\; \backslash right\backslash |\; \_\; \{2\}\; ^\; \{\backslash alpha\}\; \backslash biggr\; ]\; \backslash leq\; \backslash mathbb\; \{E\}\; \_\; \{X,\; g\}\; \backslash left[\; \backslash left|\; \backslash sum\_\; \{i\; =\; 1\}\; ^\; \{B\}\; y\; \_\; \{i\}\; \backslash right|\; ^\; \{\backslash alpha\}\; \backslash right]\; \backslash leq\; 2\; \backslash sigma^\; \{\backslash alpha\}\; B.$$

B Smoothing scheme

Lemma 8 Assumption 2 implies that $f(x)$ is $M_2$ Lipschitz on $Q$ .

Proof. For all $x_1, x_2 \in Q$

$$\begin{array}{c}\mathrm{\mid }f(x_1)-f(x_2)\mathrm{\mid }\le \mathrm{\mid }{\mathbb{E}}_{\xi }[f(x_1,\xi )-f(x_2,\xi )]\mathrm{\mid }\le {\mathbb{E}}_{\xi }[\mathrm{\mid }f(x_1,\xi )-f(x_2,\xi )\mathrm{\mid }](18)\\ \le {\mathbb{E}}_{\xi }[M_2(\xi )]\mathrm{\parallel }{x}_1-x_2{\mathrm{\parallel }}_2\le M_2\mathrm{\parallel }{x}_1-x_2{\mathrm{\parallel }}_2\mathrm{.}(19)\end{array}\backslash begin\{array\}\{l\}\; |\; f\; (x\; \_\; \{1\})\; -\; f\; (x\; \_\; \{2\})\; |\; \backslash leq\; |\; \backslash mathbb\; \{E\}\; \_\; \{\backslash xi\}\; [\; f\; (x\; \_\; \{1\},\; \backslash xi)\; -\; f\; (x\; \_\; \{2\},\; \backslash xi)\; ]\; |\; \backslash leq\; \backslash mathbb\; \{E\}\; \_\; \{\backslash xi\}\; [\; |\; f\; (x\; \_\; \{1\},\; \backslash xi)\; -\; f\; (x\; \_\; \{2\},\; \backslash xi)\; |\; ]\; (18)\; \backslash \backslash \; \backslash leq\; \backslash mathbb\; \{E\}\; \_\; \{\backslash xi\}\; \backslash left[\; M\; \_\; \{2\}\; (\backslash xi)\; \backslash right]\; \backslash |\; x\; \_\; \{1\}\; -\; x\; \_\; \{2\}\; \backslash |\; \_\; \{2\}\; \backslash leq\; M\; \_\; \{2\}\; \backslash |\; x\; \_\; \{1\}\; -\; x\; \_\; \{2\}\; \backslash |\; \_\; \{2\}.\; (19)\; \backslash \backslash \; \backslash end\{array\}$$

The following lemma gives some facts about the measure concentration on the Euclidean unit sphere for next proof.

Lemma 9 Let $f(x)$ be $M_2$ Lipschitz continuous function w.r.t $|\cdot|$ . If $e$ is random and uniformly distributed on the Euclidean sphere and $\alpha \in (1,2]$ , then

$${\mathbb{E}}_{\mathbf{e}}[(f(\mathbf{e})-{\mathbb{E}}_{\mathbf{e}}[f(\mathbf{e})]{)}^{2\alpha }]\le {\left(\frac{b{M}_2^2}{d}\right)}^{\alpha },b=\frac{1}{\sqrt{2}}\mathrm{.}\backslash mathbb\; \{E\}\; \_\; \{\backslash mathbf\; \{e\}\}\; \backslash left[\; (f\; (\backslash mathbf\; \{e\})\; -\; \backslash mathbb\; \{E\}\; \_\; \{\backslash mathbf\; \{e\}\}\; [\; f\; (\backslash mathbf\; \{e\})\; ])\; ^\; \{2\; \backslash alpha\}\; \backslash right]\; \backslash leq\; \backslash left(\backslash frac\; \{b\; M\; \_\; \{2\}\; ^\; \{2\}\}\{d\}\backslash right)\; ^\; \{\backslash alpha\},\; b\; =\; \backslash frac\; \{1\}\{\backslash sqrt\; \{2\}\}.$$

Proof. A standard result of the measure concentration on the Euclidean unit sphere implies that $\forall t > 0$

$$\begin{array}{cccc}& Pr(\mathrm{\mid }f(\mathbf{e})-\mathbb{E}[f(\mathbf{e})]\mathrm{\mid }>t)\le 2\exp (-b^{\mathrm{\prime }}d{t}^2\mathrm{/}{M}_2^2),b^{\mathrm{\prime }}=2& & \text{(20)}\end{array}P\; r\; \backslash left(|\; f\; (\backslash mathbf\; \{e\})\; -\; \backslash mathbb\; \{E\}\; [\; f\; (\backslash mathbf\; \{e\})\; ]\; |\; >\; t\backslash right)\; \backslash leq\; 2\; \backslash exp\; \backslash left(-\; b\; ^\; \{\backslash prime\}\; d\; t\; ^\; \{2\}\; /\; M\; \_\; \{2\}\; ^\; \{2\}\backslash right),\; \backslash quad\; b\; ^\; \{\backslash prime\}\; =\; 2\; \backslash tag\; \{20\}$$

(see the proof of Proposition 2.10 and Corollary 2.6 in [23]). Therefore,

$$\begin{array}{cccc}& \begin{array}{c}{\mathbb{E}}_{\mathbf{e}}[(f(\mathbf{e})-{\mathbb{E}}_{\mathbf{e}}[f(\mathbf{e})]{)}^{2\alpha }]={\int }_{t=0}^{\mathrm{\infty }}Pr(\mathrm{\mid }f(\mathbf{e})-\mathbb{E}[f(\mathbf{e})]{\mathrm{\mid }}^{2\alpha }>t)dt\\ ={\int }_{t=0}^{\mathrm{\infty }}Pr(\mathrm{\mid }f(\mathbf{e})-\mathbb{E}[f(\mathbf{e})]\mathrm{\mid }>t^{1\mathrm{/}2\alpha })dt\\ \le {\int }_{t=0}^{\mathrm{\infty }}2\exp (-b^{\mathrm{\prime }}d\cdot t^{1\mathrm{/}\alpha }\mathrm{/}{M}_2^2)dt\le {\left(\frac{b{M}_2^2}{d}\right)}^{\alpha }\mathrm{.}\end{array}& & \text{(21)}\end{array}\backslash begin\{array\}\{l\}\; \backslash mathbb\; \{E\}\; \_\; \{\backslash mathbf\; \{e\}\}\; \backslash left[\; (f\; (\backslash mathbf\; \{e\})\; -\; \backslash mathbb\; \{E\}\; \_\; \{\backslash mathbf\; \{e\}\}\; [\; f\; (\backslash mathbf\; \{e\})\; ])\; ^\; \{2\; \backslash alpha\}\; \backslash right]\; =\; \backslash int\_\; \{t\; =\; 0\}\; ^\; \{\backslash infty\}\; P\; r\; \backslash left(|\; f\; (\backslash mathbf\; \{e\})\; -\; \backslash mathbb\; \{E\}\; [\; f\; (\backslash mathbf\; \{e\})\; ]\; |\; ^\; \{2\; \backslash alpha\}\; >\; t\backslash right)\; d\; t\; \backslash \backslash \; =\; \backslash int\_\; \{t\; =\; 0\}\; ^\; \{\backslash infty\}\; P\; r\; \backslash left(|\; f\; (\backslash mathbf\; \{e\})\; -\; \backslash mathbb\; \{E\}\; [\; f\; (\backslash mathbf\; \{e\})\; ]\; |\; >\; t\; ^\; \{1\; /\; 2\; \backslash alpha\}\backslash right)\; d\; t\; \backslash \backslash \; \backslash leq\; \backslash int\_\; \{t\; =\; 0\}\; ^\; \{\backslash infty\}\; 2\; \backslash exp\; \backslash left(-\; b\; ^\; \{\backslash prime\}\; d\; \backslash cdot\; t\; ^\; \{1\; /\; \backslash alpha\}\; /\; M\; \_\; \{2\}\; ^\; \{2\}\backslash right)\; d\; t\; \backslash leq\; \backslash left(\backslash frac\; \{b\; M\; \_\; \{2\}\; ^\; \{2\}\}\{d\}\backslash right)\; ^\; \{\backslash alpha\}.\; \backslash tag\; \{21\}\; \backslash \backslash \; \backslash end\{array\}$$

Finally, below we prove Lemma 3 which states that batched gradient estimation from (5) has bounded $\alpha$ -th moment.

Proof of Lemma 3.

1. We will prove equality immediately for $g^B$ . Firstly, we notice that distribution of $\mathbf{e}$ is symmetric and by definition (5) we get

$$\begin{array}{cccc}& \begin{array}{c}{\mathbb{E}}_{\xi ,\mathbf{e}}[g^B(x,\xi ,\mathbf{e})]=\left(\frac{d}{2B\tau }\right){\sum }_{i=1}^B{\mathbb{E}}_{{\xi }_i,{\mathbf{e}}_i}[f(x+\tau {\mathbf{e}}_i,{\xi }_i){\mathbf{e}}_i-f(x-\tau {\mathbf{e}}_i,{\xi }_i){\mathbf{e}}_{\mathrm{i}}]\\ =\frac{d}{B\tau }{\sum }_{i=1}^B{\mathbb{E}}_{{\mathbf{e}}_i}[{\mathbb{E}}_{{\xi }_i}[f(x+\tau {\mathbf{e}}_i,{\xi }_i)]{\mathbf{e}}_i]\\ =\frac{d}{B\tau }{\sum }_{i=1}^B{\mathbb{E}}_{{\mathbf{e}}_i}[f(x+\tau {\mathbf{e}}_i){\mathbf{e}}_i]\mathrm{.}\end{array}& & \text{(22)}\end{array}\backslash begin\{array\}\{l\}\; \backslash mathbb\; \{E\}\; \_\; \{\backslash xi\; ,\; \backslash mathbf\; \{e\}\}\; [\; g\; ^\; \{B\}\; (x,\; \backslash xi\; ,\; \backslash mathbf\; \{e\})\; ]\; =\; \backslash left(\backslash frac\; \{d\}\{2\; B\; \backslash tau\}\backslash right)\; \backslash sum\_\; \{i\; =\; 1\}\; ^\; \{B\}\; \backslash mathbb\; \{E\}\; \_\; \{\backslash xi\_\; \{i\},\; \backslash mathbf\; \{e\}\; \_\; \{i\}\}\; \backslash left[\; f\; (x\; +\; \backslash tau\; \backslash mathbf\; \{e\}\; \_\; \{i\},\; \backslash xi\_\; \{i\})\; \backslash mathbf\; \{e\}\; \_\; \{i\}\; -\; f\; (x\; -\; \backslash tau\; \backslash mathbf\; \{e\}\; \_\; \{i\},\; \backslash xi\_\; \{i\})\; \backslash mathbf\; \{e\}\; \_\; \{\backslash mathrm\; \{i\}\}\; \backslash right]\; \backslash \backslash \; =\; \backslash frac\; \{d\}\{B\; \backslash tau\}\; \backslash sum\_\; \{i\; =\; 1\}\; ^\; \{B\}\; \backslash mathbb\; \{E\}\; \_\; \{\backslash mathbf\; \{e\}\; \_\; \{i\}\}\; [\; \backslash mathbb\; \{E\}\; \_\; \{\backslash xi\_\; \{i\}\}\; [\; f\; (x\; +\; \backslash tau\; \backslash mathbf\; \{e\}\; \_\; \{i\},\; \backslash xi\_\; \{i\})\; ]\; \backslash mathbf\; \{e\}\; \_\; \{i\}\; ]\; \backslash \backslash \; =\; \backslash frac\; \{d\}\{B\; \backslash tau\}\; \backslash sum\_\; \{i\; =\; 1\}\; ^\; \{B\}\; \backslash mathbb\; \{E\}\; \_\; \{\backslash mathbf\; \{e\}\; \_\; \{i\}\}\; [\; f\; (x\; +\; \backslash tau\; \backslash mathbf\; \{e\}\; \_\; \{i\})\; \backslash mathbf\; \{e\}\; \_\; \{i\}\; ].\; \backslash tag\; \{22\}\; \backslash \backslash \; \backslash end\{array\}$$

Using $\nabla \hat{f}{\tau}(x) = \frac{d}{\tau}\mathbb{E}{\mathbf{e}}[f(x + \tau \mathbf{e})\mathbf{e}]$ from Lemma 2 we obtain necessary result.

2. By definition (4) of estimated gradient $g$ we bound $\alpha$ -th moment as

$$\begin{array}{cccc}& \begin{array}{c}{\mathbb{E}}_{\xi ,\mathbf{e}}[\mathrm{\parallel }g(x,\xi ,\mathbf{e}){\mathrm{\parallel }}^{\alpha }]={\mathbb{E}}_{\xi ,\mathbf{e}}\left[{\mid \frac{d}{2\tau }(f(x+\tau \mathbf{e},\xi )-f(x-\tau \mathbf{e},\xi ))\mathbf{e}\parallel }_2^{\alpha }\right]\\ ={\left(\frac{d}{2\tau }\right)}^{\alpha }{\mathbb{E}}_{\xi ,\mathbf{e}}[\mathrm{\parallel }\mathbf{e}{\mathrm{\parallel }}_2^{\alpha }\mathrm{\mid }f(x+\tau \mathbf{e},\xi )-f(x-\tau \mathbf{e},\xi ){\mathrm{\mid }}^{\alpha }]\mathrm{.}\end{array}& & \text{(23)}\end{array}\backslash begin\{array\}\{l\}\; \backslash mathbb\; \{E\}\; \_\; \{\backslash xi\; ,\; \backslash mathbf\; \{e\}\}\; [\; \backslash |\; g\; (x,\; \backslash xi\; ,\; \backslash mathbf\; \{e\})\; \backslash |\; ^\; \{\backslash alpha\}\; ]\; =\; \backslash mathbb\; \{E\}\; \_\; \{\backslash xi\; ,\; \backslash mathbf\; \{e\}\}\; \backslash left[\; \backslash left|\; \backslash frac\; \{d\}\{2\; \backslash tau\}\; (f\; (x\; +\; \backslash tau\; \backslash mathbf\; \{e\},\; \backslash xi)\; -\; f\; (x\; -\; \backslash tau\; \backslash mathbf\; \{e\},\; \backslash xi))\; \backslash mathbf\; \{e\}\; \backslash right\backslash rVert\_\; \{2\}\; ^\; \{\backslash alpha\}\; \backslash right]\; \backslash \backslash \; =\; \backslash left(\backslash frac\; \{d\}\{2\; \backslash tau\}\backslash right)\; ^\; \{\backslash alpha\}\; \backslash mathbb\; \{E\}\; \_\; \{\backslash xi\; ,\; \backslash mathbf\; \{e\}\}\; \backslash left[\; \backslash |\; \backslash mathbf\; \{e\}\; \backslash |\; \_\; \{2\}\; ^\; \{\backslash alpha\}\; |\; f\; (x\; +\; \backslash tau\; \backslash mathbf\; \{e\},\; \backslash xi)\; -\; f\; (x\; -\; \backslash tau\; \backslash mathbf\; \{e\},\; \backslash xi)\; |\; ^\; \{\backslash alpha\}\; \backslash right].\; \backslash tag\; \{23\}\; \backslash \backslash \; \backslash end\{array\}$$

Considering $| \mathbf{e}| _2 = 1$ we can omit it. Next we add $\pm \delta (\xi)$ in (23) for all $\delta (\xi)$ and get

$$\begin{array}{c}{\mathbb{E}}_{\xi ,\mathbf{e}}[\mathrm{\mid }f(x+\tau \mathbf{e},\xi )-f(x-\tau \mathbf{e},\xi ){\mathrm{\mid }}^{\alpha }]\\ ={\mathbb{E}}_{\xi ,\mathbf{e}}[\mathrm{\mid }(f(x+\tau \mathbf{e},\xi )-\delta )-(f(x-\tau \mathbf{e},\xi )-\delta ){\mathrm{\mid }}^{\alpha }]\mathrm{.}\end{array}\backslash begin\{array\}\{l\}\; \backslash mathbb\; \{E\}\; \_\; \{\backslash xi\; ,\; \backslash mathbf\; \{e\}\}\; \backslash left[\; |\; f\; (x\; +\; \backslash tau\; \backslash mathbf\; \{e\},\; \backslash xi)\; -\; f\; (x\; -\; \backslash tau\; \backslash mathbf\; \{e\},\; \backslash xi)\; |\; ^\; \{\backslash alpha\}\; \backslash right]\; \backslash \backslash \; =\; \backslash mathbb\; \{E\}\; \_\; \{\backslash xi\; ,\; \backslash mathbf\; \{e\}\}\; [\; |\; (f\; (x\; +\; \backslash tau\; \backslash mathbf\; \{e\},\; \backslash xi)\; -\; \backslash delta)\; -\; (f\; (x\; -\; \backslash tau\; \backslash mathbf\; \{e\},\; \backslash xi)\; -\; \backslash delta)\; |\; ^\; \{\backslash alpha\}\; ].\; \backslash \backslash \; \backslash end\{array\}$$

Using Jensen's inequality for $|\cdot|^{\alpha}$ we bound

$$\begin{array}{c}{\mathbb{E}}_{\xi ,\mathbf{e}}[\mathrm{\mid }(f(x+\tau \mathbf{e},\xi )-\delta )-(f(x-\tau \mathbf{e},\xi )-\delta ){\mathrm{\mid }}^{\alpha }]\\ \le 2^{\alpha -1}{\mathbb{E}}_{\xi ,\mathbf{e}}[\mathrm{\mid }f(x+\tau \mathbf{e},\xi )-\delta {\mathrm{\mid }}^{\alpha }]+2^{\alpha -1}{\mathbb{E}}_{\xi ,\mathbf{e}}[\mathrm{\mid }f(x-\tau \mathbf{e},\xi )-\delta {\mathrm{\mid }}^{\alpha }]\mathrm{.}\end{array}\backslash begin\{array\}\{l\}\; \backslash mathbb\; \{E\}\; \_\; \{\backslash xi\; ,\; \backslash mathbf\; \{e\}\}\; [\; |\; (f\; (x\; +\; \backslash tau\; \backslash mathbf\; \{e\},\; \backslash xi)\; -\; \backslash delta)\; -\; (f\; (x\; -\; \backslash tau\; \backslash mathbf\; \{e\},\; \backslash xi)\; -\; \backslash delta)\; |\; ^\; \{\backslash alpha\}\; ]\; \backslash \backslash \; \backslash leq\; 2\; ^\; \{\backslash alpha\; -\; 1\}\; \backslash mathbb\; \{E\}\; \_\; \{\backslash xi\; ,\; \backslash mathbf\; \{e\}\}\; \backslash left[\; |\; f\; (x\; +\; \backslash tau\; \backslash mathbf\; \{e\},\; \backslash xi)\; -\; \backslash delta\; |\; ^\; \{\backslash alpha\}\; \backslash right]\; +\; 2\; ^\; \{\backslash alpha\; -\; 1\}\; \backslash mathbb\; \{E\}\; \_\; \{\backslash xi\; ,\; \backslash mathbf\; \{e\}\}\; \backslash left[\; |\; f\; (x\; -\; \backslash tau\; \backslash mathbf\; \{e\},\; \backslash xi)\; -\; \backslash delta\; |\; ^\; \{\backslash alpha\}\; \backslash right].\; \backslash \backslash \; \backslash end\{array\}$$

We note that distribution of $\mathbf{e}$ is symmetric and add two terms together

$$\begin{array}{c}{2}^{\alpha -1}{\mathbb{E}}_{\xi ,\mathbf{e}}[\mathrm{\mid }f(x+\tau \mathbf{e},\xi )-\delta {\mathrm{\mid }}^{\alpha }]+2^{\alpha -1}{\mathbb{E}}_{\xi ,\mathbf{e}}[\mathrm{\mid }f(x-\tau \mathbf{e},\xi )-\delta {\mathrm{\mid }}^{\alpha }]\\ \le 2^{\alpha }{\mathbb{E}}_{\xi ,\mathbf{e}}[\mathrm{\mid }f(x+\tau \mathbf{e},\xi )-\delta {\mathrm{\mid }}^{\alpha }]\mathrm{.}\end{array}\backslash begin\{array\}\{l\}\; 2\; ^\; \{\backslash alpha\; -\; 1\}\; \backslash mathbb\; \{E\}\; \_\; \{\backslash xi\; ,\; \backslash mathbf\; \{e\}\}\; \backslash left[\; |\; f\; (x\; +\; \backslash tau\; \backslash mathbf\; \{e\},\; \backslash xi)\; -\; \backslash delta\; |\; ^\; \{\backslash alpha\}\; \backslash right]\; +\; 2\; ^\; \{\backslash alpha\; -\; 1\}\; \backslash mathbb\; \{E\}\; \_\; \{\backslash xi\; ,\; \backslash mathbf\; \{e\}\}\; \backslash left[\; |\; f\; (x\; -\; \backslash tau\; \backslash mathbf\; \{e\},\; \backslash xi)\; -\; \backslash delta\; |\; ^\; \{\backslash alpha\}\; \backslash right]\; \backslash \backslash \; \backslash leq\; 2\; ^\; \{\backslash alpha\}\; \backslash mathbb\; \{E\}\; \_\; \{\backslash xi\; ,\; \backslash mathbf\; \{e\}\}\; \backslash left[\; |\; f\; (x\; +\; \backslash tau\; \backslash mathbf\; \{e\},\; \backslash xi)\; -\; \backslash delta\; |\; ^\; \{\backslash alpha\}\; \backslash right].\; \backslash \backslash \; \backslash end\{array\}$$

Let $\delta (\xi) = \mathbb{E}_{\mathbf{e}}[f(x + \tau \mathbf{e},\xi)]$ , then because of Cauchy-Schwartz inequality and conditional expectation properties we obtain

$$\begin{array}{c}{2}^{\alpha }{\mathbb{E}}_{\xi ,\mathbf{e}}[\mathrm{\mid }f(x+\tau \mathbf{e},\xi )-\delta {\mathrm{\mid }}^{\alpha }]=2^{\alpha }{\mathbb{E}}_{\xi }[{\mathbb{E}}_{\mathbf{e}}[\mathrm{\mid }f(x+\tau \mathbf{e},\xi )-\delta {\mathrm{\mid }}^{\alpha }]]\\ \le 2^{\alpha }{\mathbb{E}}_{\xi }\left[\sqrt{{\mathbb{E}}_{\mathbf{e}}[\mathrm{\mid }f(x+\tau \mathbf{e},\xi )-{\mathbb{E}}_{\mathbf{e}}[f(x+\tau \mathbf{e},\xi )]{\mathrm{\mid }}^{2\alpha }]}\right]\mathrm{.}\end{array}\backslash begin\{array\}\{l\}\; 2\; ^\; \{\backslash alpha\}\; \backslash mathbb\; \{E\}\; \_\; \{\backslash xi\; ,\; \backslash mathbf\; \{e\}\}\; [\; |\; f\; (x\; +\; \backslash tau\; \backslash mathbf\; \{e\},\; \backslash xi)\; -\; \backslash delta\; |\; ^\; \{\backslash alpha\}\; ]\; =\; 2\; ^\; \{\backslash alpha\}\; \backslash mathbb\; \{E\}\; \_\; \{\backslash xi\}\; [\; \backslash mathbb\; \{E\}\; \_\; \{\backslash mathbf\; \{e\}\}\; [\; |\; f\; (x\; +\; \backslash tau\; \backslash mathbf\; \{e\},\; \backslash xi)\; -\; \backslash delta\; |\; ^\; \{\backslash alpha\}\; ]\; ]\; \backslash \backslash \; \backslash leq\; 2\; ^\; \{\backslash alpha\}\; \backslash mathbb\; \{E\}\; \_\; \{\backslash xi\}\; \backslash left[\; \backslash sqrt\; \{\backslash mathbb\; \{E\}\; \_\; \{\backslash mathbf\; \{e\}\}\; \backslash left[\; |\; f\; (x\; +\; \backslash tau\; \backslash mathbf\; \{e\}\; ,\; \backslash xi)\; -\; \backslash mathbb\; \{E\}\; \_\; \{\backslash mathbf\; \{e\}\}\; [\; f\; (x\; +\; \backslash tau\; \backslash mathbf\; \{e\}\; ,\; \backslash xi)\; ]\; |\; ^\; \{2\; \backslash alpha\}\; \backslash right]\}\; \backslash right].\; \backslash \backslash \; \backslash end\{array\}$$

Next, we use Lemma 9 for $f(x + \tau \mathbf{e},\xi)$ with fixed $\xi$ and Lipschitz constant $M_2(\xi)\tau$

$$\begin{array}{c}{2}^{\alpha }{\mathbb{E}}_{\xi }\left[\sqrt{{\mathbb{E}}_{\mathbf{e}}[\mathrm{\mid }f(x+\tau \mathbf{e},\xi )-{\mathbb{E}}_{\mathbf{e}}[f(x+\tau \mathbf{e},\xi )]{\mathrm{\mid }}^{2\alpha }]}\right]\le 2^{\alpha }{\mathbb{E}}_{\xi }\left[\sqrt{{\left(\frac{2^{-1\mathrm{/}2}{\tau }^2{M}_2^2(\xi )}{d}\right)}^{\alpha }}\right]\\ =2^{\alpha }{\left(\frac{{\tau }^2{2}^{-1\mathrm{/}2}}{d}\right)}^{\alpha \mathrm{/}2}{\mathbb{E}}_{\xi }[M_2^{\alpha }(\xi )]\le 2^{\alpha }{\left(\sqrt{\frac{2^{-1\mathrm{/}2}}{d}}{M}_2\tau \right)}^{\alpha }\mathrm{.}\end{array}\backslash begin\{array\}\{l\}\; 2\; ^\; \{\backslash alpha\}\; \backslash mathbb\; \{E\}\; \_\; \{\backslash xi\}\; \backslash left[\; \backslash sqrt\; \{\backslash mathbb\; \{E\}\; \_\; \{\backslash mathbf\; \{e\}\}\; \backslash left[\; |\; f\; (x\; +\; \backslash tau\; \backslash mathbf\; \{e\}\; ,\; \backslash xi)\; -\; \backslash mathbb\; \{E\}\; \_\; \{\backslash mathbf\; \{e\}\}\; [\; f\; (x\; +\; \backslash tau\; \backslash mathbf\; \{e\}\; ,\; \backslash xi)\; ]\; |\; ^\; \{2\; \backslash alpha\}\; \backslash right]\}\; \backslash right]\; \backslash leq\; 2\; ^\; \{\backslash alpha\}\; \backslash mathbb\; \{E\}\; \_\; \{\backslash xi\}\; \backslash left[\; \backslash sqrt\; \{\backslash left(\backslash frac\; \{2\; ^\; \{-\; 1\; /\; 2\}\; \backslash tau^\; \{2\}\; M\; \_\; \{2\}\; ^\; \{2\}\; (\backslash xi)\}\{d\}\backslash right)\; ^\; \{\backslash alpha\}\}\; \backslash right]\; \backslash \backslash \; =\; 2\; ^\; \{\backslash alpha\}\; \backslash left(\backslash frac\; \{\backslash tau^\; \{2\}\; 2\; ^\; \{-\; 1\; /\; 2\}\}\{d\}\backslash right)\; ^\; \{\backslash alpha\; /\; 2\}\; \backslash mathbb\; \{E\}\; \_\; \{\backslash xi\}\; \backslash left[\; M\; \_\; \{2\}\; ^\; \{\backslash alpha\}\; (\backslash xi)\; \backslash right]\; \backslash leq\; 2\; ^\; \{\backslash alpha\}\; \backslash left(\backslash sqrt\; \{\backslash frac\; \{2\; ^\; \{-\; 1\; /\; 2\}\}\{d\}\}\; M\; \_\; \{2\}\; \backslash tau\backslash right)\; ^\; \{\backslash alpha\}.\; \backslash \backslash \; \backslash end\{array\}$$

Finally, we get desired bound of estimated gradient

$$\begin{array}{cccc}& {\mathbb{E}}_{\xi ,\mathbf{e}}[\mathrm{\parallel }g(x,\xi ,\mathbf{e}){\mathrm{\parallel }}_2^{\alpha }]={\left(\frac{\sqrt{d}}{2^{1\mathrm{/}4}}{M}_2\right)}^{\alpha }\mathrm{.}& & \text{(24)}\end{array}\backslash mathbb\; \{E\}\; \_\; \{\backslash xi\; ,\; \backslash mathbf\; \{e\}\}\; [\; \backslash |\; g\; (x,\; \backslash xi\; ,\; \backslash mathbf\; \{e\})\; \backslash |\; \_\; \{2\}\; ^\; \{\backslash alpha\}\; ]\; =\; \backslash left(\backslash frac\; \{\backslash sqrt\; \{d\}\}\{2\; ^\; \{1\; /\; 4\}\}\; M\; \_\; \{2\}\backslash right)\; ^\; \{\backslash alpha\}.\; \backslash tag\; \{24\}$$

Now we apply Jensen inequality to

$$\begin{array}{cccc}& {\mathbb{E}}_{\xi ,\mathbf{e}}[\mathrm{\parallel }g(x,\xi ,\mathbf{e})-{\mathbb{E}}_{\xi ,\mathbf{e}}[g(x,\xi ,\mathbf{e})]{\mathrm{\parallel }}_2^{\alpha }]\le 2^{\alpha -1}({\mathbb{E}}_{\xi ,\mathbf{e}}[\mathrm{\parallel }g(x,\xi ,\mathbf{e}){\mathrm{\parallel }}_2]+{\mathbb{E}}_{\xi ,\mathbf{e}}[\mathrm{\parallel }g(x,\xi ,\mathbf{e}){\mathrm{\parallel }}_2])& & \text{(25)}\end{array}\backslash mathbb\; \{E\}\; \_\; \{\backslash xi\; ,\; \backslash mathbf\; \{e\}\}\; [\; \backslash |\; g\; (x,\; \backslash xi\; ,\; \backslash mathbf\; \{e\})\; -\; \backslash mathbb\; \{E\}\; \_\; \{\backslash xi\; ,\; \backslash mathbf\; \{e\}\}\; [\; g\; (x,\; \backslash xi\; ,\; \backslash mathbf\; \{e\})\; ]\; \backslash |\; \_\; \{2\}\; ^\; \{\backslash alpha\}\; ]\; \backslash leq\; 2\; ^\; \{\backslash alpha\; -\; 1\}\; \backslash left(\backslash mathbb\; \{E\}\; \_\; \{\backslash xi\; ,\; \backslash mathbf\; \{e\}\}\; [\; \backslash |\; g\; (x,\; \backslash xi\; ,\; \backslash mathbf\; \{e\})\; \backslash |\; \_\; \{2\}\; ]\; +\; \backslash mathbb\; \{E\}\; \_\; \{\backslash xi\; ,\; \backslash mathbf\; \{e\}\}\; [\; \backslash |\; g\; (x,\; \backslash xi\; ,\; \backslash mathbf\; \{e\})\; \backslash |\; \_\; \{2\}\; ]\backslash right)\; \backslash tag\; \{25\}$$

And get necessary result.

For batched gradient $g^{B}$ we use Batching Lemma 7 and estimation (25).

C Missing Proofs for ZO-clipped-SSTM and R-ZO-clipped-SSTM

In this section, we provide the complete formulation of the main results for clipped-SSTM and R-clipped-SSTM and the missing proofs.

Minimization on the subset $Q$ In order to work with subsets of $Q \subseteq \mathbb{R}^d$ we must assume one more condition on $\hat{f}_{\tau}(x)$ .

Assumption 5 We assume that there exist some convex set $Q \subseteq \mathbb{R}^d$ , constants $\tau, L > 0$ such that for all $x, y \in Q$

$$\begin{array}{cccc}& {\parallel \mathrm{\nabla }{\widehat{f}}_{\tau }(x)\parallel }_2^2\le 2L({\widehat{f}}_{\tau }(x)-{\widehat{f}}_{\tau }^{\ast }),& & \text{(26)}\end{array}\backslash left\backslash |\; \backslash nabla\; \backslash hat\; \{f\}\; \_\; \{\backslash tau\}\; (x)\; \backslash right\backslash |\; \_\; \{2\}\; ^\; \{2\}\; \backslash leq\; 2\; L\; \backslash left(\backslash hat\; \{f\}\; \_\; \{\backslash tau\}\; (x)\; -\; \backslash hat\; \{f\}\; \_\; \{\backslash tau\}\; ^\; \{*\}\backslash right),\; \backslash tag\; \{26\}$$

where $\hat{f}{\tau}^{*} = \inf{x\in Q}\hat{f}_{\tau}(x) > - \infty$

When $Q = \mathbb{R}^d$ (26) follows from Lemma 2 as well. But in general case this is not true. In work [35] in Section "Useful facts" authors show that, in the worst case, to have (26) on a set $Q$ one needs to assume smoothness on a slightly larger set.

Thus, in the full version of the theorems in which we can require much smaller $Q$ , we will also require satisfying all three Assumptions 1, 2, 5.

C.1 Convex Functions

We start with the following lemma, which is a special case of Lemma 6 from [19]. This result can be seen the "optimization" part of the analysis of clipped-SSTM: the proof follows the same steps as the analysis of deterministic Similar Triangles Method [16], [12] and separates stochasticity from the deterministic part of the method.

Pay attention that in full version of Theorem 1 we require Assumptions 1, 2 to hold only on $Q = B_{3R}(x^{*})$ , however we need to require one more smoothness Assumption 5 as well.

Theorem 3 (Full version of Theorem 1) Let Assumptions 1,2, 5 with $\mu = 0$ hold on $Q = B_{3R}(x^{})$ , where $R \geq | x^0 - x^|$ , and

$$\begin{array}{cccc}& a\ge \max \{48600{\ln }^2\frac{4K}{\beta },\frac{1800\sigma (K+1)K^{\frac{1}{\alpha }}{\ln }^{\frac{\alpha -1}{\alpha }}\frac{4K}{\beta }}{B^{\frac{\alpha -1}{\alpha }}LR}\},& & \text{(27)}\end{array}a\; \backslash geq\; \backslash max\; \backslash left\backslash \{4\; 8\; 6\; 0\; 0\; \backslash ln^\; \{2\}\; \backslash frac\; \{4\; K\}\{\backslash beta\},\; \backslash frac\; \{1\; 8\; 0\; 0\; \backslash sigma\; (K\; +\; 1)\; K\; ^\; \{\backslash frac\; \{1\}\{\backslash alpha\}\}\; \backslash ln^\; \{\backslash frac\; \{\backslash alpha\; -\; 1\}\{\backslash alpha\}\}\; \backslash frac\; \{4\; K\}\{\backslash beta\}\}\{B\; ^\; \{\backslash frac\; \{\backslash alpha\; -\; 1\}\{\backslash alpha\}\}\; L\; R\}\; \backslash right\backslash \},\; \backslash tag\; \{27\}$$

$$\begin{array}{cccc}& {\lambda }_k=\frac{R}{30{\alpha }_{k+1}\ln \frac{4K}{\beta }},& & \text{(28)}\end{array}\backslash lambda\_\; \{k\}\; =\; \backslash frac\; \{R\}\{3\; 0\; \backslash alpha\_\; \{k\; +\; 1\}\; \backslash ln\; \backslash frac\; \{4\; K\}\{\backslash beta\}\},\; \backslash tag\; \{28\}$$

$$\begin{array}{cccc}& L=\frac{M_2\sqrt{d}}{\tau },& & \text{(29)}\end{array}L\; =\; \backslash frac\; \{M\; \_\; \{2\}\; \backslash sqrt\; \{d\}\}\{\backslash tau\},\; \backslash tag\; \{29\}$$

for some $K > 0$ and $\beta \in (0,1]$ such that $\ln \frac{4K}{\beta} \geq 1$ . Then, after $K$ iterations of ZO-clipped-SSTM the iterates with probability at least $1 - \beta$ satisfy

$$\begin{array}{cccc}& f\left(y^K\right)-f\left(x^{\ast }\right)\le 2M_2\tau +\frac{6aL{R}^2}{K(K+3)}and{\left\{x^k\right\}}_{k=0}^{K+1},{\left\{z^k\right\}}_{k=0}^K,{\left\{y^k\right\}}_{k=0}^K\subseteq B_{2R}\left(x^{\ast }\right)\mathrm{.}& & \text{(30)}\end{array}f\; \backslash left(y\; ^\; \{K\}\backslash right)\; -\; f\; \backslash left(x\; ^\; \{*\}\backslash right)\; \backslash leq\; 2\; M\; \_\; \{2\}\; \backslash tau\; +\; \backslash frac\; \{6\; a\; L\; R\; ^\; \{2\}\}\{K\; (K\; +\; 3)\}\; \backslash quad\; a\; n\; d\; \backslash quad\; \backslash left\backslash \{x\; ^\; \{k\}\; \backslash right\backslash \}\; \_\; \{k\; =\; 0\}\; ^\; \{K\; +\; 1\},\; \backslash left\backslash \{z\; ^\; \{k\}\; \backslash right\backslash \}\; \_\; \{k\; =\; 0\}\; ^\; \{K\},\; \backslash left\backslash \{y\; ^\; \{k\}\; \backslash right\backslash \}\; \_\; \{k\; =\; 0\}\; ^\; \{K\}\; \backslash subseteq\; B\; \_\; \{2\; R\}\; \backslash left(x\; ^\; \{*\}\backslash right).\; \backslash tag\; \{30\}$$

In particular, when parameter $a$ equals the maximum from (27), then the iterates produced by ZO-clipped-SSTM after $K$ iterations with probability at least $1 - \beta$ satisfy

$$\begin{array}{cccc}& f\left(y^K\right)-f\left(x^{\ast }\right)\le 2M_2\tau +\mathcal{O}(\max \{\frac{L{R}^2{\ln }^2\frac{K}{\beta }}{K^2},\frac{\sigma R{\ln }^{\frac{\alpha -1}{\alpha }}\frac{K}{\beta }}{(BK{)}^{\frac{\alpha -1}{\alpha }}}\}),& & \text{(31)}\end{array}f\; \backslash left(y\; ^\; \{K\}\backslash right)\; -\; f\; \backslash left(x\; ^\; \{*\}\backslash right)\; \backslash leq\; 2\; M\; \_\; \{2\}\; \backslash tau\; +\; \backslash mathcal\; \{O\}\; \backslash left(\backslash max\; \backslash left\backslash \{\backslash frac\; \{L\; R\; ^\; \{2\}\; \backslash ln^\; \{2\}\; \backslash frac\; \{K\}\{\backslash beta\}\}\{K\; ^\; \{2\}\},\; \backslash frac\; \{\backslash sigma\; R\; \backslash ln^\; \{\backslash frac\; \{\backslash alpha\; -\; 1\}\{\backslash alpha\}\}\; \backslash frac\; \{K\}\{\backslash beta\}\}\{(B\; K)\; ^\; \{\backslash frac\; \{\backslash alpha\; -\; 1\}\{\backslash alpha\}\}\}\; \backslash right\backslash \}\backslash right),\; \backslash tag\; \{31\}$$

meaning that to achieve $f(y^{K}) - f(x^{*}) \leq \varepsilon$ with probability at least $1 - \beta$ with $\tau = \frac{\varepsilon}{4M_2} ZO$ -clipped-SSTM requires

$$\begin{array}{cccc}& K=\mathcal{O}(\sqrt{\frac{M_2^2\sqrt{d}{R}^2}{{\epsilon }^2}}\ln \frac{M_2^2\sqrt{d}{R}^2}{{\epsilon }^2\beta },\frac{1}{B}{\left(\frac{\sigma R}{\epsilon }\right)}^{\frac{\alpha }{\alpha -1}}\ln \left(\frac{1}{B\beta }{\left(\frac{\sigma R}{\epsilon }\right)}^{\frac{\alpha }{\alpha -1}}\right))iterations\mathrm{.}& & \text{(32)}\end{array}K\; =\; \backslash mathcal\; \{O\}\; \backslash left(\backslash sqrt\; \{\backslash frac\; \{M\; \_\; \{2\}\; ^\; \{2\}\; \backslash sqrt\; \{d\}\; R\; ^\; \{2\}\}\{\backslash varepsilon^\; \{2\}\}\}\; \backslash ln\; \backslash frac\; \{M\; \_\; \{2\}\; ^\; \{2\}\; \backslash sqrt\; \{d\}\; R\; ^\; \{2\}\}\{\backslash varepsilon^\; \{2\}\; \backslash beta\},\; \backslash frac\; \{1\}\{B\}\; \backslash left(\backslash frac\; \{\backslash sigma\; R\}\{\backslash varepsilon\}\backslash right)\; ^\; \{\backslash frac\; \{\backslash alpha\}\{\backslash alpha\; -\; 1\}\}\; \backslash ln\; \backslash left(\backslash frac\; \{1\}\{B\; \backslash beta\}\; \backslash left(\backslash frac\; \{\backslash sigma\; R\}\{\backslash varepsilon\}\backslash right)\; ^\; \{\backslash frac\; \{\backslash alpha\}\{\backslash alpha\; -\; 1\}\}\backslash right)\backslash right)\; \backslash quad\; i\; t\; e\; r\; a\; t\; i\; o\; n\; s.\; \backslash tag\; \{32\}$$

In case when second term in max in (31) is greater total number of oracle calls is

$$K\cdot B=\mathcal{O}({\left(\frac{\sigma R}{\epsilon }\right)}^{\frac{\alpha }{\alpha -1}}\ln \left(\frac{1}{\beta }{\left(\frac{\sigma R}{\epsilon }\right)}^{\frac{\alpha }{\alpha -1}}\right))\mathrm{.}K\; \backslash cdot\; B\; =\; \backslash mathcal\; \{O\}\; \backslash left(\backslash left(\backslash frac\; \{\backslash sigma\; R\}\{\backslash varepsilon\}\backslash right)\; ^\; \{\backslash frac\; \{\backslash alpha\}\{\backslash alpha\; -\; 1\}\}\; \backslash ln\; \backslash left(\backslash frac\; \{1\}\{\backslash beta\}\; \backslash left(\backslash frac\; \{\backslash sigma\; R\}\{\backslash varepsilon\}\backslash right)\; ^\; \{\backslash frac\; \{\backslash alpha\}\{\backslash alpha\; -\; 1\}\}\backslash right)\backslash right).$$

Proof. The proof is based on the proof of Theorem F.2 from [35]. We apply first-order algorithm clipped-SSTM for $\frac{M_2\sqrt{d}}{\tau}$ -smooth function $\hat{f}_{\tau}$ with unbiased gradient estimation $g^{B}$ , that has $\alpha$ -th moment bounded with $\frac{2\sigma^{\alpha}}{B^{\alpha - 1}}$ . Additional randomization caused by smoothing doesn't affect the proof of the original Theorem.

According to it after $K$ iterations we have that with probability at least $1 - \beta$

$${\widehat{f}}_{\tau }(y^K)-{\widehat{f}}_{\tau }(x^{\ast })\le \frac{6aL{R}^2}{K(K+3)}\backslash hat\; \{f\}\; \_\; \{\backslash tau\}\; (y\; ^\; \{K\})\; -\; \backslash hat\; \{f\}\; \_\; \{\backslash tau\}\; (x\; ^\; \{*\})\; \backslash leq\; \backslash frac\; \{6\; a\; L\; R\; ^\; \{2\}\}\{K\; (K\; +\; 3)\}$$

and ${x^k}{k = 0}^{K + 1},{z^k}{k = 0}^K,{y^k}{k = 0}^K\subseteq B{2R}(x^*)$

Considering approximation properties of $\hat{f}_{\tau}$ from Lemma 2

$$f(y^K)-f(x^{\ast })\le 2M_2\tau +\frac{6aL{R}^2}{K(K+3)}\mathrm{.}f\; (y\; ^\; \{K\})\; -\; f\; (x\; ^\; \{*\})\; \backslash leq\; 2\; M\; \_\; \{2\}\; \backslash tau\; +\; \backslash frac\; \{6\; a\; L\; R\; ^\; \{2\}\}\{K\; (K\; +\; 3)\}.$$

Finally, if

$$a=\max \{48600{\ln }^2\frac{4K}{\beta },\frac{1800\sigma (K+1)K^{\frac{1}{\alpha }}{\ln }^{\frac{\alpha -1}{\alpha }}\frac{4K}{\beta }}{B^{\frac{\alpha -1}{\alpha }}LR}\},a\; =\; \backslash max\; \backslash left\backslash \{4\; 8\; 6\; 0\; 0\; \backslash ln^\; \{2\}\; \backslash frac\; \{4\; K\}\{\backslash beta\},\; \backslash frac\; \{1\; 8\; 0\; 0\; \backslash sigma\; (K\; +\; 1)\; K\; ^\; \{\backslash frac\; \{1\}\{\backslash alpha\}\}\; \backslash ln^\; \{\backslash frac\; \{\backslash alpha\; -\; 1\}\{\backslash alpha\}\}\; \backslash frac\; \{4\; K\}\{\backslash beta\}\}\{B\; ^\; \{\backslash frac\; \{\backslash alpha\; -\; 1\}\{\backslash alpha\}\}\; L\; R\}\; \backslash right\backslash \},$$

then with probability at least $1 - \beta$

$$\begin{array}{c}f(y^K)-f(x^{\ast })\le 2M_2\tau +\frac{6aL{R}^2}{K(K+3)}\\ =2M_2\tau +\max \{\frac{291600L{R}^2{\ln }^2\frac{4K}{\beta }}{K(K+3)},\frac{10800\sigma R(K+1)K^{\frac{1}{\alpha }}{\ln }^{\frac{\alpha -1}{\alpha }}\frac{4K}{\beta }}{K(K+3)B^{\frac{\alpha -1}{\alpha }}}\}\\ =2M_2\tau +\mathcal{O}(\max \{\frac{L{R}^2{\ln }^2\frac{K}{\beta }}{K^2},\frac{\sigma R{\ln }^{\frac{\alpha -1}{\alpha }}\frac{K}{\beta }}{(BK{)}^{\frac{\alpha -1}{\alpha }}}\}),\end{array}\backslash begin\{array\}\{l\}\; f\; (y\; ^\; \{K\})\; -\; f\; (x\; ^\; \{*\})\; \backslash leq\; 2\; M\; \_\; \{2\}\; \backslash tau\; +\; \backslash frac\; \{6\; a\; L\; R\; ^\; \{2\}\}\{K\; (K\; +\; 3)\}\; \backslash \backslash \; =\; 2\; M\; \_\; \{2\}\; \backslash tau\; +\; \backslash max\; \backslash left\backslash \{\backslash frac\; \{2\; 9\; 1\; 6\; 0\; 0\; L\; R\; ^\; \{2\}\; \backslash ln^\; \{2\}\; \backslash frac\; \{4\; K\}\{\backslash beta\}\}\{K\; (K\; +\; 3)\},\; \backslash frac\; \{1\; 0\; 8\; 0\; 0\; \backslash sigma\; R\; (K\; +\; 1)\; K\; ^\; \{\backslash frac\; \{1\}\{\backslash alpha\}\}\; \backslash ln^\; \{\backslash frac\; \{\backslash alpha\; -\; 1\}\{\backslash alpha\}\}\; \backslash frac\; \{4\; K\}\{\backslash beta\}\}\{K\; (K\; +\; 3)\; B\; ^\; \{\backslash frac\; \{\backslash alpha\; -\; 1\}\{\backslash alpha\}\}\}\; \backslash right\backslash \}\; \backslash \backslash \; =\; 2\; M\; \_\; \{2\}\; \backslash tau\; +\; \backslash mathcal\; \{O\}\; \backslash left(\backslash max\; \backslash left\backslash \{\backslash frac\; \{L\; R\; ^\; \{2\}\; \backslash ln^\; \{2\}\; \backslash frac\; \{K\}\{\backslash beta\}\}\{K\; ^\; \{2\}\},\; \backslash frac\; \{\backslash sigma\; R\; \backslash ln^\; \{\backslash frac\; \{\backslash alpha\; -\; 1\}\{\backslash alpha\}\}\; \backslash frac\; \{K\}\{\backslash beta\}\}\{(B\; K)\; ^\; \{\backslash frac\; \{\backslash alpha\; -\; 1\}\{\backslash alpha\}\}\}\; \backslash right\backslash \}\backslash right),\; \backslash \backslash \; \backslash end\{array\}$$

where $L = \frac{M_2\sqrt{d}}{\tau}$ by Lemma 2.

To get $f(y^{K}) - f(x^{*}) \leq \varepsilon$ with probability at least $1 - \beta$ it is sufficient to choose $\tau = \frac{\varepsilon}{4M_2}$ and $K$ such that both terms in the maximum above are $\mathcal{O}(\varepsilon)$ . This leads to

$$K=\mathcal{O}(\sqrt{\frac{M_2^2\sqrt{d}{R}^2}{{\epsilon }^2}}\ln \frac{M_2^2\sqrt{d}{R}^2}{{\epsilon }^2\beta },\frac{1}{B}{\left(\frac{\sigma R}{\epsilon }\right)}^{\frac{\alpha }{\alpha -1}}\ln \left(\frac{1}{B\beta }{\left(\frac{\sigma R}{\epsilon }\right)}^{\frac{\alpha }{\alpha -1}}\right))K\; =\; \backslash mathcal\; \{O\}\; \backslash left(\backslash sqrt\; \{\backslash frac\; \{M\; \_\; \{2\}\; ^\; \{2\}\; \backslash sqrt\; \{d\}\; R\; ^\; \{2\}\}\{\backslash varepsilon^\; \{2\}\}\}\; \backslash ln\; \backslash frac\; \{M\; \_\; \{2\}\; ^\; \{2\}\; \backslash sqrt\; \{d\}\; R\; ^\; \{2\}\}\{\backslash varepsilon^\; \{2\}\; \backslash beta\},\; \backslash frac\; \{1\}\{B\}\; \backslash left(\backslash frac\; \{\backslash sigma\; R\}\{\backslash varepsilon\}\backslash right)\; ^\; \{\backslash frac\; \{\backslash alpha\}\{\backslash alpha\; -\; 1\}\}\; \backslash ln\; \backslash left(\backslash frac\; \{1\}\{B\; \backslash beta\}\; \backslash left(\backslash frac\; \{\backslash sigma\; R\}\{\backslash varepsilon\}\backslash right)\; ^\; \{\backslash frac\; \{\backslash alpha\}\{\backslash alpha\; -\; 1\}\}\backslash right)\backslash right)$$

that concludes the proof.

C.2 Strongly Convex Functions

In the strongly convex case, we consider the restarted version of ZO-clipped-SSTM (R-ZO-clipped-SSTM). The main result is summarized below.

Pay attention that in full version of Theorem 2 we require Assumptions 1, 2 to hold only on $Q = B_{3R}(x^{*})$ , however we need to require one more smoothness Assumption 5 as well.

Theorem 4 (Full version of Theorem 2) Let Assumptions 1, 2, 5 with $\mu > 0$ hold for $Q = B_{3R}(x^{})$ , where $R \geq | x^0 - x^|^2$ and R-ZO-clipped-SSTM runs ZO-clipped-SSTM $N$ times. Let

$$\begin{array}{cccc}& L_t=\frac{\sqrt{d}{M}_2}{{\tau }_k},{\tau }_k=\frac{{\epsilon }_k}{M_2},& & \text{(33)}\end{array}L\; \_\; \{t\}\; =\; \backslash frac\; \{\backslash sqrt\; \{d\}\; M\; \_\; \{2\}\}\{\backslash tau\_\; \{k\}\},\; \backslash quad\; \backslash tau\_\; \{k\}\; =\; \backslash frac\; \{\backslash varepsilon\_\; \{k\}\}\{M\; \_\; \{2\}\},\; \backslash tag\; \{33\}$$

$$K_t=[\max \{1080\sqrt{\frac{L_t{R}_{t-1}^2}{{\epsilon }_t}}\ln \frac{2160\sqrt{L_t{R}_{t-1}^2}N}{\sqrt{{\epsilon }_t}\beta },\frac{2}{B_t}{\left(\frac{10800\sigma R_{t-1}}{{\epsilon }_t}\right)}^{\frac{\alpha }{\alpha -1}}\ln \left(\frac{4N}{B_t\beta }{\left(\frac{5400\sigma R_{t-1}}{{\epsilon }_t}\right)}^{\frac{\alpha }{\alpha -1}}\right)\}],K\; \_\; \{t\}\; =\; \backslash left.\; \backslash right.\backslash left[\; \backslash max\; \backslash left\backslash \{1\; 0\; 8\; 0\; \backslash sqrt\; \{\backslash frac\; \{L\; \_\; \{t\}\; R\; \_\; \{t\; -\; 1\}\; ^\; \{2\}\}\{\backslash varepsilon\_\; \{t\}\}\}\; \backslash ln\; \backslash frac\; \{2\; 1\; 6\; 0\; \backslash sqrt\; \{L\; \_\; \{t\}\; R\; \_\; \{t\; -\; 1\}\; ^\; \{2\}\}\; N\}\{\backslash sqrt\; \{\backslash varepsilon\_\; \{t\}\}\; \backslash beta\},\; \backslash frac\; \{2\}\{B\; \_\; \{t\}\}\; \backslash left(\backslash frac\; \{1\; 0\; 8\; 0\; 0\; \backslash sigma\; R\; \_\; \{t\; -\; 1\}\}\{\backslash varepsilon\_\; \{t\}\}\backslash right)\; ^\; \{\backslash frac\; \{\backslash alpha\}\{\backslash alpha\; -\; 1\}\}\; \backslash ln\; \backslash left(\backslash frac\; \{4\; N\}\{B\; \_\; \{t\}\; \backslash beta\}\; \backslash left(\backslash frac\; \{5\; 4\; 0\; 0\; \backslash sigma\; R\; \_\; \{t\; -\; 1\}\}\{\backslash varepsilon\_\; \{t\}\}\backslash right)\; ^\; \{\backslash frac\; \{\backslash alpha\}\{\backslash alpha\; -\; 1\}\}\backslash right)\backslash right\backslash \}\backslash right],$$

$$\begin{array}{cccc}& {\epsilon }_t=\frac{\mu R_{t-1}^2}{4},R_{t-1}=\frac{R}{2^{(t-1)\mathrm{/}2}},N=\lceil {\log }_2\frac{\mu R^2}{2\epsilon }\rceil ,\ln \frac{4K_{t}N}{\beta }\ge 1,& & \text{(35)}\end{array}\backslash varepsilon\_\; \{t\}\; =\; \backslash frac\; \{\backslash mu\; R\; \_\; \{t\; -\; 1\}\; ^\; \{2\}\}\{4\},\; \backslash quad\; R\; \_\; \{t\; -\; 1\}\; =\; \backslash frac\; \{R\}\{2\; ^\; \{(t\; -\; 1)\; /\; 2\}\},\; \backslash quad\; N\; =\; \backslash left\backslash lceil\; \backslash log\_\; \{2\}\; \backslash frac\; \{\backslash mu\; R\; ^\; \{2\}\}\{2\; \backslash varepsilon\}\; \backslash right\backslash rceil\; ,\; \backslash quad\; \backslash ln\; \backslash frac\; \{4\; K\; \_\; \{t\}\; N\}\{\backslash beta\}\; \backslash geq\; 1,\; \backslash tag\; \{35\}$$

$$\begin{array}{cccc}& a_t=\max \{48600{\ln }^2\frac{4K_{t}N}{\beta },\frac{1800\sigma (K_t+1)K_t^{\frac{1}{\alpha }}{\ln }^{\frac{\alpha -1}{\alpha }}\frac{4K_{t}N}{\beta }}{B_t^{\frac{\alpha -1}{\alpha }}{L}_t{R}_t}\},& & \text{(36)}\end{array}a\; \_\; \{t\}\; =\; \backslash max\; \backslash left\backslash \{4\; 8\; 6\; 0\; 0\; \backslash ln^\; \{2\}\; \backslash frac\; \{4\; K\; \_\; \{t\}\; N\}\{\backslash beta\},\; \backslash frac\; \{1\; 8\; 0\; 0\; \backslash sigma\; \backslash left(K\; \_\; \{t\}\; +\; 1\backslash right)\; K\; \_\; \{t\}\; ^\; \{\backslash frac\; \{1\}\{\backslash alpha\}\}\; \backslash ln^\; \{\backslash frac\; \{\backslash alpha\; -\; 1\}\{\backslash alpha\}\}\; \backslash frac\; \{4\; K\; \_\; \{t\}\; N\}\{\backslash beta\}\}\{B\; \_\; \{t\}\; ^\; \{\backslash frac\; \{\backslash alpha\; -\; 1\}\{\backslash alpha\}\}\; L\; \_\; \{t\}\; R\; \_\; \{t\}\}\; \backslash right\backslash \},\; \backslash tag\; \{36\}$$

$$\begin{array}{cccc}& {\lambda }_k^t=\frac{R_t}{30{\alpha }_{k+1}^t\ln \frac{4K_{t}N}{\beta }}& & \text{(37)}\end{array}\backslash lambda\_\; \{k\}\; ^\; \{t\}\; =\; \backslash frac\; \{R\; \_\; \{t\}\}\{3\; 0\; \backslash alpha\_\; \{k\; +\; 1\}\; ^\; \{t\}\; \backslash ln\; \backslash frac\; \{4\; K\; \_\; \{t\}\; N\}\{\backslash beta\}\}\; \backslash tag\; \{37\}$$

for $t = 1,\dots ,\tau$ . Then to guarantee $f(\hat{x}^{\tau}) - f(x^{*})\leq \varepsilon$ with probability $\geq 1 - \beta$ R-clipped-SSTM requires

$$\begin{array}{cccc}& \mathcal{O}(\max \{\sqrt{\frac{M_2^2\sqrt{d}}{\epsilon \mu }}\ln \left(\frac{\mu R^2}{\epsilon }\right)\ln (\frac{M_2{d}^{\frac{1}{4}}}{\sqrt{\mu \epsilon }\beta }\ln \left(\frac{\mu R^2}{\epsilon }\right)),{\left(\frac{{\sigma }^2}{\mu \epsilon }\right)}^{\frac{\alpha }{2(\alpha -1)}}\ln (\frac{1}{\beta }{\left(\frac{{\sigma }^2}{\mu \epsilon }\right)}^{\frac{\alpha }{2(\alpha -1)}}\ln \left(\frac{\mu R^2}{\epsilon }\right))\})& & \text{(38)}\end{array}\backslash mathcal\; \{O\}\; \backslash left(\backslash max\; \backslash left\backslash \{\backslash sqrt\; \{\backslash frac\; \{M\; \_\; \{2\}\; ^\; \{2\}\; \backslash sqrt\; \{d\}\}\{\backslash varepsilon\; \backslash mu\}\}\; \backslash ln\; \backslash left(\backslash frac\; \{\backslash mu\; R\; ^\; \{2\}\}\{\backslash varepsilon\}\backslash right)\; \backslash ln\; \backslash left(\backslash frac\; \{M\; \_\; \{2\}\; d\; ^\; \{\backslash frac\; \{1\}\{4\}\}\}\{\backslash sqrt\; \{\backslash mu\; \backslash varepsilon\}\; \backslash beta\}\; \backslash ln\; \backslash left(\backslash frac\; \{\backslash mu\; R\; ^\; \{2\}\}\{\backslash varepsilon\}\backslash right)\backslash right),\; \backslash left(\backslash frac\; \{\backslash sigma^\; \{2\}\}\{\backslash mu\; \backslash varepsilon\}\backslash right)\; ^\; \{\backslash frac\; \{\backslash alpha\}\{2\; (\backslash alpha\; -\; 1)\}\}\; \backslash ln\; \backslash left(\backslash frac\; \{1\}\{\backslash beta\}\; \backslash left(\backslash frac\; \{\backslash sigma^\; \{2\}\}\{\backslash mu\; \backslash varepsilon\}\backslash right)\; ^\; \{\backslash frac\; \{\backslash alpha\}\{2\; (\backslash alpha\; -\; 1)\}\}\; \backslash ln\; \backslash left(\backslash frac\; \{\backslash mu\; R\; ^\; \{2\}\}\{\backslash varepsilon\}\backslash right)\backslash right)\; \backslash right\backslash \}\backslash right)\; \backslash tag\; \{38\}$$

oracle calls. Moreover, with probability $\geq 1 - \beta$ the iterates of R-clipped-SSTM at stage $t$ stay in the ball $B_{2R_{t-1}}(x^*)$ .

Proof. The proof itself repeats of the proof Theorem F.3 from [35]. In this theorem authors prove convergence of restarted clipped-SSTM. In our case it is sufficient to change clipped-SSTM to ZO-clipped-SSTM and put results of Theorem 3 in order to guarantee $\varepsilon$ -solution after $\sum_{t=1}^{N} K_t$ successive iterations.

It remains to calculate the overall number of oracle calls during all runs of clipped-SSTM. We have

$$∑t=1NBtKt==O(∑t=1Nmax⁡{M22dRt−12εt2ln⁡(M22dRt−12Nεtβ),1B(σRt−1εt)αα−1ln⁡(NBβ(σRt−1εt)αα−1)})=O(∑t=1Nmax⁡{M22dRt−12μ2ln⁡(M22dNμRt−1β),(σμRt−1)αα−1ln⁡(Nβ(σμRt−1)αα−1)})=O(max⁡{∑t=1N2t/2M22dR2μ2ln⁡(2t/2M22dNμRβ),∑t=1N(σ⋅2t/2μR)αα−1ln⁡(Nβ(σ⋅2t/2μR)αα−1)})=O(max⁡{M22dR2μ22N/2ln⁡(2N/2M22dμRβln⁡(μR2ε)),(σμR)αα−1ln⁡(Nβ(σ⋅2N/2μR)αα−1)∑t=1N2αt2(α−1)})=O(max⁡{M22dεμln⁡(M22dεμβln⁡(μR2ε)),(σμR)αα−1ln⁡(Nβ(σμR)αα−1⋅2α2(α−1))2αN2(α−1)})=O(max⁡{M22dεμln⁡(M22dεμβln⁡(μR2ε)),(σ2με)α2(α−1)ln⁡(1β(σ2με)α2(α−1)ln⁡(μR2ε))}), \begin{array}{l} \sum_ {t = 1} ^ {N} B _ {t} K _ {t} = \\ = \mathcal {O} \left(\sum_ {t = 1} ^ {N} \max \left\{\sqrt {\frac {M _ {2} ^ {2} \sqrt {d} R _ {t - 1} ^ {2}}{\varepsilon_ {t} ^ {2}}} \ln \left(\frac {\sqrt {M _ {2} ^ {2} \sqrt {d} R _ {t - 1} ^ {2}} N}{\varepsilon_ {t} \beta}\right), \frac {1}{B} \left(\frac {\sigma R _ {t - 1}}{\varepsilon_ {t}}\right) ^ {\frac {\alpha}{\alpha - 1}} \ln \left(\frac {N}{B \beta} \left(\frac {\sigma R _ {t - 1}}{\varepsilon_ {t}}\right) ^ {\frac {\alpha}{\alpha - 1}}\right) \right\}\right) \\ = \mathcal {O} \left(\sum_ {t = 1} ^ {N} \max \left\{\sqrt {\frac {M _ {2} ^ {2} \sqrt {d}}{R _ {t - 1} ^ {2} \mu^ {2}}} \ln \left(\frac {\sqrt {M _ {2} ^ {2} \sqrt {d}} N}{\mu R _ {t - 1} \beta}\right), \left(\frac {\sigma}{\mu R _ {t - 1}}\right) ^ {\frac {\alpha}{\alpha - 1}} \ln \left(\frac {N}{\beta} \left(\frac {\sigma}{\mu R _ {t - 1}}\right) ^ {\frac {\alpha}{\alpha - 1}}\right) \right\}\right) \\ = \mathcal {O} \left(\max \left\{\sum_ {t = 1} ^ {N} 2 ^ {t / 2} \sqrt {\frac {M _ {2} ^ {2} \sqrt {d}}{R ^ {2} \mu^ {2}}} \ln \left(2 ^ {t / 2} \frac {\sqrt {M _ {2} ^ {2} \sqrt {d}} N}{\mu R \beta}\right), \sum_ {t = 1} ^ {N} \left(\frac {\sigma \cdot 2 ^ {t / 2}}{\mu R}\right) ^ {\frac {\alpha}{\alpha - 1}} \ln \left(\frac {N}{\beta} \left(\frac {\sigma \cdot 2 ^ {t / 2}}{\mu R}\right) ^ {\frac {\alpha}{\alpha - 1}}\right) \right\}\right) \\ = \mathcal {O} \left(\max \left\{\sqrt {\frac {M _ {2} ^ {2} \sqrt {d}}{R ^ {2} \mu^ {2}}} 2 ^ {N / 2} \ln \left(2 ^ {N / 2} \frac {\sqrt {M _ {2} ^ {2} \sqrt {d}}}{\mu R \beta} \ln \left(\frac {\mu R ^ {2}}{\varepsilon}\right)\right), \left(\frac {\sigma}{\mu R}\right) ^ {\frac {\alpha}{\alpha - 1}} \ln \left(\frac {N}{\beta} \left(\frac {\sigma \cdot 2 ^ {N / 2}}{\mu R}\right) ^ {\frac {\alpha}{\alpha - 1}}\right) \sum_ {t = 1} ^ {N} 2 ^ {\frac {\alpha t}{2 (\alpha - 1)}} \right\}\right) \\ = \mathcal {O} \left(\max \left\{\sqrt {\frac {M _ {2} ^ {2} \sqrt {d}}{\varepsilon \mu}} \ln \left(\frac {\sqrt {M _ {2} ^ {2} \sqrt {d}}}{\sqrt {\varepsilon \mu} \beta} \ln \left(\frac {\mu R ^ {2}}{\varepsilon}\right)\right), \left(\frac {\sigma}{\mu R}\right) ^ {\frac {\alpha}{\alpha - 1}} \ln \left(\frac {N}{\beta} \left(\frac {\sigma}{\mu R}\right) ^ {\frac {\alpha}{\alpha - 1}} \cdot 2 ^ {\frac {\alpha}{2 (\alpha - 1)}}\right) 2 ^ {\frac {\alpha N}{2 (\alpha - 1)}} \right\}\right) \\ = \mathcal {O} \left(\max \left\{\sqrt {\frac {M _ {2} ^ {2} \sqrt {d}}{\varepsilon \mu}} \ln \left(\frac {\sqrt {M _ {2} ^ {2} \sqrt {d}}}{\sqrt {\varepsilon \mu} \beta} \ln \left(\frac {\mu R ^ {2}}{\varepsilon}\right)\right), \left(\frac {\sigma^ {2}}{\mu \varepsilon}\right) ^ {\frac {\alpha}{2 (\alpha - 1)}} \ln \left(\frac {1}{\beta} \left(\frac {\sigma^ {2}}{\mu \varepsilon}\right) ^ {\frac {\alpha}{2 (\alpha - 1)}} \ln \left(\frac {\mu R ^ {2}}{\varepsilon}\right)\right) \right\}\right), \\ \end{array} $$

which concludes the proof.