Unifying mirror descent and dual averaging

Abstract

We introduce and analyze a new family of first-order optimization algorithms which generalizes and unifies both mirror descent and dual averaging. Within the framework of this family, we define new algorithms for constrained optimization that combines the advantages of mirror descent and dual averaging. Our preliminary simulation study shows that these new algorithms significantly outperform available methods in some situations.

Appendices

Convex analysis tools

Definition 9

(Lower-semicontinuity) A function $g:{\mathbb {R}}^n\rightarrow {\mathbb {R}}\cup \{+\infty \}$ is lower-semicontinuous if for all $c\in {\mathbb {R}}$, the sublevel set $\left\{ x\in {\mathbb {R}}^n\,:f(x)\leqslant c \right\}$ is closed.

One can easily check that the sum of two lower-semicontinuous functions is lower-semicontinuous. Continuous functions and characteristic functions $I_{{{\mathcal {X}}}}$ of closed sets ${{\mathcal {X}}}\subset {\mathbb {R}}^n$ are examples of lower-semicontinuous functions.

Definition 10

(Strong-convexity) Let $g:{\mathbb {R}}^n\rightarrow {\mathbb {R}}\cup \{+\infty \}$, $\left\| \,\cdot \,\right\| _{}$ be a norm in ${\mathbb {R}}^n$ and $K > 0$. Function g is said to be strongly convex with modulus $\kappa$ with respect to norm $\left\| \,\cdot \,\right\| _{}$ if for all $x,x'\in {\mathbb {R}}^n$ and $\lambda \in \left[ 0,1 \right]$,

$$\begin{aligned} g(\lambda x+(1-\lambda )x')\leqslant \lambda g(x)+(1-\lambda )g(x')-\frac{\kappa \lambda (1-\lambda )}{2}\left\| x'-x \right\| _{}^2. \end{aligned}$$

Proposition 12

(Theorem 23.5 in [38]) Let $g:{\mathbb {R}}^n\rightarrow {\mathbb {R}}\cup \{+\infty \}$ be a lower-semicontinuous convex function with nonempty domain. Then for all $x,y\in {\mathbb {R}}^n$, the following statements are equivalent.

• (i) $x\in \partial g^*(y)$;

• (ii) $y\in \partial g(x)$;

• (iii) $\langle y | x \rangle =g(x)+g^*(y)$;

• (iv) $x\in {{\,\mathrm{Arg\,max}\,}}_{x'\in {\mathbb {R}}^n}\left\{ \langle y | x' \rangle - g(x')\right\}$;

• (v) $y\in {{\,\mathrm{Arg\,max}\,}}_{y'\in {\mathbb {R}}^n}\left\{ \langle y' | x \rangle - g^*(y')\right\}$.

Postponed proofs

1.1 Proofs for Section 2

1.1.1 Proof of Proposition 1

Let ${\vartheta }\in {\mathbb {R}}^n$. By property (iii) from Definition 1, there exists $x_1\in {\mathcal {D}}_F$ such that $\nabla F(x_1)={\vartheta }$. Therefore, function $\varphi _{{\vartheta }}:x\mapsto \langle {\vartheta }| x \rangle -F(x)$ is differentiable at $x_1$ and $\nabla \varphi _{{\vartheta }}(x_1)=0$. Moreover, $\varphi _{{\vartheta }}$ is strictly concave as a consequence of property (i) from Definition 1. Therefore, $x_1$ is the unique maximizer of $\varphi _{{\vartheta }}$ and:

$$\begin{aligned} F^*({\vartheta })=\max _{x\in {\mathbb {R}}^n}\left\{ \langle {\vartheta }| x \rangle -F(x) \right\} <+\infty , \end{aligned}$$

which proves property (i).

Besides, we have

$$\begin{aligned} x_1\in \partial F^*({\vartheta }) \quad \Longleftrightarrow \quad {\vartheta }=\nabla F(x_1) \quad \Longleftrightarrow \quad x_1\text { minimizer of }\phi _{{\vartheta }}, \end{aligned}$$

(18)

where the first equivalence comes from Proposition 12. Point $x_1$ being the unique maximizer of $\varphi _{{\vartheta }}$, we have that $\partial F^*({\vartheta })$ is a singleton. In other words, $F^*$ is differentiable in ${\vartheta }$ and

$$\begin{aligned} \nabla F^*({\vartheta })=x_1\in {\mathcal {D}}_F. \end{aligned}$$

(19)

First, the above (19) proves property (ii). Second, this equality combined with the equality from (18) gives the second identity from property (iv). Third, this proves that $\nabla F^*({\mathbb {R}}^n)\subset {\mathcal {D}}_F$.

It remains to prove the reverse inclusion to get property (iii). Let $x\in {\mathcal {D}}_F$. By property (ii) from Definition 1, F is differentiable in x. Consider

$$\begin{aligned} {\vartheta }:=\nabla F(x), \end{aligned}$$

(20)

and all the above holds with this special point ${\vartheta }$. In particular, $x_1=x$ by uniqueness of $x_1$. Therefore (19) gives

$$\begin{aligned} \nabla F^*({\vartheta })=x, \end{aligned}$$

(21)

and this proves $\nabla F^*({\mathbb {R}}^n)\supset {\mathcal {D}}_F$ and thus property (iii). Combining (20) and (21) gives the first identity from property (iv).

1.1.2 Proof of Theorem 1

Let $x_0\in {\mathcal {D}}_F$. By definition of the mirror map, F is differentiable at $x_0$. Therefore, $D_F(x,x_0)$ is well-defined for all $x\in {\mathbb {R}}^n$.

For all real value $\alpha \in {\mathbb {R}}$, consider the sublevel set $S_{{{\mathcal {X}}}}(\alpha )$ of function $x\mapsto D_F(x,x_0)$ associated with value $\alpha$ and restricted to ${{\mathcal {X}}}$:

$$\begin{aligned} S_{{{\mathcal {X}}}}(\alpha ):=\left\{ x\in {{\mathcal {X}}}\,:D_F(x,x_0)\leqslant \alpha \right\} . \end{aligned}$$

Inheriting properties from F, function $D_F(\,\cdot \,,x_0)$ is lower-semicontinuous and strictly convex: consequently, the sublevel sets $S_{{{\mathcal {X}}}}(\alpha )$ are closed and convex.

Let us also prove that the sublevel sets $S_{{{\mathcal {X}}}}(\alpha )$ are bounded. For each value $\alpha \in {\mathbb {R}}$, we write

$$\begin{aligned} S_{{{\mathcal {X}}}}(\alpha )\subset S_{{\mathbb {R}}^n}(\alpha ):=\left\{ x\in {\mathbb {R}}^n\,:\,D_F(x,x_0)\leqslant \alpha \right\} \end{aligned}$$

and aim at proving that the latter set is bounded. By contradiction, let us suppose that there exists an unbounded sequence in $S_{{\mathbb {R}}^n}(\alpha )$: let $(x_k)_{k\geqslant 1}$ be such that $0<\left\| x_k-x_0 \right\| _{}\xrightarrow [k \rightarrow +\infty ]{}+\infty$ and $D_F(x_k,x_0)\leqslant \alpha$ for all $k\geqslant 1$. Using the Bolzano–Weierstrass theorem, there exists $v\ne 0$ and a subsequence $(x_{\phi (k)})_{k\geqslant 1}$ such that

$$\begin{aligned} \frac{x_{\phi (k)}-x_0}{\left\| x_{\phi (k)}-x_0 \right\| }\xrightarrow [k \rightarrow +\infty ]{}v. \end{aligned}$$

The point $x_0+\frac{x_{\phi (k)}-x_0}{\left\| x_{\phi (k)}-x_0 \right\| }$ being a convex combination of $x_0$ and $x_{\phi (k)}$, we can write the corresponding convexity inequality for function $D_F(\,\cdot \,,x_0)$:

$$\begin{aligned} D_F\left( x_0+\lambda _k(x_{\phi (k)}-x_0),x_0 \right)&\leqslant (1-\lambda _k)D_F(x_0,x_0) +\lambda _kD_F(x_{\phi (k)},x_0 )\\&\leqslant \lambda _k\alpha \xrightarrow [k \rightarrow +\infty ]{}0, \end{aligned}$$

where we used shorthand $\lambda _k:=\left\| x_{\phi (k)}-x_0 \right\| ^{-1}$. For the first above inequality, we used $D_F(x_0,x_0)=0$ and that $D_F(x_{\phi (k)},x_0)\leqslant \alpha$ by definition of $(x_k)_{k\geqslant 1}$. Then, using the lower-semicontinuity of $D_F(\,\cdot \,,x_0)$ and the fact that $x_0+\lambda _k(x_{\phi (k)}-x_0) \xrightarrow [k \rightarrow +\infty ]{}x_0+v$, we have

$$\begin{aligned} D_F(x_0+v,x_0)\leqslant \liminf _{k\rightarrow +\infty }D_F(x_0+\lambda _k(x_{\phi (k)}-x_0),x_0)\leqslant \liminf _{k\rightarrow +\infty }\lambda _k\alpha =0. \end{aligned}$$

The Bregman divergence of a convex function being nonnegative, the above implies $D_F(x_0+v,x_0)=0$. Thus, function $D_F(\,\cdot \,,x_0)$ attains its minimum (0) at two different points (at $x_0$ and at $x_0+v$): this contradicts its strong convexity. Therefore, sublevel sets $S_{{{\mathcal {X}}}}(\alpha )$ are bounded and thus compact.

We now consider the value $\alpha _{\text {inf}}$ defined as

$$\begin{aligned} \alpha _{\text {inf}}:=\inf \left\{ \alpha \,:S_{{{\mathcal {X}}}}(\alpha )\ne \emptyset \right\} . \end{aligned}$$

In other words, $\alpha _{\text {inf}}$ is the infimum value of $D_F(\,\cdot \,,x_0)$ on ${{\mathcal {X}}}$, and thus the only possible value for the minimum (if it exists). We know that $\alpha _{\text {inf}} \geqslant 0$ because the Bregman divergence is always nonnegative. From the definition of the sets $S_{{{\mathcal {X}}}}(\alpha )$, it easily follows that:

$$\begin{aligned} S_{{{\mathcal {X}}}}(\alpha _{\text {inf}})=\bigcap _{\alpha > \alpha _{\text {inf}}}^{}S_{{{\mathcal {X}}}}(\alpha ). \end{aligned}$$

Naturally, the sets $S_{{{\mathcal {X}}}}(\alpha )$ are increasing in $\alpha$ with respect to the inclusion order. Therefore, $S_{{{\mathcal {X}}}}(\alpha _{\text {inf}})$ is the intersection of a nested sequence of nonempty compact sets. It is thus nonempty as well by Cantor’s intersection theorem. Consequently, $D_F(\,\cdot \,,x_0)$ does admit a minimum on ${{\mathcal {X}}}$, and the minimizer is unique because of the strong convexity.

Let us now prove that the minimizer $x_*:=\mathop {{{\,\mathrm{arg\,min}\,}}}\limits _{x\in {{\mathcal {X}}}}D_F(x,\ x_0)$ also belongs to ${\mathcal {D}}_F$. Let us assume by contradiction that $x_*\in {{\mathcal {X}}}{\setminus } {\mathcal {D}}_F$. By definition of the mirror map, ${{\mathcal {X}}}\cap {\mathcal {D}}_F$ is nonempty; let $x_1\in {{\mathcal {X}}}\cap {\mathcal {D}}_F$. The set ${\mathcal {D}}_F$ being open by definition, there exists $\varepsilon > 0$ such that the closed Euclidean ball ${\overline{B}}(x_1,\varepsilon )$ centered in $x_1$ and of radius $\varepsilon$ is a subset of ${\mathcal {D}}_F$. We consider the convex hull

$$\begin{aligned} {\mathcal {C}}:={\text {co}}\left( \left\{ x_* \right\} \cup {\overline{B}}(x_1,\varepsilon ) \right) , \end{aligned}$$

which is clearly is a compact set.

Consider function G defined by:

$$\begin{aligned} G(x):=D_F(x,x_0)=F(x)-F(x_0)-\left\langle \nabla F(x_0) \vert x-x_0 \right\rangle , \end{aligned}$$

so that $x_*$ is the minimizer of G on ${{\mathcal {X}}}$. In particular, G is finite in $x_*$. G inherits strict convexity, lower-semicontinuity, and differentiability on ${\mathcal {D}}_F$ from function F. G is continuous on the compact set ${\overline{B}}(x_1,\varepsilon )$ because G is convex on the open set ${\mathcal {D}}_F\supset {\overline{B}}(x_1,\varepsilon )$. Therefore, G is bounded on ${\overline{B}}(x_1,\varepsilon )$. Let us prove that G is also bounded on ${\mathcal {C}}$. Let $x\in {\mathcal {C}}$. By definition of ${\mathcal {C}}$, there exists $\lambda \in [0,1]$ and $x'\in {\overline{B}}(x_1,\varepsilon )$ such that $x=\lambda x_*+(1-\lambda )x'$. By convexity of G, we have:

$$\begin{aligned} G(x)\leqslant \lambda G(x_*)+(1-\lambda )G(x')\leqslant G(x_*)+G(x'). \end{aligned}$$

We know that $G(x_*)$ is finite and that $G(x')$ is bounded for $x'\in {\overline{B}}(x_1,\varepsilon )$. Therefore G is bounded on ${\mathcal {C}}$: let us denote $G_{\text {max}}$ and $G_{\text {min}}$ some upper and lower bounds for the value of G on ${\mathcal {C}}$.

Because ${{\mathcal {X}}}$ is a convex set, the segment $[x_*,x_1]$ (in other words the convex hull of $\left\{ x_*,x_1 \right\}$) is a subset of ${{\mathcal {X}}}$. Besides, let us prove that the set

$$\begin{aligned} (x_*,x_1]:=\left\{ (1-\lambda )x_*+\lambda x_1\,:\lambda \in (0,1] \right\} \end{aligned}$$

is a subset of ${\mathcal {D}}_F$. Let $x_{\lambda }:=(1-\lambda )x_*+\lambda x_1$ (with $\lambda \in (0,1]$) a point in the above set, and let us prove that it belongs to ${\mathcal {D}}_F$. By definition of the mirror map, we have ${{\mathcal {X}}}\subset {\text {cl}}{\mathcal {D}}_F$, and besides $x_*\in {{\mathcal {X}}}$ by definition. Therefore, there exists a sequence $(x_k)_{k\geqslant 1}$ in ${\mathcal {D}}_F$ such that $x_k\rightarrow x_*$ as $k\rightarrow +\infty$. Then, we can write

$$\begin{aligned} x_{\lambda }&=(1-\lambda )x_*+\lambda x_1\\&=(1-\lambda )x_k + (1-\lambda )(x_*-x_k)+\lambda x_1\\&=(1-\lambda )x_k+\lambda \left( x_1+\frac{1-\lambda }{\lambda }(x_*-x_k) \right) . \end{aligned}$$

Since $x_k\rightarrow x_*$, for high enough k, the point $x_1+(1-\lambda )\lambda ^{-1}(x_*-x_k)$ belongs to ${\overline{B}}(x_1,\varepsilon )$ and therefore to ${\mathcal {D}}_F$. Then, the point $x_{\lambda }$ belongs to the convex set¹¹${\mathcal {D}}_F$ as the convex combination of two points in ${\mathcal {D}}_F$. Therefore, $(x_*,x_1]$ is indeed a subset of ${\mathcal {D}}_F$.

G being differentiable on ${\mathcal {D}}_F$ by definition of the mirror map, the gradient of G exists at each point of $(x_*,x_1]$. Let us prove that $\nabla G$ is bounded on $(x_*,x_1]$. Let $x_{\lambda }\in (x_*,x_1]$, where $\lambda \in (0,1]$ is such that

$$\begin{aligned} x_{\lambda }= (1-\lambda )x_*+\lambda x_1, \end{aligned}$$

and let $u\in {\mathbb {R}}^n$ such that $\left\| u \right\| _2=1$. The point $x_1+\varepsilon u$ belongs to ${\mathcal {C}}$ because it belongs to ${\overline{B}}(x_1,\varepsilon )$. The following point also belongs to convex set ${\mathcal {C}}$ as the convex combination of $x_*$ and $x_1+\varepsilon u$ which both belong to ${\mathcal {C}}$:

$$\begin{aligned} x_{\lambda }+\lambda \varepsilon u = (1-\lambda )x_*+\lambda (x_1+\varepsilon u)\in {\mathcal {C}}. \end{aligned}$$

(22)

Let $h\in (0,\varepsilon ]$. The following point also belongs to ${\mathcal {C}}$ as a convex combination of $x_{\lambda }$ and the above point $x_{\lambda }+\lambda \varepsilon u$:

$$\begin{aligned} x_{\lambda }+\lambda hu = \left( 1-\frac{h}{\varepsilon } \right) x_{\lambda }+\frac{h}{\varepsilon }\left( x_{\lambda }+\lambda \varepsilon u \right) \in {\mathcal {C}}. \end{aligned}$$

(23)

Now using for G the convexity inequality associated with the convex combination from (23), we write:

$$\begin{aligned} G(x_{\lambda }+h\lambda u)-G(x_{\lambda })&\leqslant \frac{h}{\varepsilon }\left( G(x_{\lambda }+\lambda \varepsilon u)-G(x_{\lambda }) \right) \nonumber \\&=\frac{h}{\varepsilon }\left( G(x_{\lambda }+\lambda \varepsilon u)-G(x_*)+G(x_*)-G(x_{\lambda }) \right) \nonumber \\&\leqslant \frac{h}{\varepsilon }\left( G(x_{\lambda }+\lambda \varepsilon u)-G(x_*) \right) , \end{aligned}$$

(24)

where for the last line we used $G(x_*)\leqslant G(x_{\lambda })$ which is true because $x_{\lambda }$ belongs to ${{\mathcal {X}}}$ and $x_*$ is by definition the minimizer of G on ${{\mathcal {X}}}$. Using the convexity inequality associated with the convex combination from (22), we also write

$$\begin{aligned} G(x_{\lambda }+\lambda \varepsilon u)-G(x_*)&\leqslant \lambda \left( G(x_1+\varepsilon u)-G(x_*) \right) \nonumber \\&\leqslant \lambda \left( G_{\text {max}}-G_{\text {min}} \right) . \end{aligned}$$

(25)

Combining (24) and (25) and dividing by $h\lambda$, we get

$$\begin{aligned} \frac{G(x_{\lambda }+h\lambda u)-G(x_{\lambda })}{h\lambda }\leqslant \frac{G_{\text {max}}-G_{\text {min}}}{\varepsilon }. \end{aligned}$$

Taking the limit as $h\rightarrow 0^+$, we get that $\langle \nabla G(x_{\lambda }) | u \rangle \leqslant (G_{\text {max}}-G_{\text {min}})/\varepsilon$. This being true for all vector u such that $\left\| u \right\| _2=1$, we have

$$\begin{aligned} \left\| \nabla G(x_{\lambda }) \right\| _{2}=\max _{\left\| u \right\| _{2}=1 }\langle \nabla G(x_{\lambda }) | u \rangle \leqslant \frac{G_{\text {max}}-G_{\text {min}}}{\varepsilon }. \end{aligned}$$

As a result, $\nabla G$ is bounded on $(x_*,x_1]$.

Let us deduce that $\partial G(x_*)$ is nonempty. The sequence $(\nabla G(x_{1/k}))_{k\geqslant 1}$ is bounded. Using the Bolzano–Weierstrass theorem, there exists a subsequence $(\nabla G(x_{1/\phi (k)}))_{k\geqslant 1}$ which converges to some vector ${\vartheta }_*\in {\mathbb {R}}^n$. For each $k\geqslant 1$, the following is satisfied by convexity of G:

$$\begin{aligned} \langle \nabla G(x_{1/\phi (k)}) | x-x_{1/\phi (k)} \rangle \leqslant G(x)-G(x_{1/\phi (k)}),\quad x\in {\mathbb {R}}^n. \end{aligned}$$

Taking the limsup on both sides for each $x\in {\mathbb {R}}^n$ as $k\rightarrow +\infty$, we get (because obviously $x_{1/\phi (k)}\rightarrow x_*$):

$$\begin{aligned} \langle {\vartheta }_* | x-x_* \rangle \leqslant G(x)-\liminf _{k\rightarrow +\infty }G(x_{1/\phi (k)})\leqslant G(x)-G(x_*),\quad x\in {\mathbb {R}}^n, \end{aligned}$$

where the second inequality follows from the lower-semicontinuity of G. Consequently, ${\vartheta }_*$ belongs to $\partial G(x_*)$.

But by definition of the mirror map $\nabla F$ takes all possible values and so does $\nabla G$, because it follows from the definition of G that $\nabla G=\nabla F-\nabla F(x_0)$. Therefore, there exists a point ${\tilde{x}}\in {\mathcal {D}}_F$ (thus ${\tilde{x}} \ne x_*$) such that $\nabla G({\tilde{x}})={\vartheta }_*$. Considering the point $x_{\text {mid}}=\frac{1}{2}(x_*+{\tilde{x}})$, we can write the following convexity inequalities:

$$\begin{aligned} \langle {\vartheta }_* | x_{\text {mid}}-x_* \rangle&\leqslant G(x_{\text {mid}})-G(x_*)\\ \langle {\vartheta }_* | x_{\text {mid}}-{\tilde{x}} \rangle&\leqslant G(x_{\text {mid}})-G({\tilde{x}}). \end{aligned}$$

We now add both inequalities and use the fact that $x_{\text {mid}}-{\tilde{x}}=x_*-x_{\text {mid}}$ by definition of $x_{\text {mid}}$ to get $0\leqslant 2G(x_{\text {mid}})-G(x_*)-G({\tilde{x}})$, which can also be written

$$\begin{aligned} G\left( \frac{x_*+{\tilde{x}}}{2} \right) \geqslant \frac{G(x_*)+G({\tilde{x}})}{2}, \end{aligned}$$

which contradicts the strong convexity of G. We conclude that $x_*\in {\mathcal {D}}_F$.

1.1.3 Proof of Proposition 2

Let ${\vartheta }\in {\mathbb {R}}^n$. For each of the three assumptions, let us prove that $h^*({\vartheta })$ is finite. This will prove that ${\text {dom}}h^*={\mathbb {R}}^n$.

• (i) Because ${\text {cl}}{\text {dom}}h={{\mathcal {X}}}$ by definition of a pre-regularizer, we have:

  $$\begin{aligned} h^*({\vartheta })=\max _{x\in {\mathbb {R}}^n}\left\{ \left\langle {\vartheta } \vert x \right\rangle -h(x) \right\} =\max _{x\in {{\mathcal {X}}}}\left\{ \left\langle {\vartheta } \vert x \right\rangle -h(x) \right\} . \end{aligned}$$

  Besides, the function $x\mapsto \left\langle {\vartheta } \vert x \right\rangle -h(x)$ is upper-semicontinuous and therefore attains a maximum on ${{\mathcal {X}}}$ because ${{\mathcal {X}}}$ is assumed to be compact. Therefore $h^*({\vartheta })<+\infty$.

• (ii) Because $\nabla h({\mathcal {D}}_h)={\mathbb {R}}^n$ by assumption, there exists $x\in {\mathcal {D}}_h$ such that $\nabla h(x)={\vartheta }$. Then, by Proposition 12, $h^*({\vartheta })=\left\langle {\vartheta } \vert x \right\rangle -h(x)<+\infty$.

• (iii) The function $x\mapsto \left\langle {\vartheta } \vert x \right\rangle -h(x)$ is strongly concave on ${\mathbb {R}}^n$ and therefore admits a maximum. Therefore, $h^*({\vartheta })<+\infty$.

1.1.4 Proof of Proposition 3

Let ${\vartheta }\in {\mathbb {R}}^n$. Because ${\text {dom}}h^*={\mathbb {R}}^n$, the subdifferential $\partial h^*({\vartheta })$ is nonempty—see e.g. [38,  Theorem 23.4]. By Proposition 12, $\partial h^*({\vartheta })$ is the set of maximizers of function $x\mapsto \left\langle {\vartheta } \vert x \right\rangle -h(x)$, which is strictly concave. Therefore, the maximizer is unique and $h^*$ is differentiable at ${\vartheta }$.

Let $x\in {\mathcal {D}}_F$ and let us prove that $\nabla F(x)\in \partial h(x)$. By convexity of F, the following is true

$$\begin{aligned} \forall x'\in {\mathbb {R}}^n,\quad F(x')-F(x)\geqslant \langle \nabla F(x) | x'-x \rangle . \end{aligned}$$

By definition of h, we obviously have $h(x')\geqslant F(x')$ for all $x'\in {\mathbb {R}}^n$, and $h(x)=F(x)+I_{{{\mathcal {X}}}}(x)=F(x)$ because $x\in {{\mathcal {X}}}$. Therefore, the following is also true

$$\begin{aligned} \forall x'\in {\mathbb {R}}^n,\quad h(x')-h(x)\geqslant \langle \nabla F(x) | x'-x \rangle . \end{aligned}$$

In other words, $\nabla F(x)\in \partial f(x)$.

1.1.5 Proof of Proposition 4

h is strictly convex as the sum of two convex functions, one of which (F) is strictly convex. h is lower-semicontinuous as the sum of two lower-continuous functions.

Let us now prove that ${\text {cl}}{\text {dom}}h={{\mathcal {X}}}$. First, we write

$$\begin{aligned} {\text {dom}}h={\text {dom}}(F+I_{{{\mathcal {X}}}})={\text {dom}}F\cap {\text {dom}}I_{{{\mathcal {X}}}}={\text {dom}}F\cap {{\mathcal {X}}}. \end{aligned}$$

Let $x\in {\text {cl}}{\text {dom}}h={\text {cl}}({\text {dom}}F\cap {{\mathcal {X}}})$. There exists a sequence $(x_k)_{k\geqslant 1}$ in ${\text {dom}}F\cap {{\mathcal {X}}}$ such that $x_k\rightarrow x$. In particular, each $x_k$ belongs to closed set ${{\mathcal {X}}}$, and so does the limit: $x\in {{\mathcal {X}}}$.

Conversely, let $x\in {{\mathcal {X}}}$ and let us prove that $x\in {\text {cl}}({\text {dom}}F\cap {{\mathcal {X}}})$ by constructing a sequence $(x_k)_{k\geqslant 1}$ in ${\text {dom}}F\cap {{\mathcal {X}}}$ which converges to x. By definition of the mirror map, we have ${{\mathcal {X}}}\subset {\text {cl}}{\mathcal {D}}_F$, where ${\mathcal {D}}_F:={\text {int}}{\text {dom}}F$. Therefore, there exists a sequence $(x_l')_{l\geqslant 1}$ in ${\mathcal {D}}_F$ such that $x_l'\rightarrow x$ as $l\rightarrow +\infty$. From the definition of the mirror map, we also have that ${{\mathcal {X}}}\cap {\mathcal {D}}_F\ne \emptyset$. Let $x_0\in {{\mathcal {X}}}\cap {\mathcal {D}}_F$. In particular, $x_0$ belongs ${\mathcal {D}}_F$ which is an open set by definition. Therefore, there exists a neighborhood $U\subset {\mathcal {D}}_F$ of point $x_0$. We now construct the sequence $(x_k)_{k\geqslant 1}$ as follows:

$$\begin{aligned} x_k:=\left( 1-\frac{1}{k} \right) x+\frac{1}{k}x_0,\quad k\geqslant 1. \end{aligned}$$

$x_k$ belongs to ${{\mathcal {X}}}$ as the convex combination of two points in the convex set ${{\mathcal {X}}}$, and obviously converges to x. Besides, $x_k$ can also be written, for any $k,l\geqslant 1$,

$$\begin{aligned} x_k&= \left( 1-\frac{1}{k} \right) x'_l+\left( 1-\frac{1}{k} \right) (x-x_l')+\frac{1}{k}x_0 \\&=\left( 1-\frac{1}{k} \right) x_l'+ \frac{1}{k}\left( x_0+(k-1)(x-x_l') \right) \\&=\left( 1-\frac{1}{k} \right) x_l'+\frac{1}{k}x_{0}^{(kl)}, \end{aligned}$$

where we set $x_0^{(kl)}:=x_0+(k-1)(x-x_l')$. For a given $k\geqslant 1$, we see that $x_0^{(kl)}\rightarrow x_0$ as $l\rightarrow +\infty$ because $x_l'\rightarrow x$ by definition of $(x_l')_{l\geqslant 1}$. Therefore, for large enough l, $x_0^{(kl)}$ belongs to the neighborhood U and therefore to ${\mathcal {D}}_F$. $x_k$ then appears as the convex combination of $x_l'$ and $x_0^{(kl)}$ which both belong to the convex set ${\mathcal {D}}_F\subset {\text {dom}}F$. $(x_k)$ is thus a sequence in ${\text {dom}}F\cap {{\mathcal {X}}}$ which converges to x. Therefore, $x\in {\text {cl}}({\text {dom}}F\cap {{\mathcal {X}}})$ and h is an ${{\mathcal {X}}}$-pre-regularizer.

Finally, we have $F\leqslant h$ by definition of h. One can easily check that this implies $h^*\leqslant F^*$ and we know from Proposition 1 that ${\text {dom}}F^*={\mathbb {R}}^n$, in other words that $F^*$ only takes finite values. Therefore, so does $h^*$ and h is an ${{\mathcal {X}}}$-regularizer.

1.2 Proofs for Section 4

1.2.1 Proof of Proposition 11

Let $t\geqslant 2$. It follows from the definition of the iterates that $x_t-y_t=(\nu _{t-1}^{-1}-1)(y_t-y_{t-1})$. Therefore, utilizing the convexity of f, we get

$$\begin{aligned} \langle \gamma _tf'(y_t) | x_t-x_* \rangle&=\gamma _t\langle f'(y_t) | y_t-x_* \rangle +\gamma _t\langle f'(y_t) | x_t-y_t \rangle \\&=\gamma _t\langle f'(y_t) | y_t-x_* \rangle +\gamma _t(\nu _{t-1}^{-1}-1)\langle f'(y_t) | y_t-y_{t-1} \rangle \\&\geqslant \gamma _t\left( f(y_t)-f_* \right) +\gamma _t(\nu _{t-1}^{-1}-1)\left( f(y_t)-f(y_{t-1}) \right) \\&= \gamma _t\nu _{t-1}^{-1}f(y_t)-\gamma _t(\nu _{t-1}^{-1}-1)f(y_{t-1})-\gamma _tf_*. \end{aligned}$$

Besides this, for $t=1$, we have $\gamma _1\langle f'(y_1) | x_1-x_* \rangle \geqslant \gamma _1(f(y_1)-f_*)$ because $x_1=y_1$ by definition. Then, summing over $t=1,\dots ,T$, we obtain after simplifications:

$$\begin{aligned}&(\gamma _1-\gamma _2(\nu _1^{-1}-1))f(y_1)+\sum _{t=2}^{T-1}(\gamma _t\nu _{t-1}^{-1}-\gamma _{t+1}(\nu _t^{-1}-1))f(y_t)+\gamma _T\nu _{t-1}^{-1}f(y_T)\\&\quad -\left( \sum _{t=1}^T\gamma _t \right) f_*\leqslant \sum _{t=1}^T\langle \gamma _tf'(y_t) | x_t-x_* \rangle . \end{aligned}$$

Using the definition of coefficients $\nu _t$, the above left-hand side simplifies to result in the inequality

$$\begin{aligned} \left( \sum _{t=1}^T\gamma _t \right) \left( f(y_T)-f_* \right) \leqslant \sum _{t=1}^T\langle \gamma _tf'(y_t) | x_t-x_* \rangle . \end{aligned}$$

Finally, because $(x_t,{\vartheta }_t)_{t\geqslant 1}$ is a sequence of UMD$(h,\xi )$ iterates with dual increments $\xi :=(-\gamma _tf'(y_t))_{t\geqslant 1}$, the result then follows by applying inequality (8) from Corollary 2 and dividing by $\sum _{t=1}^T\gamma _t$. $\square$

1.2.2 Proof of Theorem 2

First, observe that whenever $\gamma _t\leqslant 1/L$, due to (11),

$$\begin{aligned} f(x_{t+1})-f(x_{t})-\left\langle \nabla f(x_t) \vert x_{t+1}-x_t \right\rangle\leqslant & {} {L\over 2}\Vert x_{t+1}-x_t\Vert ^2\nonumber \\ {}\leqslant & {} (2\gamma _t)^{-1}\Vert x_{t+1}-x_t\Vert ^2. \end{aligned}$$

(26)

Thus,

$$\begin{aligned} \gamma _t D_f(x_{t+1},x_t)&=\gamma _t[f(x_{t+1})-f(x_{t})-\left\langle f'(x_t) \vert x_{t+1}-x_t \right\rangle ]\\&\leqslant \frac{1}{2}\Vert x_{t+1}-x_t\Vert ^2\\&\leqslant D_h(x_{t+1},x_t;\ {\vartheta }_t). \end{aligned}$$

by the strong convexity of $D_h$. On the other hand, by (6) of Lemma 1, for any $x\in {{\mathcal {X}}}\cap {\text {dom}}h$,

$$\begin{aligned} D_h(x,x_{t+1}; {\vartheta }_{t+1})&\leqslant D_h(x,x_t;{\vartheta }_t)+\gamma _t\left\langle \nabla f(x_t) \vert x-x_{t+1} \right\rangle -D_h(x_{t+1},x_t; {\vartheta }_t)\\ [\text{ by } (12)]&\leqslant D_h(x,x_t;{\vartheta }_t)+\gamma _t\left\langle \nabla f(x_t) \vert x-x_{t} \right\rangle \\&\qquad -\gamma _t\left\langle \nabla f(x_t) \vert x_{t+1}-x_{t} \right\rangle -D_f(x_{t+1},x_t)\\ [\text{ due } \text{ to } (16)]&\leqslant D_h(x,x_t;{\vartheta }_t)+\gamma _t\left\langle \nabla f(x_t) \vert x-x_{t} \right\rangle -\gamma _t[f(x_{t+1}-f(x_t)]\\ [\text{ by } \text{ convexity } \text{ of } f]&\leqslant D_h(x,x_t;{\vartheta }_t)-\gamma _t(f(x_{t+1})-f(x)). \end{aligned}$$

Consequently, $\forall x\in {{\mathcal {X}}}\cap {\text {dom}}h$,

$$\begin{aligned} \gamma _t(f(x_{t+1})-f(x_t))\leqslant D_h(x,x_t;{\vartheta }_t)-D_h(x,x_{t+1}; {\vartheta }_{t+1}). \end{aligned}$$

When applying the above inequality to $x=x_t$ we conclude that

$$\begin{aligned} \gamma _t(f(x_{t+1})-f(x_t))\leqslant -D_h(x_t,x_{t+1}; {\vartheta }_{t+1})\leqslant 0. \end{aligned}$$

Finally, when setting $x=x_*$, we obtain

$$\begin{aligned} \left( \sum _{t=1}^T\gamma _t\right) (f(x_{T+1})-f_*)\leqslant \sum _{t=1}^T\gamma _t(f(x_{t+1})-f(x_*))\leqslant D_h(x_*,x_1;{\vartheta }_1) \end{aligned}$$

which implies (13). $\square$

1.2.3 Proof of Theorem 3

We start with the following technical result.

Lemma 2

Assume that positive step-sizes $\nu _t\in (0,1]$ and $\gamma _t>0$ are such that the relationship

$$\begin{aligned} f({z}_{t+1})\leqslant f(y_t)+\nu _t\left\langle \nabla f(y_t) \vert {x}_{t+1}-x_t \right\rangle +{\nu _t\over \gamma _t} D_h({x}_{t+1},{x}_t;{\vartheta }_t), \end{aligned}$$

(27)

holds for all t which is certainly the case if $\nu _t\gamma _t\leqslant L^{-1}$. Denote $s_t=f(z_t)-f_*$; then

$$\begin{aligned} {\gamma _t\nu _t^{-1}}(s_{t+1}-s_t)+\gamma _ts_t&\leqslant D_h(x_*,{x}_t;{\vartheta }_t)-D_h(x_*,{x}_{t+1};{\vartheta }_{t+1}). \end{aligned}$$

(28)

Proof of the lemma

Observe first that by construction,

$$\begin{aligned} {z}_{t+1}-y_t= (1-\nu _t){z}_t+\nu _t{x}_{t+1}-[(1-\nu _t){z}_t+\nu _t{x}_{t}]=\nu _t({x}_{t+1}-{x}_t) \end{aligned}$$

By strong convexity of h, for $\nu _t\gamma _t\leqslant L^{-1}$ we have

$$\begin{aligned} f({z}_{t+1})&\leqslant f(y_t)+\langle \nabla f(y_t),{z}_{t+1}-y_t\rangle +{L\over 2}\Vert {z}_{t+1}-y_t\Vert ^2\nonumber \\&= f(y_t)+\nu _t\langle \nabla f(y_t),{x}_{t+1}-x_t\rangle +{L\nu _t^2\over 2}\Vert {x}_{t+1}-{x}_t\Vert ^2\nonumber \\&\leqslant f(y_t)+\nu _t\langle \nabla f(y_t),{x}_{t+1}-x_t\rangle +{\nu _t\over \gamma _t}D_h({x}_{t+1},{x}_t;{\vartheta }_t), \end{aligned}$$

what is (27).

Next, observe that by (14a),

$$\begin{aligned} \nu _t(x_*-x_t)=(\nu _tx_*+(1-\nu _t)z_t)-y_t, \end{aligned}$$

whence, by convexity of f,

$$\begin{aligned} \nu _t\left\langle \nabla f(y_t) \vert x_*-x_t \right\rangle&=\left\langle \nabla f(y_t) \vert (\nu _tx_*+(1-\nu _t)z_t)-y_t \right\rangle \\&\leqslant f(\nu _tx_*+(1-\nu _t)z_t)-f(y_t)\\&\leqslant \nu _t(f(x_*)-f(y_t))+(1-\nu _t)(f(z_t)-f(y_t)). \end{aligned}$$

When substituting the latter bound into (27) we get

$$\begin{aligned} f({z}_{t+1})&\leqslant f(y_t)+\nu _t\left\langle \nabla f(y_t) \vert {x}_{t+1}-x_* \right\rangle +\nu _t(f(x_*)-f(y_t))\\ {}&\qquad +(1-\nu _t)(f(z_t)-f(y_t)) +{\nu _t\over \gamma _t} D_h({x}_{t+1},{x}_t;{\vartheta }_t), \end{aligned}$$

or

$$\begin{aligned} f({z}_{t+1})-f(z_t)\leqslant \nu _t\left\langle \nabla f(y_t) \vert {x}_{t+1}-x_* \right\rangle +\nu _t(f_*-f(z_t)) +{\nu _t\over \gamma _t} D_h({x}_{t+1},{x}_t;{\vartheta }_t). \end{aligned}$$

Now, because $(x_t,{\vartheta }_t)_{t\geqslant 1}$ is a sequence of UMD iterates, by (6) of Lemma 1,

$$\begin{aligned} \gamma _t\left\langle \nabla f(y_t) \vert {x}_{t+1}-x_* \right\rangle \leqslant D_h(x_*,{x}_t;{\vartheta }_t)-D_h(x_*,{x}_{t+1};{\vartheta }_{t+1})-D_h({x}_{t+1},{x}_t;{\vartheta }_t), \end{aligned}$$

and we arrive at

$$\begin{aligned} {\gamma _t\nu _t^{-1}}(f({z}_{t+1})-f(z_t))\leqslant D_h(x_*,{x}_t;{\vartheta }_t)-D_h(x_*,{x}_{t+1};{\vartheta }_{t+1})+\gamma _t(f_*-f(z_t)), \end{aligned}$$

what is (28). $\square$

Proof of the Theorem

Assume that $\nu _t$ and $\gamma _t$ satisfy

$$\begin{aligned} \nu _1=1,\quad \nu _t\in (0,1], \quad \gamma _{t+1}(\nu _{t+1}^{-1}-1)\leqslant \gamma _t\nu _{t}^{-1}. \end{aligned}$$

(29)

When summing (28) up from 1 to T we get

$$\begin{aligned} D_h(x_*,x_t;{\vartheta }_1)&\geqslant \sum _{t=1}^T[{\gamma _t\nu _t^{-1}}(s_{t+1}-s_t)+\gamma _ts_t]\\&={\gamma _T\nu _T^{-1}}s_{T+1}+\sum _{t=2}^Ts_t\left( {\gamma _{t-1}\nu _{t-1}^{-1}}-\gamma _t(\nu _t^{-1}-1)\right) -\gamma _1(\nu _1^{-1}-1)s_1\\&\quad \underbrace{\geqslant }_{[\text{ by } (29)]} {\gamma _T\nu _T^{-1}}s_{T+1}={\gamma _T\nu _T^{-1}}(f(z_{T+1})-f_*). \end{aligned}$$

It is clear that the choice of $\gamma _1=L^{-1}$, $\nu _1=1$ and $\nu _t=(\gamma _tL)^{-1}$ satisfies the relationship $\gamma _t\nu _t\leqslant L^{-1}$. In this case, when choosing step-sizes $(\gamma _t)_{t\geqslant 1}$ to saturate recursively the last relation in (29), specifically,

$$\begin{aligned} \gamma ^2_{t+1}L-\gamma _{t+1}= \gamma _t^2L \end{aligned}$$

we come to celebrated Nesterov step-sizes (15) which satisfy $\gamma _t\nu _t^{-1}\geqslant {(t+1)^2\over 4L}$, and we arrive at (16). $\square$