# Basics of Convex Analysis

We introduce here briefly a few important building blocks of convex optimization:

1. the notion of subgradient (aka. generalized gradient) and subdifferential, the first-order optimality condition

2. strict and strong convexity, positive-definite functions and a hint at Bregman divergences,

3. the convex conjugate.

## 1. Subgradient and First-order Optimality Condition

We say that $$y\in X$$ is a subgradient of the function $$f\in\Gamma_{0}(X)$$ at $$x\in X$$ and belongs to the subdifferential of $$f$$ at that point (denoted $$\partial f(x)$$) if it verifies the following inequality:

$\begin{eqnarray} f(z) &\ge & f(x) + \langle z-x, y \rangle, \qquad \forall z\in X.\label{subgradient inequality} \end{eqnarray}$

The inequality above just indicates that the graph of the function $$f$$ is supported by the hyperplane defined by the right-hand side. A subgradient is thus the ‘‘slope’’ of one such supporting hyperplane. If the function is differentiable at $$x$$ then there is only one such subgradient (the standard gradient) and correspondingly only one supporting hyperplane. However, if the function is not differentiable at $$x$$ (e.g., if there is a kink at $$x$$) then there is an infinity of hyperplanes supporting the function and correspondingly the subdifferential at that point is a set with a continuum of subgradients.
A classical example is the absolute value function $$f(x)=|x|$$ which is not differentiable at $$0$$. It is however supported at that point by all lines of the form $$\ell(x)=\alpha x$$ with $$\alpha\in [-1,1]$$. This set is precisely the subdifferential of the function at $$0$$, denoted $$\partial f(0)$$.

 Illustration of the function $$f(x)=|x|$$ (thick line) and of two supporting lines at the origin (dashed lines). Each of these supporting lines has slope in the subdifferential $$\partial f(0)$$. Note that the horizontal line is also a supporting hyperplane, illustrating that $$0\in\partial f(0)$$ and hence that the function has a minimizer at the origin by the first-order condition (cf. below).

Now, by definition of an optimal point $$x^{\sharp}$$ for the unconstrained problem, we must have $$f(z)\ge f(x^{\sharp})$$ for all $$z\in X$$. This can be written as

$\begin{eqnarray} f(z) &\ge& f(x^{\sharp}) + \langle z-x,0 \rangle, \qquad \forall z \in X,\nonumber \end{eqnarray}$

and hence $$0$$ must be a subgradient of $$f$$ at $$x^\sharp$$.

This is the first-order optimality condition (FOC):

$\begin{eqnarray} x^\sharp \,\in\, \arg\min_x \, f(x) &\Longleftrightarrow& 0\,\in\, \partial f(x^\sharp). \end{eqnarray}$

If we take the subdifferential as an operator then, intuitively, looking for a minimizer amounts to ‘‘inverting’’ the subdifferential and evaluating it at $$0$$, i.e.: $$x^\sharp = (\partial f)^{-1}(0)$$. We shall come back to this in more details but this idea of inverting an operator involving the subdifferential to find the minimizer is very important.

Before moving on, it is useful to note (and not too hard to convince oneself) that the following inclusion holds for the subdifferential of a sum:

$\begin{eqnarray} \sum_i \partial f_i &\subseteq& \partial \sum_i f_i. \end{eqnarray}$

For most problems of interest, it holds as an equality but note that if $$0\in \sum_i \partial f_i(x^\dagger)$$ then the above inclusion implies that $$0\in\partial \sum_i f_i(x^\dagger)$$ which is sufficient to show that $$x^\dagger$$ is a minimizer (which is what we are mostly interested in).

## 2. Strict and Strong convexity

The function $$\psi$$ is said to be strictly convex at $$x$$ if the subgradient inequality $$\eqref{subgradient inequality}$$ holds strictly for all $$z\in X\backslash \{x\}$$ i.e.,:

$\begin{eqnarray} \psi(z) &>& \psi(x) + \langle z-x , y\rangle, \quad\forall z\in X,\, z\neq x \end{eqnarray}$

where $$y\in\partial \psi(x)$$.

Recalling that a function $$d: X\times X$$ is said to be positive-definite if $$d(u,u)=0$$ and $$d(u,v)>0$$ for $$u\neq v$$, we can observe that we can use such a strictly convex function to define a positive-definite function $$B_{\psi}:X\times X\to \mathbb R^{+}$$ as:

$\begin{eqnarray} B_{\psi}(z,x) &:=& \psi(z)-\psi(x) - \langle z-x,y \rangle, \quad \text{with}\quad y\in\partial \psi(x). \label{Bregman A} \end{eqnarray}$

Equation $$\eqref{Bregman A}$$ actually defines Bregman divergences. We shall come back to that (e.g., in the notes on the Mirror Descent Algorithm).

The function $$\varphi$$ is said to be $$\mu$$-strongly convex with parameter $$\mu>0$$ if it is strictly convex and if the Bregman divergence associated to it is lower bounded by $$\mu$$ times the squared Euclidean ($$\ell^{2}$$) distance, i.e.:

$\begin{eqnarray} B_{\varphi}(x,y) &\ge& {\mu\over 2} {\|x-y\|^{2}_{2}}, \quad\forall x,y\in X. \end{eqnarray}$

The factor $$1/2$$ might seem irrelevant but makes other developments look nicer. For now, observe that if we take the derivative of the right-hand side, then we are left with $$(x-y)$$ without a spurious factor $$2$$…

Remark: note that, obviously, a strongly-convex function is also strictly convex.

## 3. Legendre-Fenchel convex conjugate

Let us consider again the definition of the subgradient of a convex function $$f$$ at a point $$x$$:

$\begin{eqnarray} \partial f(x) &=& \{y\,|\, f(z)\,\ge\, f(x)+\langle z-x,y\rangle, \,\forall z\}.\nonumber \end{eqnarray}$

Note that we can then rearrange terms in the condition as follows:

$\begin{eqnarray} \partial f(x) &=& \{y\,|\, \langle z,y\rangle - f(z)\,\le\, \langle x,y\rangle - f(x), \,\forall z\},\nonumber \end{eqnarray}$

but since the condition must hold for all $$z$$, it must equivalently hold for all $$z$$ that maximizes the lower bound. Note that the maximum of that lower bound tightens the inequality exactly (just take $$z=x$$). We can thus write the subgradient as

$\begin{eqnarray} \partial f(x) &=& \{y \,|\, \max_{z} \,\, [\langle z,y\rangle -f(z)] \,=\, \langle x,y\rangle -f(x)\}.\nonumber \end{eqnarray}$

This justifies the definition of the convex conjugate $$f^\star(y):X\to \overline{\mathbb R}$$ as follows:

$\begin{eqnarray} f^\star(y) &:=& \max_z \,\,[\langle z,y \rangle - f(z)]. \end{eqnarray}$

where $$\overline{\mathbb R}=\mathbb R \cup \{\pm\infty\}$$ is the extended real line.

The subgradient can then be expressed in terms of the convex conjugate as:

$\begin{eqnarray} \partial f(x) &=& \{ y \,|\, f^{\star}(y) \,=\, \langle x,y\rangle - f(x)\}. \end{eqnarray}$

A useful property of the convex conjugate for nice convex functions is that $$f^{\star\star}=f$$ (again, by ‘‘nice’’ we mean $$f\in \Gamma_0(X)$$). We give a sketch of a proof for this in more convex analysis.

We can consider a simple (and yet quite useful) example for the convex conjugate: if we define $$\psi(x):=\|x\|^2/2$$, its convex conjugate is then

$\begin{eqnarray} \psi^\star(y) &=& \arg \max_z\,\, \langle z,y\rangle - \frac12\langle z,z\rangle, \end{eqnarray}$

and the FOC leads immediately to $$\psi^\star(y)=\psi(y)$$.

The definition of the convex conjugate also directly implies Fenchel's inequality:

$\begin{eqnarray} f(x) + f^\star(y) &\ge & \langle x,y\rangle, \quad \forall x,y\in X. \end{eqnarray}$

## 4. A couple of comments

We use $$\max$$ and $$\min$$ everywhere but it would be more correct to use $$\sup$$ and $$\inf$$. In practice this point is usually irrelevant.
The notion of duality is rather important in convex analysis and, to be a bit more precise, we should note that a subgradient belongs to the dual space $$X^{\star}$$ and that the convex conjugate is actually defined on $$X^{\star}$$ as well. However, in most cases, $$X$$ is $$\mathbb R^n$$ which is self-dual and one can afford to drop making the distinction which makes the notations less cumbersome. Generalization to arbitrary Banach spaces is not difficult but requires a bit of care.

For more about all these technical details (and much more), standard phrase: please refer to Rockafellar's book :-).