Convex analysis

Convex analysis is the branch of mathematics devoted to the study of properties of convex functions and convex sets, often with applications in convex minimization, a subdomain of optimization theory.

Convex sets

A subset $C\subseteq X$ of some vector space $X$ is called convex if it satisfies any of the following equivalent conditions:

If $0\leq r\leq 1$ is real and $x,y\in C$ then $rx+(1-r)y\in C.$ ^[1]
If $0<r<1$ is real and $x,y\in C$ with $x\neq y,$ then $rx+(1-r)y\in C.$
$(r+s)C=rC+sC$ for all positive real $r>0$ and $s>0.$ ^[2]

Convex functions

A convex function is any function $f:X\to [-\infty ,\infty ]$ valued in the extended real numbers $[-\infty ,\infty ]=\mathbb {R} \cup \{\pm \infty \}$ and defined on a convex set $X$ such that for any $x,y\in X$ and any real $0<r<1,$

f(rx+(1-r)y)\leq rf(x)+(1-r)f(y)

^[1]

which shows that these functions are exactly those that satisfies the hypothesis of Jensen's inequality. Equivalently, a convex function is any extended real valued function such that its epigraph

\operatorname {epi} f:=\left\{(x,r)\in X\times \mathbb {R} ~:~f(x)\leq r\right\}

is a convex set.^[3]

A function $f:X\to [-\infty ,\infty ]$ defined on a convex set is called strictly convex if for all $x,y\in X$ with $x\neq y$ and any real $0<r<1,$

f(rx+(1-r)y)<rf(x)+(1-r)f(y).

The domain of a function $f:X\to [-\infty ,\infty ]$ is denoted by $\operatorname {domain} f$ while its effective domain is the set

\operatorname {dom} f:=\{x\in X~:~f(x)<\infty \}.

^[3]

The function $f:X\to [-\infty ,\infty ]$ is called proper if $\operatorname {dom} f\neq \varnothing$ and $f(x)>-\infty$ for all $x\in \operatorname {domain} f.$ ^[3] Alternatively, this means that there exists some $x\in \operatorname {domain} f$ at which $f(x)\in \mathbb {R}$ and $f$ is also never equal to $-\infty .$ In words, a function is proper if its domain is not empty, it never takes on the value $-\infty ,$ and it also is not identically equal to $+\infty .$ If $f:\mathbb {R} ^{n}\to [-\infty ,\infty ]$ is a proper convex function then there exist some vector $b\in \mathbb {R} ^{n}$ and some $r\in \mathbb {R}$ such that

f(x)\geq x\cdot b-r

for every

x

where $x\cdot b$ denotes the dot product of these vectors.

Convex conjugate

The convex conjugate of an extended real-valued function $f:X\to [-\infty ,\infty ]$ (not necessarily convex) is the function $f^{*}:X^{*}\to [-\infty ,\infty ]$ from the (continuous) dual space $X^{*}$ of $X,$ and^[4]

f^{*}\left(x^{*}\right)=\sup _{z\in X}\left\{\left\langle x^{*},z\right\rangle -f(z)\right\}

where the brackets $\left\langle \cdot ,\cdot \right\rangle$ denote the canonical duality $\left\langle x^{*},z\right\rangle :=x^{*}(z).$ The biconjugate of $f$ is the map $f^{**}=\left(f^{*}\right)^{*}:X\to [-\infty ,\infty ]$ defined by $f^{**}(x):=\sup _{z^{*}\in X^{*}}\left\{\left\langle x,z^{*}\right\rangle -f\left(z^{*}\right)\right\}$ for every $x\in X.$ If $\operatorname {Func} (X;Y)$ denotes the set of $Y$ -valued functions on $X,$ then the map $\operatorname {Func} (X;[-\infty ,\infty ])\to \operatorname {Func} \left(X^{*};[-\infty ,\infty ]\right)$ defined by $f\mapsto f^{*}$ is called the Legendre-Fenchel transform.

Subdifferential set

If $f:X\to [-\infty ,\infty ]$ and $x\in X$ then the subdifferential set is

{\begin{alignedat}{4}\partial f(x):&=\left\{x^{*}\in X^{*}~:~f(z)\geq f(x)+\left\langle x^{*},z-x\right\rangle {\text{ for all }}z\in X\right\}&&({\text{“}}z\in X{\text{''}}{\text{ can be replaced with: }}{\text{“}}z\in X{\text{ such that }}z\neq x{\text{''}})\\&=\left\{x^{*}\in X^{*}~:~\left\langle x^{*},x\right\rangle -f(x)\geq \left\langle x^{*},z\right\rangle -f(z){\text{ for all }}z\in X\right\}&&\\&=\left\{x^{*}\in X^{*}~:~\left\langle x^{*},x\right\rangle -f(x)\geq \sup _{z\in X}\left\langle x^{*},z\right\rangle -f(z)\right\}&&{\text{ The right hand side is }}f^{*}\left(x^{*}\right)\\&=\left\{x^{*}\in X^{*}~:~\left\langle x^{*},x\right\rangle -f(x)=f^{*}\left(x^{*}\right)\right\}&&{\text{ Taking }}z:=x{\text{ in the }}\sup {}{\text{ gives the inequality }}\leq .\\\end{alignedat}}

For example, in the important special case where $f=\|\cdot \|$ is a norm on $X$ , it can be shown^{[proof 1]} that if $0\neq x\in X$ then this definition reduces down to:

\partial f(x)=\left\{x^{*}\in X^{*}~:~\left\langle x^{*},x\right\rangle =\|x\|{\text{ and }}\left\|x^{*}\right\|=1\right\}

and

\partial f(0)=\left\{x^{*}\in X^{*}~:~\left\|x^{*}\right\|\leq 1\right\}.

For any $x\in X$ and $x^{*}\in X^{*},$ $f(x)+f^{*}\left(x^{*}\right)\geq \left\langle x^{*},x\right\rangle ,$ which is called the Fenchel-Young inequality. This inequality is an equality (i.e. $f(x)+f^{*}\left(x^{*}\right)=\left\langle x^{*},x\right\rangle$ ) if and only if $x^{*}\in \partial f(x).$ It is in this way that the subdifferential set $\partial f(x)$ is directly related to the convex conjugate $f^{*}\left(x^{*}\right).$

Biconjugate

The biconjugate of a function $f:X\to [-\infty ,\infty ]$ is the conjugate of the conjugate, typically written as $f^{**}:X\to [-\infty ,\infty ].$ The biconjugate is useful for showing when strong or weak duality hold (via the perturbation function).

For any $x\in X,$ the inequality $f^{**}(x)\leq f(x)$ follows from the Fenchel–Young inequality. For proper functions, $f=f^{**}$ if and only if $f$ is convex and lower semi-continuous by Fenchel–Moreau theorem.^[4]^[5]

Convex minimization

A convex minimization (primal) problem is one of the form

find

\inf _{x\in M}f(x)

when given a convex function

f:X\to [-\infty ,\infty ]

and a convex subset

M\subseteq X.

Dual problem

In optimization theory, the duality principle states that optimization problems may be viewed from either of two perspectives, the primal problem or the dual problem.

In general given two dual pairs separated locally convex spaces $\left(X,X^{*}\right)$ and $\left(Y,Y^{*}\right).$ Then given the function $f:X\to [-\infty ,\infty ],$ we can define the primal problem as finding $x$ such that

\inf _{x\in X}f(x).

If there are constraint conditions, these can be built into the function $f$ by letting $f=f+I_{\mathrm {constraints} }$ where $I$ is the indicator function. Then let $F:X\times Y\to [-\infty ,\infty ]$ be a perturbation function such that $F(x,0)=f(x).$ ^[6]

The dual problem with respect to the chosen perturbation function is given by

\sup _{y^{*}\in Y^{*}}-F^{*}\left(0,y^{*}\right)

where $F^{*}$ is the convex conjugate in both variables of $F.$

The duality gap is the difference of the right and left hand sides of the inequality^[7]^[6]^[8]

\sup _{y^{*}\in Y^{*}}-F^{*}\left(0,y^{*}\right)\leq \inf _{x\in X}F(x,0).

This principle is the same as weak duality. If the two sides are equal to each other, then the problem is said to satisfy strong duality.

There are many conditions for strong duality to hold such as:

$F=F^{**}$ where $F$ is the perturbation function relating the primal and dual problems and $F^{**}$ is the biconjugate of $F$ ;^{[citation needed]}
the primal problem is a linear optimization problem;
Slater's condition for a convex optimization problem.^[9]^[10]

Lagrange duality

For a convex minimization problem with inequality constraints,

\min {}_{x}f(x)

subject to

g_{i}(x)\leq 0

for

i=1,\ldots ,m.

the Lagrangian dual problem is

\sup {}_{u}\inf {}_{x}L(x,u)