Perturbation function

In

In some texts the value function is called the perturbation function, and the perturbation function is called the bifunction.^[2]

Definition

Given two

${\displaystyle \left(X,X^{*}\right)}$ and ${\displaystyle \left(Y,Y^{*}\right)}$. Then given the function ${\displaystyle f:X\to \mathbb {R} \cup \{+\infty \}}$, we can define the primal problem by

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

If there are constraint conditions, these can be built into the function ${\displaystyle f}$ by letting ${\displaystyle f\leftarrow f+I_{\mathrm {constraints} }}$ where ${\displaystyle I}$ is the characteristic function. Then ${\displaystyle F:X\times Y\to \mathbb {R} \cup \{+\infty \}}$ is a perturbation function if and only if ${\displaystyle F(x,0)=f(x)}$.^[1][3]

Use in duality

The duality gap is the difference of the right and left hand side of the inequality

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

where ${\displaystyle F^{*}}$ is the convex conjugate in both variables.^[3][4]

For any choice of perturbation function F

with ${\displaystyle 0\in \operatorname {core} ({\Pr }_{Y}(\operatorname {dom} F))}$ (where ${\displaystyle \operatorname {core} }$ is the algebraic interior and ${\displaystyle {\Pr }_{Y}}$ is the projection onto Y defined by ${\displaystyle {\Pr }_{Y}(x,y)=y}$) and X, Y are Fréchet spaces then strong duality holds.^[1]

Examples

Lagrangian

Let ${\displaystyle (X,X^{*})}$ and ${\displaystyle (Y,Y^{*})}$ be dual pairs. Given a primal problem (minimize f(x)) and a related perturbation function (F(x,y)) then the Lagrangian ${\displaystyle L:X\times Y^{*}\to \mathbb {R} \cup \{+\infty \}}$ is the negative conjugate of F with respect to y (i.e. the concave conjugate). That is the Lagrangian is defined by

${\displaystyle L(x,y^{*})=\inf _{y\in Y}\left\{F(x,y)-y^{*}(y)\right\}.}$

In particular the weak duality minmax equation can be shown to be

${\displaystyle \sup _{y^{*}\in Y^{*}}-F^{*}(0,y^{*})=\sup _{y^{*}\in Y^{*}}\inf _{x\in X}L(x,y^{*})\leq \inf _{x\in X}\sup _{y^{*}\in Y^{*}}L(x,y^{*})=\inf _{x\in X}F(x,0).}$

If the primal problem is given by

${\displaystyle \inf _{x:g(x)\leq 0}f(x)=\inf _{x\in X}{\tilde {f}}(x)}$

where ${\displaystyle {\tilde {f}}(x)=f(x)+I_{\mathbb {R} _{+}^{d}}(-g(x))}$. Then if the perturbation is given by

${\displaystyle \inf _{x:g(x)\leq y}f(x)}$

then the perturbation function is

${\displaystyle F(x,y)=f(x)+I_{\mathbb {R} _{+}^{d}}(y-g(x)).}$

Thus the connection to Lagrangian duality can be seen, as L can be trivially seen to be

${\displaystyle L(x,y^{*})={\begin{cases}f(x)-y^{*}(g(x))&{\text{if }}y^{*}\in \mathbb {R} _{-}^{d},\\-\infty &{\text{else}}.\end{cases}}}$

Fenchel duality

Let ${\displaystyle (X,X^{*})}$ and ${\displaystyle (Y,Y^{*})}$ be dual pairs. Assume there exists a linear map ${\displaystyle T:X\to Y}$ with

${\displaystyle T^{*}:Y^{*}\to X^{*}}$. Assume the primal ${\displaystyle f(x)}$ (including the constraints by way of the indicator function) can be written as ${\displaystyle f(x)=J(x,Tx)}$ such that ${\displaystyle J:X\times Y\to \mathbb {R} \cup \{+\infty \}}$. Then the perturbation function is given by

${\displaystyle F(x,y)=J(x,Tx-y).}$

In particular if the primal objective is ${\displaystyle f(x)+g(Tx)}$ then the perturbation function is given by ${\displaystyle F(x,y)=f(x)+g(Tx-y)}$, which is the traditional definition of

References