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I am studying convex analysis and of course, one needs the Fenchel conjugate (or polar function) and also the $\Gamma$-regularization of a function (I'm going to write it as $F^\Gamma$).

In this question, $V$ is a topological vector space, $V'$ its dual (topological dual). If $x^*\in V'$, I denote $x^*(x)$ by $\left\langle x,x^*\right\rangle$.

In order to understand my question, I attach an image of the book Nonsmooth analysis written by Winfried Schirotzek:

$\Gamma$ regularization of $F$ in two points:

enter image description here

Let $z_0$ be the point of the image given by the intersection of $F$ and $a_2$.
That is, $z_0=F^\Gamma \left(x_0\right)$ for some $x_0$.

My question is, I think that

$$\exists x^{*}\in V':\ F^{*}\left(x^{*}\right)=a_{2}\left(x_{0}\right)=z_{0}\,.$$

In other words, If you have $F^\Gamma \left(x_0\right)$ then you have $F^*$ since $F^{\Gamma}\left(x_{0}\right)=a_2\left(x_{0}\right)$ and thus, you have the affine function $a_2$. I.e., both functions return the same result, $z_0$, but under different points of view.

Is this correct? If it is, what is the difference between $F^*$ and $F^{**}$ which is always $F^\Gamma$?

Otherwise, how can I understand the geometric difference between $F^*$ and $F^\Gamma$?

Thank you so much in advance!

Definitions: The definition of the polar function is:

Let $F:V\to \overline{\mathbb{R}}$. Then, the functional, $F^{*}:V'\to \overline{\mathbb{R}}$ is defined by, $$F^*\left(x^*\right)={\displaystyle \sup_{x\in V}\left\{ \left\langle x,x^*\right\rangle -F\left(x\right)\right\} }.$$

On the other hand, the definition of the $F^\Gamma$ is:

Let $$\mathcal{A}\left(F\right)=\sup\left\{ a:E\to\mathbb{R\ }|\ a\text{ is continuous and affine},\ a\leq F\right\}$$ So, the functional $F^\Gamma$ defined by, $$F^{\Gamma}\left(x\right)=\sup\left\{ a\left(x\right)\ |\ a\in\mathcal{A}\right\}$$ is called the $\Gamma$ regularization of $F$.

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I solved the question. I will use the example above, with $a_2$, to explain it

Background

First. Obviously $a_2$ is the tangent line to $F$ at $z_0$.

Second. At page 30 of the book Nonsmooth analysis, Winfried Schirotzek one can find the geometric interpretation of $F^{*}$. That is, we consider an affine function $$\alpha \left(x\right)=\left\langle x,x^{*}\right\rangle -F^{*}\left(x^{*}\right),$$ s.t. $$\alpha \left(0\right)=-F\left(x^*\right).$$ Since $\alpha$ is parallel to the linear functional $\left\langle x,x^{*}\right\rangle$ they have the same slope.

So, If we choose the function $$\widetilde{\alpha}\left(x\right)=\left\langle x,x_{0}^{*}\right\rangle -F^{*}\left(x_{0}^{*}\right),$$ where $x_0$ is none other than the $y$-coordinate of the point $z$, i.e., $z=\left(x_0,x^*_0\right)$ (in general, $x^*_0$ is an element of $V'$), we have, $$\widetilde{\alpha}\left(x_{0}\right)=\left\langle x_{0},x_{0}^{*}\right\rangle -F^{*}\left(x_{0}^{*}\right)=\left\langle x_{0},x_{0}^{*}\right\rangle -\left\langle x_{0},x_{0}^{*}\right\rangle -F\left(x_{0}\right)=-F\left(x_{0}\right).$$ Thus, $\widetilde{\alpha}\left(x\right)$ is also tangent line to $z$ and, since there is only one tangent (if $F$ is "good enough") to one point, we conclude that, $$\widetilde{\alpha}\left(x\right)=a_2.$$ Then, $$F^{\Gamma}\left(x_{0}\right)=\left\langle x_{0},x_{0}^{*}\right\rangle -F^{*}\left(x_{0}^{*}\right)=a_{2}\left(x_{0}\right)=\sup\left\{ a\left(x\right)|\ a\in\mathcal{A}\right\} .$$

Answer

So, $F^*$ is just the constant of the affine function $a_j$ (an affine function is defined by the sum of one linear function and one constant) which is the supremum of all affine functions $\in \mathcal{A}$ evaluated at point $x_0$. In other words, $a_j\left(x_{0}\right)$ is $F^\Gamma \left(x_{0}\right)$ and $a_j\left(x\right) = \left\langle x,x_{0}^{*}\right\rangle -F^{*}\left(x_{0}^{*}\right)$ for some $x^{*}_0$.

Observe that, $$F^{**}\left(x\right)={\displaystyle \sup_{x^{*}\in V'}\left\{ \left\langle x,x_{0}^{*}\right\rangle -F^{*}\left(x_{0}^{*}\right)\right\} ={\displaystyle \sup\left\{ a\left(x\right)|\ a\in\mathcal{A}\right\} =F^{\Gamma}\left(x\right)}}$$

so that is the relation between $F^*$ and $F^{**}$

Curiosity

We have deduced a fast way to compute $F^\Gamma$. If we know the explicit expression of $F^{*}$, $F^\Gamma \left(x_0\right)$ is only $$\left\langle x_{0},x_{0}^{*}\right\rangle -F^{*}\left(x_{0}^{*}\right),$$ for the convenient $x_0 ^*.$

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  • $\begingroup$ If you do the calculations with concrete numbers for example, $x_0= \sqrt{2}, y_0=x^*_0=2$ and $F=x^2$ it is easier to understand the answer. $\endgroup$ – Alejandro Mus Jan 10 '17 at 9:48

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