# Legendre-Fenchel Transform [closed]

For any function $g:\mathbb{R}\rightarrow\mathbb{R}\cup\{\infty\}$, the Legendre-Fenchel transform is

$f(y)=\sup_{x\in\mathbb{R}}\{xy-g(x)\}.$

Prove that $f$ is convex.

## closed as off-topic by Crostul, E. Joseph, Leucippus, Shailesh, Zain PatelNov 29 '16 at 1:31

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• @Crostul et al: Why is this off-topic? It is perfectly well defined and a math question. – max_zorn Jan 19 '18 at 8:58

The Legendre-Fenchel transform naturally arises in the context of functions which are allowed to take values in $\Bbb R\cup\{+\infty\}$, provided that they are not constantly $g(x)\equiv+\infty$ and that there are $y$ and $b$ such that $g(x)\ge yx+b$ for all $x$.

Now, consider the family of functions $\{f_x\,:\, g(x)\ne+\infty\}$, defined by $f_x(y)=xy-g(x)$. Each $f_x$ is an affine map $\Bbb R\to\Bbb R$, hence convex.

You have that $f=\sup\limits_x f_x$.

$\sup$ of convex functions is convex. There are two ways to see it.

1. Convexity is equivalent to $$\operatorname{epi} f:=\{(x,r)\,:\, f(x)\le r<+\infty\}\subseteq \Bbb R\times\Bbb R$$ being convex. Once you know this, you just observe that $$\operatorname{epi}\sup_\alpha f_\alpha=\bigcap_\alpha \operatorname{epi}f_\alpha$$ and you use the fact that intersection of convex sets is convex. All these claims can be verified directly using the definitions.

2. It can be done directly by observing that, for $t\in (0,1)$, $$\left[\sup_\alpha f_\alpha\right](tx+(1-t)y)=\sup_\alpha (f_\alpha(tx+(1-t)y))\le \sup_\alpha (tf_\alpha(x)+(1-t)f_\alpha(y))\le\\\le t\sup_\alpha(f_\alpha(x))+(1-t)\sup_\alpha(f_\alpha(y))=t\left[\sup_\alpha f_\alpha\right](y)+(1-t)\left[\sup_\alpha f_\alpha\right](y)$$

• Sorry, for down voting your answer, I just don't understand this solution – BronchoX Nov 29 '16 at 3:19