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Let a function $f:\mathbb{R}^n \rightarrow \mathbb{R}$ be convex and satisfies $$ \lim_{|q| \rightarrow \infty}\frac{f(q)}{|q|}=+\infty.$$ The Legendre's transsformation of $L$ is defined by $$L^*(p)=sup_{q \in\mathbb{R}^n} [p\cdot q-L(q)]$$ for $p \in \mathbb{R}^n$. How to prove that this supremum is realized, that is for each $p \in \mathbb{R}^n$ there exists a $q_0 \in \mathbb{R}^n$ such that $$sup_{q \in \mathbb{R}^n} [p\cdot q -L(q)]=p\cdot q_0-L(q_0).$$


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$f$ is convex on all of $\mathbb{R}$ so it's continuous. The idea will be the following: we know that the supremum is taken over a continuous function, so if we can restrict $q$ to a closed set, we'll be guaranteeing the supremum is achieved since a continuous function on a closed set achieves a min and max.

First, $f$ has a minimum at some $x_0$. To see this, notice that for all $|p|$ greater than some $\delta>0$, $f(p)/|p|>=1$.Using the fact that $f$ is convex gives $\inf_{q\in\mathbb{R}^n} f(q)=\inf_{q\in[-\delta,\delta]^n}f(q)$ which is an infimum over a closed set and is therefore achieved at some $x_0$.

If $p=0$, then the supremum is achieved at the minimum of $f$ at $q=x_0$, so fix $|p|>0$. Also, fix an $M>0$. By the limit property, there is some $\delta$ such that $f(q)/|q|>(1+M)|p|$ for all $|q|>\delta$. If we let $([-\delta,\delta]^n)^c$ refer to the complement of the box $[-\delta,\delta]^n$, then in particular,

$\sup_{q\in ([-\delta,\delta]^n)^c} [p\cdot q-f(q)]\leq\sup_{q\in ([-\delta,\delta]^n)^c} |q|[|p|-f(q)/|q|]<-M|p|$.

Notice that I can pick $\delta$ to make $M$ anything I want. Pick ANY $q_1$. That will give a value $C$ for $p\cdot q - f(q)$. By picking $\delta$ large enough, we get that $\sup_{q\in ([-\delta,\delta]^n)^c}[p\cdot q-f(q)]<C$. So that implies that $\sup_{q\in \mathbb{R}^n}[p\cdot q-f(q)]=\sup_{q\in[-\delta,\delta]^n}[p\cdot q-f(q)]$. Since the latter supremum is over a closed set of a continuous function, it must be achieved at some $q_0\in[-\delta,\delta]^n$.

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