How to prove that continuous convex functional on normed vector space must be lower bounded by some continuous affine functional?

Question

Let $X$ be a normed vector space. Let $f:X\mapsto \mathbb{R}$ be a continuous convex function. How to show that there exists a continuous linear functional $l$ and a constant $c\in\mathbb{R}$ such that $f(x)>l(x)+c$ holds for all $x\in X$? I think it could be proved by hyperplane separation theorem. But I don't know how to start it.

Answer 1

The epigraph $\operatorname{epi}f$ is closed and convex set in $X\times\Bbb R$ and $(x,f(x)-\epsilon)\not\in\operatorname{epi}f$ is compact for any $\epsilon>0$ and $x\in X$. By the separation theorem there exists a non-zero $(\phi,\psi)\in(X\times\Bbb R)'=X'\times\Bbb R$ such that the hyperplane $(\phi,\psi)\cdot(x,y)=\phi(x)+\psi y=\alpha$ separates $\operatorname{epi}f$ and the point. We can choose the sign such that $\operatorname{epi}f$ belongs to the half-space $\phi(x)+\psi y>\alpha$. Clearly $\psi\ne 0$ since $\phi(x)>\alpha$ is not possible for all $x\in X$ and $\phi\ne 0$. Hence, $\psi>0$ otherwise $y\to+\infty$ (still in $\operatorname{epi}f$) will give us a contradiction with $\phi(x)+\psi y>\alpha$. Finally $(x,f(x))\in\operatorname{epi}f$ then $\phi(x)+\psi f(x)>\alpha$. Divide by $\psi$ and move the things around.