[Math] Show that any convex function is locally bounded

Question

Show that a convex function $f:\mathbb{R}^n \rightarrow \overline{\mathbb{R}}$ is bounded in a neighborhood of $x\in \text{ri}(\text{dom}(f))$. Showing that it has an upper bound is not difficult using Jensen's inequality. To show that it has a lower bound I showed that if it wasn't bound in $B(x,\delta)$, then the epigraph in this region would be $B(x,\delta)\times \langle-\infty,+\infty \rangle$, but to do this I made a weird assumption which might or might not be true. If $f$ didn't had a lower bound in $B(x,\delta)$, then for every $M\in \mathbb{R}$ it exists an $x_M$ such that $f(x_M)<M$. Naturally $(x_M,M)\in \text{epi}(f_{|B(x_0,\delta)})$ and since $\text{epi}(f_{|B(x_0,\delta)})$ is convex for any $(x,\lambda) \in \text{epi}(f_{|B(x_0,\delta)})$ it would follow that $(tx+(1-t)x_M,t\lambda+(1-t)M)\in \text{epi}(f_{|B(x_0,\delta)})$ for any $t\in ]0,1[$. My assumption is that with this convex combination I can reach any point in $B(x,\delta)\times \langle-\infty,+\infty \rangle$. Any help regarding my assumption or another way to show that there's a real lower bound would be welcome, thanks in advance.

For this problem $\overline{\mathbb{R}}=\mathbb{R}\cup\{+\infty\}$,$\text{dom}(f)=\{x\in \mathbb{R}^n/ f(x)<+\infty\}$ and $\text{ri}(C)$ is the relative interior of $C$.

After reading some books I found that not only is the function bound but it's uniformly continuous, the proof wasn't that simple. I feel there must be an easy way to find the lower bound maybe showing that it's lower semicontinuous at a neighborhood of $x$?

Answer 1

Here is one way to prove that $f$ is lower semicontinuous at every point $x$ in the relative interior of its effective domain.

1. Get rid of "relative" by restricting attention to an appropriate affine subspace.
2. You already know that $f$ has an upper bound $M$ in some closed neighborhood $\overline B(x,\delta)$.
3. Suppose, to the contrary, that there is $\epsilon>0$ and a sequence $x_n\to x$ such that $f(x_n)\le f(x)-\epsilon$ for all $n$.
4. Let $y_n=x-\delta\dfrac{x_n-x}{\|x_n-x\|}$. Observe that $x=(1-\lambda_n)x_n+\lambda_n y_n$ where $\lambda_n=\dfrac{\|x_n-x\|}{\|x_n-x\|+\delta}\to 0$ as $n\to\infty$.
5. By virtue of convexity, $f(x)\le (1-\lambda_n)( f(x)-\epsilon)+\lambda_n M$. Let $n\to\infty$ to reach a contradiction.