# Show that any convex function is locally bounded

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 $x$?

• What is $\overline{\mathbb{R}}$? What is $\operatorname{ri}$? (If the notation is coming from some text you're using, it will be helpful to name the text.) – Pete L. Clark May 30 '13 at 2:50
• Hope things are a little more clear after the edit. – Julio Cáceres May 30 '13 at 2:55
• It's a little more clear. But if the domain of $f$ is $\mathbb{R}^n$, isn't $\operatorname{ri}(\operatorname{dom}(f))$ just $\mathbb{R}$? Also if the function is allowed to be infinite-valued, how can it be bounded in any sense in a neighborhood of a point at which it is infinite?? – Pete L. Clark May 30 '13 at 3:00
• Woops forgot to add something else about how we define $\text{dom}(f)$, for this problem we're just concerned with the effective domain – Julio Cáceres May 30 '13 at 4:34
• In my opinion this assertion doesn't hold. Let us assume for example a function $I_{M}:\mathbb{R}^{2}\rightarrow \overline{\mathbb{R}}$ where $I_{M}$ is the indicator function defined as follows: $I_{M}(x)=0$ for $x\in M$ and $I_{M}(x)=\infty$ for $x\notin M$. Here $M$ denotes the $x$ axis, i.e. $M=\mathbb{R}\times \{0\}$. Then $ri\left( dom~f\right) =ri\left( M\right) =M$. $I_{M}$ is clearly convex. On the other hand, this function is not bounded on any ball $B\left( x;\delta \right)$ for $x\in M$, $\delta >0$. – Dusan May 31 '13 at 16:55

Here is one way to prove that $f$ is lower semicontinuous at every point $x$ in the relative interior of its effective domain.
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.
• Well, you restrict topology to an appropriate affine subspace generated by the effective domain of the function. Right? But why you can assume that $f$ is bounded above on some closed ball? – Dusan Jun 1 '13 at 8:20
• @Dusan Right. Also, the OP said "Showing that it has an upper bound is not difficult using Jensen's inequality", so I took that part for granted. To prove it, I would pick a cube with $x$ as its center, let $M$ be the maximum of the values of $f$ at its vertices, and argue by convexity that $f\le M$ on the cube. – ˈjuː.zɚ79365 Jun 1 '13 at 8:33
• Now I see in Rockafellar's book "Convex Analysis" the following even stronger result (Theorem 10.1, p.82) that $f$ must be continuous relative to $ri(dom f).$ – Dusan Jun 1 '13 at 10:09