# Show that the support function of a bounded set is continuous.

The support function of a set $$A \in \mathbb{R}^n$$ is defined as the following

$$S_A(x)=\sup_{y \in A} x^Ty$$ where $$x \in \mathbb{R}^n$$.

Show that the support function of a bounded set is continuous.

I tried the following:

Let $$A$$ be a bounded set in $$\mathbb{R}^n$$ and $$y \in A$$. So $$\|y\| \leq M$$ where $$M>0$$. (If $$M=0 \rightarrow \|y\|=0 \rightarrow y=0 \rightarrow S_A(x)=0\rightarrow S_A(x)$$ is continuous $$\forall x$$).

Let $$\|x-x_c\|<\delta=\frac{\epsilon}{M}, \,\,\,\forall \epsilon>0$$ be a neighborhood of $$x_c$$ where $$x_c \in \mathbb{R}^n$$.

I need to show that $$|S_A(x)-S_A(x_c)|=|\sup_{y \in A} x^Ty-\sup_{y \in A} x_c^Ty|<\epsilon$$

How can I proceed?

Here's a solution which reduces to another problem on this site: Supremum is continuous over equicontinuous family of functions

Step 1: define for each $$a\in A$$ the function $$f_a(x) = x \cdot a$$. Each of these is continuous.

Step 2: Show that $$S_A$$ is the supremum of the family $$\{f_a\}$$

Step 3: Show that because $$A$$ is bounded, the family is not only continuous but equicontinuous.

Then conclude using the link above!

• I want to proof it in a way that I explained. Could you help me to do that. – Saeed Nov 8 at 21:25
• You will need to do an $\epsilon/3$-style argument. Basically rewrite $|S_A(x)-S_A(x_c)|$ as $|S_A(x)-x \cdot y +x \cdot y -x_c\cdot y +x_c\cdot y -S_A(x_c)|$ for some $y$ which almost attains the supremum $S_A(x_c)$ and use the triangle inequality and equicontinuity. – user25959 Nov 8 at 22:42
• Let me I change your $y$ to $z$. Then how can I treat $|\sup_{y \in A}x^Ty-x^Tz|$? – Saeed Nov 8 at 23:57