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

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