Show that supremum of over a bounded set is continuous.

Question

Let $A \in \mathbb{R}^n$ $S_A(x)$ is defined as the following

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

Show that if $A$ is a bounded set, then $S_A(x)$ is a continuous function.

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? Please follow my proof and complete it.

Answer 1

Write $x^{T}y$ as $(x^{T}-x_c^{T})y+x_c^{T}y$. So $x_c^{T}y \leq \delta \|y\|+x_c^{T}y$. Take sup over $y$ to get $S_A(x)\leq \epsilon +S_A(x_c)$. Repeat the same argument with $x$ and $x_c$ interchanged to get $S_A(x_c)\leq \epsilon +S_A(x)$. Put these two together to get $|S_A(x)-S_A(x_c)|<\epsilon$.

Answer 2

Using Cauchy-Schwarz inequality we can bound $(x^{T}-x_c^{T})y$

$(x^{T}-x_c^{T})y\leq \|x^{T}-x_c^{T}\|\|y\| $

Since $\|x^{T}-x_c^{T}\| < \delta$, we have the following

So

$$(x^{T}-x_c^{T})y \leq \|x^{T}-x_c^{T}\|\|y\| < \delta \|y\|$$

So $x_c^{T}y < \delta \|y\|+x_c^{T}y$ for all $y$ in $A$.

Take sup over $y$

$$\sup_{y \in A} x_c^{T}y < \sup_{y \in A} (\delta \|y\|+x_c^{T}y) \leq \sup_{y \in A} \delta \|y\|+ \sup_{y \in A}x_c^{T}y$$

So

$$S_A(x) < \delta M + S_A(x_c)$$

Do the same for $(x_c^{T}-x^{T})y$ to get

$$S_A(x_c) < \delta M + S_A(x)$$

Combine them to get the following

$|S_A(x)-S_A(x_c)|<\delta M=\epsilon$.