# Show that the set formed by the supporting function is closed.

\begin{align*} A = \{a \in \Bbb R^n\mid\sup_x\langle a, x\rangle \leq b\}. \end{align*}

where $x \in C$, a nonempty set in $\Bbb R^n$ and $b$ is some scalar.

I'm supposed to show the set $A$ is convex and closed. I showed it was convex by showing the function $\sup_x\langle a, x\rangle$ was convex and then applying the relationship between convex functions and convex sets. I'm having trouble showing that this set is closed however.

Does the inequality not already guarantee the set will be closed, as there should be a set of $a$ that will cause the function to equal $b$ exactly and therefore includes the bounds of the set? I feel like I have far too simple of an understanding of it though and am missing something important. Any help is appreciated.

• Show that function f is continuous and notice A is the inverse image of a closed set by f. – William Elliot Mar 2 '18 at 4:06

First notice that $f_{x_{0}}:\mathbb{R}^{n}\to \mathbb{R}$ defined as $f_{x_{0}}(x)=\langle x,x_{0}\rangle$ if continuous for each $x_{0}\in \mathbb{R}^{n}$.
Then, if $x\in C$ and $(a_{n})_{n\in\mathbb{N}}$ is a sequence in $A$ that converges to $a$, then $(f_{x}(a_{n}))$ converges to $f_{x}(a)$, this implies that $f_{x}(a)\leq b$ for each $x\in C$ (bacause $f_{x}(a_{n})\leq b$ for each $x$ and each $n$), and therefore $\sup_{x\in C}\langle a,x\rangle\leq b$.
Hint:$$A=\bigcap_{x \in R^n}A_x$$
Where $$A_x = \{a \in \Bbb R^n :~~~ \langle a, x\rangle \leq b\}$$