If $K \subset \Bbb R^d$ is not bounded, is $h_K(x) = \sup_{y\in K} \langle x,y\rangle$ not Lipschitz?

Question

Question: If $K \subset \Bbb R^d$ is not bounded, is $h_K(x) = \sup_{y\in K} \langle x,y\rangle$ not Lipschitz?

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Background.
I came across the following proposition.

If $K\subset\Bbb R^d$ is a compact and convex subset of $\Bbb R^d$, then the support function $h_K(x) = \sup_{y\in K} \langle x,y\rangle$ is Lipschitz.

The author left the proof to the reader. My proof (as follows) uses the compactness (in particular, the boundedness) of $K$.

Proof. Let $x,y\in \Bbb R^d$. For any $z\in K$, we have $$\langle x,z\rangle - \langle y,z\rangle \le |\langle x,z\rangle - \langle y,z\rangle| \le \|x-y\|\|z\| \le \|x-y\| \sup_{z\in K} \|z\|$$ Since $K$ is compact, $\sup_{z\in K} \|z\| < \infty$. Say, $M:= \sup_{z\in K} \|z\|$. Then, $$\langle x,z\rangle - \langle y,z\rangle \le M\|x-y\|$$ so that $$\langle x,z\rangle \le M\|x-y\| + \langle y,z\rangle \le M\|x-y\| + h_K(y)$$ Taking the supremum over all $z\in K$, $$h_K(x) - h_K(y) \le M\|x-y\|$$ Swapping the roles of $x$ and $y$, we get $$|h_K(x) - h_K(y)| \le M\|x-y\|$$ i.e., $h_K$ is Lipschitz.

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Clearly, the boundedness of $K$ is a vital part of showing that $h_K$ is Lipschitz. If $K$ is not bounded, is $h_K$ necessarily not Lipschitz? If not, I'm hoping to get some help with constructing examples of unbounded $K\subset\Bbb R^d$ (convex if need be) for which $h_K$ is Lipschitz. It might also be interesting to see unbounded $K$ for which $h_K$ is not Lipschitz.

Thanks a lot!

Answer 1

No, $h_K(x)$ is not even well-defined when $K$ is unbounded.

PhoemueX observed the following fact.

$h_K(x)$ is well-defined (i.e., $h_K(x)<\infty$ for all $x$) if and only if $K$ is bounded.

Proof.
"$\impliedby$". Suppose for some constant $c>0$, $|y|\le c$ for all $y\in K$.
$|\langle x, y\rangle|\le\|x\|\|y\|\le\|x\|c$ for all $y\in K$.
Hence $|h_K(x)|\le \|x\|c\lt\infty$.

"$\implies$". Suppose $h_K(x)<\infty$ for all $x\in\Bbb R^d$.
Let $e_1, e_2, \cdots, e_d$ be the standard orthnormal basis of $\Bbb R^d$, i.e., $e_1=(1,0,0,\cdots, 0)$, etc.

Let $x=(x_1, x_2, \cdots, x_d)\in K$.
Then $$\begin{aligned}\|x\|^2&=\langle x,x\rangle\\ &=\langle x_1e_1 +\cdots+ x_de_d,x\rangle\\ &=x_1\langle e_1, x\rangle + \cdots + x_d\langle e_d,x\rangle\\ &\le|x_1|\,|\langle e_1, x\rangle| + \cdots + |x_d|\,|\langle e_d,x\rangle|\\ &\le|x_1|\max(h_K(e_1), h_K(-e_1)) + \cdots + |x_d|\max(h_K(e_d), h_K(-e_d))\\ &\le\|x\| dM \end{aligned}$$ where $M=\max\left(\max(h_K(e_1), h_K(-e_1)),\cdots,\max(h_K(e_d), h_K(-e_d))\right)$. The second last inequality uses the inequality $$|\langle z,x\rangle|\le\max(h_K(z), h_K(-z))$$ for all $z\in\Bbb Z$ and $x\in K$. If $\langle z,x\rangle\ge0$, then $|\langle z,x\rangle|=\langle z,x\rangle\le h_K(z)$; otherwise $\langle z,x\rangle\lt0$, we have $|\langle z,x\rangle|=-\langle z,x\rangle=\langle-z,x\rangle\le h_K(-z)$.

Hence $\|x\|\le dM$.

Can $h_K(x)$ be Lipschitz on an open set when $K$ is unbounded?

Yes. Let $d=1$, $K=\Bbb R^+$. Then $h_K(x)=0$ for $x<0$. So $h_K(x)$ is Lipschitz on $(-\infty, 0)$.