Is $\phi(h)=\displaystyle \sup_{x\in X}{\|h-x\|}$ continuous on Hilbert space when $X$ is bounded?

Question

Let $H$ be a hilbert space, and let $X \subset H$ be a bounded subset of $H$. Let define the function $\phi:H \to H$ by the rule $\phi(h)=\displaystyle \sup_{x\in X}{\|h-x\|}$.

I want to know if this function is continuous.

Clearly I tried to bound the term $\displaystyle \|\phi(h)-\phi(h')\|= |(\sup_{x\in X}{\|h-x\|} - \sup_{y\in Y}{\|h'-y} \| )|$

I would like to use the inequality $| \|a\|-\|b\|| \le \|a-b\|$ But I'm not sure how to work with the term $|(\sup_{x\in X}{\|h-x\|} - \sup_{y\in Y}{\|h'-y} \| )|$

Thanks!

Answer 1

Boundedness of $X$ is needed just to conclude that $\phi$ is finite.

$\|h-x\| \le \|h'-x\| + \|h-h'\|$. Hence $\|h-x\| \le \phi(h') + \|h-h'\|$, and so $\phi(h) \le \phi(h') + \|h-h'\|$. Reversing the roles of $h,h'$ gives $|\phi(h)- \phi(h')| \le \|h-h'\|$.