Why is the function, $f(x) = \sup \lbrace \| x-y \| : y\in K \rbrace$, continuous?

Question

If $K$ is a no empty subset, weakly compact and convex on a Banach space, which is minimal for a non-expansive application $T: K \longrightarrow K$

Why is the function $f : K \longrightarrow \mathbb{R}^{+}$, defined by $f(x) = \sup \lbrace \| x-y \| : y\in K \rbrace$, for all $x\in K$, continuous?

Answer 1

Here is a partial proof of semicontinuity.

Take a sequence $x_n\to x$ in $X$. Take $y\in K$ arbitrary. Then $$ f(x_n) \ge \|x_n -y\|. $$ Passing to the limit yields $$ \operatorname{liminf}_{n\to\infty} f(x_n) \ge \|x-y\|. $$ Taking the supremum over $y\in K$ gives $$ \operatorname{liminf}_{n\to\infty} f(x_n) \ge f(x). $$ I could not prove the other half of the claim, as the weak lower semicontiuity of the norm somehow goes the wrong direction.

Answer 2

Completion of daw's proof: the other part is even easier: we have $||x_n -y|| \leq ||x_n -x|| +||x-y|| < ||x-y|| + \epsilon$ for all n sufficiently large. Take sup over $y \in K$ to get $f(x_n) \leq f(x) +\epsilon$. Hence $\limsup f(x_n) \leq f(x)$.