given a normed reflexive space $V$, then for every closed subset $U \subset V$ and $x \in V$ $\inf_{y \in U} \|x-y\| \in U$.
In some paper this is given als "clear" or "trivial". But I am a bit confused on the influence of reflexive(dual space) on this. Would be nice if someone can give me a link to a proof or something. Thank you.
Hint : Let $\alpha > 0 $ such that $\inf_{y \in U} \|x-y\| < \alpha$
Then the set $$ A: = \{ y \in U ~ | \quad \|x-y\| \leq \alpha \} $$
is closed and bounded , hence weakly compact in $V$
And note that $$ \inf_{y \in U} \|x-y\| =\inf_{y \in A} \|x-y\| $$
and $\|\|$ is weakly-lower semicontinuous. i.e., $ \| x \| \leq \liminf_{x_n \overset{w}{\rightarrow} x} \| x_n\| $
Now you should be able to deduce why $\inf$ is attained .