# distance it's attained on closed and finite dimensional subspaces of a Banach space

Let $X$ be a Banach space. And let $F\subset X$ be a closed and linear subspace (in particular is Banach). I want to prove the following:

Let $d(x,F)=\displaystyle \inf_{y\in F} ||x-y||$. Is it true that there exist $y \in F$ such that $d(x)=||x-y||$?

What if $F$ it's also finite dimensional?

-

In general, there need not exist a $y \in F$ that realises the distance. If $F$ is finite-dimensional, or if $X$ is reflexive, then there always exists such a $y$.

In the case of finite-dimensional $F$, the intersection of the closed balls around $x$ and $F$ is compact, since $F$ is locally compact. Thus the intersection

$$\bigcap_{r > d(x,F)} F\cap \overline{B_r(x)}$$

is not empty, since all finite intersections are non-empty. All elements of that intersection realise the distance.

If $X$ is reflexive, the closed balls are weakly compact, and the same argument of compactness guarantees that the above intersection is non-empty.

-
Very, very nice! I didn't know such a simple argument. –  Norbert Nov 12 '13 at 8:52