In Hilbert space every non empty, closed, convex subset contains a unique element of smallest norm. Why is that true also in Banach space which is uniformly convex?
(normed space which is uniformly convex is a space in which for all sequences $\{x_n\}$, $\{y_n\}$ s.t $||x_n||,||y_n||\leq 1$ exists: if $\lim_n||x_n+y_n||=2$ then $\lim_n||x_n-y_n||=0$.)
I thought of defining $a=\inf\{||x|| : x\in X\}$, so exists a sequence $\{x_n\}$ s.t $\lim_n||x_n||=\inf||x_n||=a$ and then to show that $\{x_n\}$ is a Cauchy sequence and it's limit is the element we are looking for, but i didn't manage to prove that it is Cauchy.
Best Answer
Your question is a particular case Ex 3.32 in Brezis's book of Functional Analysis. In fact, consider the projection of $0\in E$ into $C$.