[Math] Why in uniformly convex Banach space every non empty, closed, convex subset contains a unique element of smallest norm

functional-analysis

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 $C$.

Let $(E, |\cdot|)$ be a uniformly convex Banach space and $C \subset E$ a nonempty.

  1. Prove that for every $x \in E$, $$ \inf _{y \in C}|x-y| $$ is achieved by some unique point in $C$, denoted by $P x$.
  2. Prove that every minimizing sequence $\left(y_{n}\right)$ in $C$ converges strongly to $P x$.
  3. Prove that the map $x \mapsto P x$ is continuous from $E$ strong into $E$ strong.
  4. More precisely, prove that $P$ is uniformly continuous on bounded subsets of $E$.
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