Let $E$ be a uniformly convex banach space, $K$ convex and closed in $E$, and $x\in B\setminus K$.
Can someone give a concise proof that there is a unique $y\in K$ s.t. $dist(x,K) = \Vert x-y\Vert$ using the notion of uniform convexity exactly as given on the relevant wikipedia page and without using weak compactness of reflexive spaces?
Best Answer
Larsen's Functional Analysis, an Introduction, Thm 8.2.2 has an elementary proof of this, that relies on a lemma, which he leaves to the reader. Here is the statement and proof of the lemma:
Let $E$ be a uniformly convex normed linear space over $F$. If $(x_n)\subseteq E$ is a sequence such that
$(i) \underset{n\to \infty}\lim \|x_n\| = 1,\ (ii) \ \underset{n,m\to \infty}\lim \|x_n + x_m\| = 2.\ \text{Then},\ (x_n)\ \text{is Cauchy.}$
$(i)$ implies that for each integer $k$, there is an $x_{n_k}$ such that $\|x_{n_k}\|<1+1/k$.
Now, if $(x_n)$ is not Cauchy, then we may choose $(x_{n_k})$ so that the subsequence is not Cauchy either. Then, there is an $\epsilon>0$ such that for all integers $K$, there are integers $n_k,n_l$ such that $\|x_{n_k}-x_{n_l}\|>\epsilon$ with $k,l>K.$
Without loss of generality, assume $l>k$ so that $1+1/l<1+1/k$ and therefore $\|x_{n_k}\|<1+1/k$ and $\|x_{n_l}\|<1+1/l<1+1/k.$
Then, the hypothesis of uniform convexity gives us a $\delta>0$ such that $\frac{\|x_{n_k}+x_{n_l}\|}{2(1+1/k)}<1-\delta.$ But now, on taking limits, and applying $(ii)$,we get a contradiction: $1=\underset{k,l\to \infty}\lim\frac{\|x_{n_k}+x_{n_l}\|}{2(1+1/k)}\le 1-\delta$