Projection on Convex Set in uniformly convex Banach spaces

Question

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?

Answer 1

HINT:

We may assume that $x=0$ and $d(0, K) = 1$.

Consider $y_n \in K$ so that $\|y_n \| < 1 + \frac{1}{n}$. Since $K$ is convex we have $$1\le \|\frac{y_m+y_n}{2}\|$$ Let $y_n'=\frac{y_n}{\|y\|}$. We have $\|y_n - y_n'\|< \frac{1}{n}$. If follows that $$\|\frac{y_m+y_n}{2}- \frac{y'_m+y'_n}{2}\|< \frac{1/m+1/n}{2}$$ and so $$ \|\frac{y'_m+y'_n}{2}\|> 1-\frac{1/m+1/n}{2}$$

Now uniformly convex means:$\|x\|=\|y\|=1$, $\|\frac{x+y}{2}\|>1-\delta$ implies $\|x-y\|< \epsilon$. Therefore, since $$1- \|\frac{y'_m+y'_n}{2}\|\to 0$$ as $m,n\to \infty$ we get $$\|y'_m-y'_n\|\to 0$$ and so $$\|y_m - y_n\|\to 0$$ and so $y_n$ is convergent to $y\in K$ so that $\|y\|=1$.

Obs: The proof would also show that if $d(x,y_n) \to d(x, K)$ then $y_n$ converges to the unique closest point in $K$ to $x$.

Answer 2

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: $1=\underset{k,l\to \infty}\lim\frac{\|x_{n_k}+x_{n_l}\|}{2(1+1/k)}\le 1-\delta$