Best approximation in a reflexive normed space

Question

Let $(E, \| \cdot \|)$ be a reflexive normed space and let $\emptyset\ne K\subseteq E$ be closed and convex. Show that for every $x\in E$ there exists a "best approximation" in $K$, that is, a $y\in K$ with $\|x-y\|=d(x,K):=\inf\{\|x-z\| : z\in K\}$.

Is additionally $(E,\|\cdot\|)$ uniformly convex, then the "best approximation" is unique.

For the first part it is given as a hint, to show the existence of a bounded sequence $(y_n)_n\in K$ with $\|x-y_n\|\to d(x,K)$.

I am stuck here. Can you help me out, or give me a hint?

Answer 1

Let $d := d(x, K)$. By definition of $\inf$, you can construct a minimizing sequence $(y_n)\subset K$ such that $\|x - y_n\| \to d$. Namely, for every $n\in\mathbb{N}$ you can find a point $y_n\in K$ such that $d\leq \|x- y_n\| \leq d + 1/n$.

By construction, $(y_n) \in \overline{B}_{d+1}(x)$, hence $(y_n)$ is bounded.

Since $E$ is a reflexive Banach space, there exist a point $y_0\in E$ and a subsequence $(y_{n_j})$ of $(y_n)$ such that $y_{n_j} \rightharpoonup y_0$. Moreover, since $K$ is convex and closed, you have that $y_0\in K$.

Let us prove that $\|x - y_0\| = d$. Namely, by the lower semicontinuity of the norm with respect to the weak convergence, we have that $$ d \leq \|x - y_0\| \leq \liminf_{j\to +\infty} \|x - y_{n_j}\| \leq \liminf_{j\to +\infty} \left(d + \frac{1}{n_j}\right) = d. $$

Let us prove that the minimizer is unique if $E$ is uniformly convex. Clearly, if $d=0$ (i.e., if $x\in K$), then the unique minimizer is $x$ itself.

Assume now that $x\not\in K$ (i.e. $d>0$), and assume by contradiction that there exist two points $y_0, y_1\in K$ such that $$ \|x-y_0\| = \|x - y_1\| = d. $$ Let us consider the normalized vectors $$ \xi_i := \frac{x- y_i}{d}, \qquad i = 1,2. $$ Setting $\epsilon := \|\xi_0 - \xi_1\| = \|y_0 - y_1\| / d > 0$, by the definition of uniform convexity there exists $\delta > 0$ such that $$ \left\|\frac{\xi_0+\xi_1}{2}\right\| \leq 1 - \delta. $$ On the other hand $$ \left\|\frac{\xi_0+\xi_1}{2}\right\| = \frac{1}{d} \left\|x - \frac{y_0 + y_1}{2}\right\|\geq 1, $$ since $K$ is convex and so the point $(y_0+y_1)/2$ belongs to $K$.