# Best approximation in a convex subset of strictly convex normed space.

The full problem is:

There is a normed space $$(X,\|\cdot\|)$$ with Clarkson's property, which means: exist a pair of conjugated numbers $$(p,q),1, then $$\forall\{x_1,x_2\}\subseteq X$$ meets following conditions:$$(\|x_1+x_2\|^q + \|x_1-x_2\|^q)^{1/q} \le 2^{1/q}(\|x_1\|^p + \|x_2\|^p)^{1/p} \\ (\|x_1+x_2\|^p + \|x_1-x_2\|^p)^{1/p} \ge 2^{1/q}(\|x_1\|^p + \|x_2\|^p)^{1/p}$$

Prove that:

$$(i)$$ Let $$E$$ be a convex subset of $$X$$, then $$\forall x\in X$$, there is at most one best approximation $$Bx\in E$$.

$$(ii)$$ $$Bx$$ (when exist) continuously depends on $$x$$

$$(iii)$$ When $$E$$ is complete, $$Bx$$ always exists.

Through Clarkson's property, I have proved that $$(X,\|\cdot\|)$$ is strictly convex, and with $$E$$ convex, the first part is done.

But I got stuck in $$(ii)$$ and $$(iii),$$ and totally don't know where to go.

Any suggestion is great!

Let $$d$$ be the least distance between $$x$$ and $$E$$.
(i) Suppose there are two points of approximation, $$a$$ and $$b$$, both in $$E$$. Then by the Clarkson property, $$\|a-b\|^q+2^q\|\frac{a+b}{2}-x\|^q\le 2(\|a-x\|^p+\|b-x\|^p)^{q/p}$$ $$\therefore\ \|a-b\|^q+2^qd^q\le2(2d^p)^{q/p}=2^{1+q/p}d^q=2^qd^q$$ hence $$a=b$$.
(ii) Let $$Bx$$ be the best approximation to $$x$$, and $$m=(Bx+Bx')/2\in E$$. \begin{align}\|Bx-Bx'\|^q+2^q\|x-m\|^q&\le2(\|x-Bx\|^p+\|x-Bx'\|^p)^{q/p}\\ &\le2(\|x-Bx\|^q+\|x-Bx'\|^q)\\ \|Bx-Bx'\|^q+2(2^{q/p}-1)\|x-m\|^q&\le2\|x-Bx'\|^q\quad\textrm{since} \|x-m\|\ge\|x-Bx\|\end{align} But $$\|x-Bx'\|\le\|x'-Bx'\|+\|x-x'\|$$ and $$\|x-m\|\ge\|x'-m\|-\|x-x'\|$$, so $$\|Bx-Bx'\|^q\le2\|x'-Bx'\|^q-2\|x'-m\|^q+O(\|x-x'\|)=O(\|x-x'\|)$$
(iii) Let $$y_n$$ be a sequence $$\|y_n-x\|\to d$$, and let $$u_{n,m}:=(y_n+y_m)/2$$. Then \begin{align} \|y_n-y_m\|^q+2^q\|u_{n,m}-x\|^q&\le2(\|y_n-x\|^p+\|y_m-x\|^p)^{q/p}\to2^qd^q\\ \|y_n-y_m\|^q+2^qd^q\le2^qd^q \end{align} so $$\|y_n-y_m\|\to0$$. Since $$E$$ is complete, $$y_n\to y$$. By continuity of the norm, $$\|y-x\|=d$$, so $$y=Bx$$.
• Should be $\le$. Commented Sep 29, 2020 at 15:46