1
$\begingroup$

The full problem is:

There is a normed space $(X,\|\cdot\|)$ with Clarkson's property, which means: exist a pair of conjugated numbers $(p,q),1<p\le 2$, 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!

$\endgroup$

1 Answer 1

1
$\begingroup$

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 $y=Bx$.

$\endgroup$
2
  • $\begingroup$ Nice solution! But in part(ii) equation, why the first '' = '' is true? $\endgroup$
    – robothead
    Commented Sep 29, 2020 at 15:20
  • $\begingroup$ Should be $\le$. $\endgroup$ Commented Sep 29, 2020 at 15:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .