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Consider an inner product space $X$ with the induced metric $d$ (induced by the inner product). Suppose that the induced metric space $(X,d)$ is complete. Moreover, for all $x,y,z\in X$,

$$[d(x,y)]^2+[d(x,z)]^2=2\left[d\left(x,\frac{y+z}{2}\right)\right]^2+\frac{1}{2}[d(y,z)]^2$$

Let $C$ be a nonempty, closed, convex subset of $X$ and $v\in X$. I want to show that the minimization problem

$$\min_{x\in C}d(x,v)$$

has a solution. How do I do this?

What I'm trying to do is to define $m\equiv \inf_{x\in C}d(x,v)$. Consider the sequence $d(x_n,v)\to m$. I'm having problems showing this sequence is Cauchy.

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Take $x=v$, $y=x_n$, and $z=x_m$, and let $d(v,x_n)<m+\epsilon$ and $d(v,x_m)<m+\epsilon$:

$$d(v,x_n)^2+d(v,x_m)^2=2d\left(v,\frac{x_n+x_m}{2}\right)^2+\frac{1}{2}d(x_n,x_m)^2$$

$$d(x_n,x_m)^2=2d(v,x_n)^2+2d(v,x_m)^2-4d\left(v,\frac{x_n+x_m}{2}\right)^2.$$

Since $C$ is convex, $(x_n+x_m)/2\in C$, which implies

$$d\left(v,\frac{x_n+x_m}{2}\right)\geq m.$$

Therefore, $$d(x_n,x_m)^2\leq 2d(v,x_n)^2+2d(v,x_m)^2-4m^2$$ $$\leq 2(m^2+\epsilon^2+2m\epsilon)+2(m^2+\epsilon^2+2m\epsilon)-4m^2=4\epsilon^2+8m\epsilon.$$ It follows that the sequence $(x_n)$ is Cauchy.

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  • $\begingroup$ Hi! I wanna ask why you took $x=v$ in the first place. If you do this, doesn't it mean that $\min d(x,v)=0$? $\endgroup$
    – user45416
    Nov 5, 2023 at 14:48
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    $\begingroup$ I mean that I replaced $x$ with $v$ in your parallelogram law. In the minimization problem, $x$ isn't a fixed element. $\endgroup$ Nov 5, 2023 at 14:59

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