# How do I show that the minimization problem has a solution?

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.

Take $$x=v$$, $$y=x_n$$, and $$z=x_m$$, and let $$d(v,x_n) and $$d(v,x_m):

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

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

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.

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