Assume $C$ is a convex subset of a Hilbert space $H$ ($C$ is not necessarily close) and $x_0\notin C$.Let $r=d(x_0,C)$. Prove that $\{y\in H\mid\|y-x_0\|\leq r\}\cap C$Has at most 1 element.
I want to use this post to show that for:$\|x-x_0\|=\|y-x_0\|=r$
We can say $\|\frac{x+y}{2}-x_0\|=\|\frac{x-x_0+y-x_0}{2}\| <r$ (i remember something similar from class)
But this is as far as i could get. Any ideas regarding this approach?
Use the parallelogram law: $$2\|a\|^2 + 2\|b\|^2 = \|a+b\|^2 + \|a-b\|^2.$$ Suppose that $y_1$ and $y_2$ are both elements of $C$ with $\|y_1 - x_0\| \le r$ and $\|y_2 - x_0\| \le r$. Then $$\|(y_1 + y_2) - 2x_0\|^2 + \|y_1 - y_2\|^2 = 2\|y_1 - x_0\|^2 + 2\|y_2 - x_0\|^2 \le 4r^2.$$ Since $C$ is convex you have $\dfrac{y_1 + y_2}{2} \in C$, so that $$4r^2 \le 4 \left\| \frac{y_1 + y_2}{2} - x_0 \right\|^2 = \| (y_1 + y_2) - 2x_0\|^2.$$ You can put these together to obtain $$ 4r^2 + \|y_1 - y_2\|^2 \le 4r^2$$ from which $y_1 = y_2$ follows.
