# How did this formula use convexity and parallelogram law?

Given a closed convex subset $$M$$ of a Hilbert space $$H$$, let $$x \in H$$ be a point s.t. it is not in $$M$$, i.e. $$x \not \in M$$.

Let $$d = \text{inf}_{y \in M}||x-y||$$, where $$||.||$$ is the inner product and $$d$$ is the smallest distance from $$x$$ to the unique point $$y$$ in $$M$$..

The text am reading says that there is a sequence of vectors $$y_n \in M$$ s.t. $$||x-y_n|| \rightarrow d$$.

Now, it says that using the convexity of $$M$$ and the parallelogram law, we can show that $$y_n$$ is a Cauchy sequence, i.e.

$$||y_n-y_m||^2 = 2||y_n-x||^2 + 2||y_m-x||^2 - ||(y_n+y_m)-2x||^2 \rightarrow 0 \quad \text{as n,m \rightarrow \infty}$$

But I could not figure out how the above formula was derived....

how does one apply (1) convexity of $$M$$ and (2) the parallelogram law to get this formula?

The parallelogram law is $$2||x||^2 + 2||y||^2 = ||x+y||^2 + ||x-y||^2.$$ So $$2||y_n-x||^2 + 2||y_m-x||^2 = ||(y_n+y_m)-2x||^2+||y_n-y_m||^2.$$
• @skim $$2||y_n-x||^2 + 2||y_m-x||^2 - ||(y_n+y_m)-2x||^2=2||y_n-x||^2 + 2||y_m-x||^2 - 4||(y_n+y_m)/2-x||^2.$$ Note that $(y_n+y_m)/2\in M$ by convexity of $M$. – Shivering Soldier Feb 8 '20 at 3:15
• @Skim: Since $(y_n+y_m)/2\in M$, we have $||(y_n+y_m)/2-x||\geq d$. So, $$||y_n-y_m||^2 = 2||y_n-x||^2 + 2||y_m-x||^2 - 4||(y_n+y_m)/2-x||^2\leq2||y_n-x||^2 + 2||y_m-x||^2-4d.$$ As $n$ goes to infinity, $2||y_n-x||^2 + 2||y_m-x||^2-4d\to0$. – Shivering Soldier Feb 8 '20 at 3:25