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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?

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The parallelogram law is $$ 2||x||^2 + 2||y||^2 = ||x+y||^2 + ||x-y||^2.$$ So $$ 2||y_n-x||^2 + 2||y_m-x||^2 = ||(y_n+y_m)-2x||^2+||y_n-y_m||^2.$$

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  • $\begingroup$ where was convexity applied? $\endgroup$ – sma Feb 8 '20 at 2:30
  • $\begingroup$ @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 $M$. $\endgroup$ – Shivering Soldier Feb 8 '20 at 3:15
  • $\begingroup$ @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, $2||y_n-x||^2 + 2||y_m-x||^2-4d\to0$. $\endgroup$ – Shivering Soldier Feb 8 '20 at 3:25

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