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 $$ 2||y_n-x||^2 + 2||y_m-x||^2 = ||(y_n+y_m)-2x||^2+||y_n-y_m||^2.$$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.