Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $H$ be a Hilbert space and $\left\{ e_{i}\right\} _{i=1}^{\infty}$ an orthonormal system. I need to prove that the following set is a convex set:

$$C=\left\{ x\in H\,:\,\sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)^{2}\cdot\left|\left\langle x,e_{n}\right\rangle \right|^{2}\leq1\right\}.$$

I thought to define: $C_{k}=\left\{ x\in H\,:\,\sum_{n=1}^{k}\left(1+\frac{1}{n}\right)^{2}\cdot\left|\left\langle x,e_{n}\right\rangle \right|^{2}\leq1\right\}$ and to prove that $C_{k}$ is convex for all $k$ and then we have $C=\bigcap_k C_{k}$ which will be convex too, but i couldn't manage to prove it. I would be glad to get some help.

Thanks!

share|improve this question
    
thanks a lot for the hint I solved it :) –  user18217 Nov 29 '11 at 16:36

1 Answer 1

As Davide comments, it suffices to show that the function $$ x \mapsto \sum_{n=1}^{\infty} \left( 1 + \frac1n \right)^2 |\langle x, e_n \rangle|^2 $$ is convex, since the sublevel sets of a convex function are convex. To prove this, it is sufficient to show that for any $e_n$, the map $x \mapsto |\langle x, e_n\rangle|^2$ is convex.

The function $|\langle x, e_n \rangle|^2$ is the composition of a linear function $x \mapsto \langle x, e_n \rangle$ with the convex function $\mathbf C \to \mathbf R : u \mapsto |u|^2$, and hence it is convex. Alternatively, we can establish convexity as follows. Fix vectors $x, y$ and $t \in [0, 1]$. Then $$ |\langle tx+(1-t)y, e_n \rangle|^2 = (t \langle x, e_n \rangle + (1-t) \langle y, e_n \rangle)^2 \leqslant t |\langle x, e_n \rangle|^2 + (1-t) |\langle y, e_n \rangle|^2, $$ where the last step is by the convexity of the function $\mathbf C \to \mathbf R : u \mapsto |u|^2$. Thus the map $x \mapsto |\langle x,e_n \rangle|^2$ is convex.

Edit: I had originally assumed real Hilbert space, but the proof works exactly the same for complex Hilbert spaces.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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