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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.


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

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.

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