# Show that there exists constant $C$ such that $\sum_{n=1}^{\infty}|\langle f,x_n\rangle |^2\le C \|f\|^2$

Let $H$ be Hilbert space.

I have to show that if $\sum_{n=1}^{\infty}|\langle f,x_n\rangle|^2 < \infty, \:\:\: f\in H$

then there exists constant $C\ge 0$ such that $\sum_{n=1}^{\infty}|\langle f,x_n\rangle|^2\le C \|f\|^2, \:\:\: f\in H$

Is this somehow connected with Bessel inequality? Could you give my any tips?

• Finite for all $f$ or for some $f$? I guess for all $f$... Jun 14 '14 at 10:13
• Obviously for all $f$ Jun 14 '14 at 10:15
• You want to use the open mapping theorem, or closed graph theorem, in some manner. Jun 14 '14 at 10:17
• Yes otherwise it's trivial =P Jun 14 '14 at 10:17
• What are the $x_n$? Some Hilbert basis? Some family of pairwise orthogonal vectors? Jun 14 '14 at 10:24

Let $\{e_n\}$ be an orthonormal basis in $H$. Define an operator $A:H\to H$ by $$Af=\sum_{n=1}^\infty\langle f,x_n\rangle e_n$$ and apply the Banach-Steinhaus theorem to prove that $A$ is bounded. Indeed, set $$A_mf=\sum_{n=1}^m\langle f,x_n\rangle e_n.$$ Then $A_mf\to Af$ for every $f\in H$ and hence $A_m$ are uniformly bounded by Banach-Steinhaus and $A$ is bounded.
• Because $||Af||^2$ is just the sum of squares you need to estimate in terms of $||f||^2$, so your problem is reworded as "prove that $A$ is bounded". And for this sort of task, there are various theorems. In your case, Banach-Steinhaus applies. Jun 14 '14 at 10:42
• Exactly where we 1) see that $Af$ is well defined (the series converges); 2) see that $A_mf\to Af$ (again because the series converges). Jun 14 '14 at 10:57
• Since the vectors $e_n$ are orthonormal, we can use Pythagoras' formula to compute the norm of the sum: $||\sum_{n=n_1}^{n_2}\langle f,x_n\rangle e_n||^2=\sum_{n=n_1}^{n_2}|\langle f,x_n\rangle|^2$. Jun 14 '14 at 11:50