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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?

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  • $\begingroup$ Finite for all $f$ or for some $f$? I guess for all $f$... $\endgroup$ Jun 14 '14 at 10:13
  • $\begingroup$ Obviously for all $f$ $\endgroup$
    – luka5z
    Jun 14 '14 at 10:15
  • $\begingroup$ You want to use the open mapping theorem, or closed graph theorem, in some manner. $\endgroup$ Jun 14 '14 at 10:17
  • $\begingroup$ Yes otherwise it's trivial =P $\endgroup$ Jun 14 '14 at 10:17
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    $\begingroup$ What are the $x_n$? Some Hilbert basis? Some family of pairwise orthogonal vectors? $\endgroup$ Jun 14 '14 at 10:24
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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.

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  • $\begingroup$ Could you explain me why you are constructing such an operator and how do we know it is bounded (in order to apply banach-steinhaus)? $\endgroup$
    – luka5z
    Jun 14 '14 at 10:40
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    $\begingroup$ 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. $\endgroup$
    – Vladimir
    Jun 14 '14 at 10:42
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    $\begingroup$ See the edited answer where I added how to use B-S $\endgroup$
    – Vladimir
    Jun 14 '14 at 10:46
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    $\begingroup$ 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). $\endgroup$
    – Vladimir
    Jun 14 '14 at 10:57
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    $\begingroup$ Since the vectors $e_n$ are orthonormal, we can use Pythagoras' formula to compute the norm of the sum: $||\sum_{n=n_1}^{n_2}\langle f,x_n\rangle e_n||^2=\sum_{n=n_1}^{n_2}|\langle f,x_n\rangle|^2$. $\endgroup$
    – Vladimir
    Jun 14 '14 at 11:50

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