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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?

Stephen Montgomery-Smith
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luka5z
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1 Answers1

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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.

Vladimir
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  • 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)? – luka5z Jun 14 '14 at 10:40
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    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. – Vladimir Jun 14 '14 at 10:42
  • COuld you also explain how $||Af||^2=\sum_{n=1}^{\infty}|\langle f,x_n\rangle|^2$ ? EDIT: sorry I know, $e_n$ are orthonormal. everything is okay – luka5z Jun 14 '14 at 10:45
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    See the edited answer where I added how to use B-S – Vladimir Jun 14 '14 at 10:46
  • Sorry again, where we use the fact that $\sum_{n=1}^{\infty}|\langle f,x_n\rangle|^2 < \infty, ::: f\in H$? – luka5z Jun 14 '14 at 10:55
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    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). – Vladimir Jun 14 '14 at 10:57
  • And we only use the fact that $|\langle f,x_n\rangle| \rightarrow 0$ – luka5z Jun 14 '14 at 10:59
  • I just can;t see how we can conclude that $Af=\sum_{n=1}^\infty\langle f,x_n\rangle e_n < \infty$ from $\sum_{n=1}^{\infty}|\langle f,x_n\rangle|^2 < \infty$. Could you explain? It is my last question. I would be extremely grateful. – luka5z Jun 14 '14 at 11:17
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    Since the vectors $e_n$ are orthonormal, we can use Pythagoras' formula to compute the norm of the sum: $||\sum_{n=n_1}^{n_2}\langle f,x_n\rangle e_n||^2=\sum_{n=n_1}^{n_2}|\langle f,x_n\rangle|^2$. – Vladimir Jun 14 '14 at 11:50