I have a similar question to definition of Bessel sequence, where it was solved using Banach-Steinhaus.

A sequence $\{f_k\}_{k=1}^{\infty}$ is called a Bessel sequence in a Hilbert space $\mathscr{H}$, if $\sum_{k=1}^{\infty}|\langle f,f_k\rangle|^2 < \infty \, \forall f \in \mathscr{H}$.

Statement: Given a Bessel sequence use the close graphing theorem to show that $\sum_{k=1}^{\infty}|\langle f,f_k \rangle|^2 \leq B\|h\|^2$ where $B < \infty$

Attempted proof: Let $\{f_i\}_{i=1}^{\infty} \, \in \, \mathscr{H}$ where $\lim_{i\rightarrow \infty} f_i = f$ and define $T(f) = \{\langle f, f_n\rangle\} = \{F_n\}_{n=1}^{\infty}\, \in \, l_2$

Now for a fixed $m$ we have:

$|F_m - \langle f_i, f_m\rangle| \leq |\sum_{k=1}^{\infty}|F_k -\langle f_i,f_k\rangle|^2|^{\frac{1}{2}} = \|{F_k} - T(f_i)\|_2 \rightarrow 0 \, \mbox{as} \, i\rightarrow \infty$


$\lim_{i\rightarrow \infty} \langle f_i, f_m\rangle = F_m$

So we have convergence hence we are closed and by the CGT

$\sum_{k=1}^{\infty}|\langle f,f_k \rangle|^2 \leq B\|h\|^2$ where $B < \infty$



It looks like you're using the Banach-Steinhaus Theorem, not the Closed Graph Theorem.

First note that since $(f_i)$ is a Bessel sequence, the map $T$ is well-defined. Clearly, $T$ is linear. We wish to show that $T$ is bounded, which is precisely what the inequality in your Statement says.

To use the Closed Graph Theorem, you can show the following holds:

$\ \ \ $Suppose $(x_n)$ is a sequence in $H$, $x_n\rightarrow x$, and $Tx_n\rightarrow y$, where $x\in H$ and $y\in\ell_2$.

$\ \ \ $Then $y=Tx$.

If you can do this, the Closed Graph Theorem will tell you $T$ is bounded;

So suppose $(x_n)$ is a sequence in $H$ that converges to $x\in H$, and that $Tx_n\rightarrow y\in \ell_2$. In the following, we denote the $j$'th coordinate of a vector $x$ by $x(j)$.

Fix a coordinate $j$. We'll show $(Tx)(j)=y(j)$. As, $j$ is arbitrary, once we've done this, we can conclude $Tx=y$.

The $j$'th coordinate of $Tx_n$ is $f_j(x_n)$. Since $x_n\rightarrow x$, as $n\rightarrow\infty$, and since $f_j$ is a continuous linear functional on $H$, we have $$(Tx_n)(j)=f_j(x_n)\ \buildrel{n\rightarrow\infty}\over\longrightarrow \ f_j(x) =(Tx)(j).$$ In short, $Tx_n$ converges to $Tx$ coordinatewise. As mentioned above, this implies $y=Tx$, as desired.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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