Prove that if $\left\{f_k\right\}_{k=1}^{\infty}$ is a Bessel sequence in $\mathcal{H}$, then $\sum_{k=1}^{\infty} c_k f_k$ converges unconditionally.

Question

I would like to prove the following:

Let $\mathcal{H}$ be a Hilbert space. If $\left\{f_k\right\}_{k=1}^{\infty}$ is a Bessel sequence in $\mathcal{H}$, then $\sum_{k=1}^{\infty} c_k f_k$ converges unconditionally for all $\left\{c_k\right\}_{k=1}^{\infty} \in \ell^2(\mathbb{N})$.

My attempt. Recall that if $\sum_{k=1}^{\infty} f_{\sigma(k)}$ is convergent for all permutations $\sigma$, we say that $\sum_{k=1}^{\infty} f_k$ is unconditionally convergent. Moreover, if $\left\{f_k\right\}_{k=1}^{\infty}$ is a Bessel sequence in $\mathcal{H}$, then the operator $$ T:\left\{c_k\right\}_{k=1}^{\infty} \rightarrow \sum_{k=1}^{\infty} c_k f_k $$ is a well-defined bounded operator from $\ell^2(\mathbb{N})$ into $\mathcal{H}$. Let us consider then $T\left\{c_{\sigma(k)}\right\}_{k=1}^{\infty} $. Hence, \begin{aligned} \left\| T\left\{c_k\right\}_{k=1}^{n}-T\left\{c_{\sigma(k)}\right\}_{k=1}^{m} \right\| &=\sup _{\|g\|=1}\left|\left\langle T\left\{c_k\right\}_{k=1}^{n}- T\left\{c_{\sigma(k)}\right\}_{k=1}^{m},g\right\rangle\right| \\ &=\sup _{\|g\|=1}\left|\left\langle\sum_{k=1}^n c_{k} f_{k}-\sum_{k=1}^m c_{\sigma(k)} f_{\sigma(k)}, g\right\rangle\right| \end{aligned} It seemed like a good idea but I don't know now what to do with the last step.

Can anyone help me out with this? Thank you!

Answer 1

It is enough to show that $\sum_kc_k f_k$ converges in $H$ since for any permutation $\sigma$ on $\mathbb{N}$, $(f_{\sigma(k)}:k\in\mathbb{N})$ is a Bessel sequence whenever $(f_k)$ is. Indeed, \begin{align} \sum_k|\langle x,f_k\rangle|^2=\sum_k|\langle x,f_{\sigma(k)}\rangle|^2\tag{1}\label{one} \end{align}

Suppose then that $$ \Big(\sum_k|\langle x,f_k\rangle|^2\Big)^{1/2}\leq B\|x\|_H, \qquad x\in H.$$

For any $m<n$ \begin{align}\Big\|\sum^n_{k=m}c_kf_k\Big\|_H&=\sup_{g\in H: \|g\|_H=1}\Big|\langle\sum^n_{k=m}c_kf_k,g\rangle\Big|\leq\sup_{g\in H: \|g\|_H=1}\sum^n_{k=m}|c_k||\langle f_k,g\rangle|\\ &\leq\sup_{g\in H:\|g\|_H=1} \Big(\sum^n_{k=m}|c_k|^2\Big)^{1/2}\Big(\sum^n_{k=m}|\langle f_k,g\rangle|^2\Big)^{1/2}\leq B\Big(\sum^n_{k=m}|c_k|^2\Big)^{1/2}\tag{2}\label{two} \end{align}

From this, it follows that $\sum^n_{k=1}c_kf_k$ defines a Cauchy sequence in $H$ and so, it converges. A similar computation shows that $$\|\sum_k c_kf_k\|_H\leq B\|c\|_{\ell_2}$$

Finally, fix a permutation $\sigma$ on $\mathbb{N}$. To show that $$x=\sum_kc_kf_k=\sum_jc_{\sigma(k)}f_{\sigma(k)}$$ set $s_n=\sum^n_{k=1}c_kf_k$ and $t_n=\sum^n_{k=1}c_{\sigma(k)}f_{\sigma(k)}$ We have shown that $\lim_n\|x-s_n\|_H=0$. From \eqref{two}, given $\varepsilon>0$ there is $N$ so that $$\|x-s_n\|^2_H\leq B^2\sum_{k>n}|c_k|^2<\varepsilon^2/4, \qquad n\geq N$$ Then $$\|x-t_n\|^2_H\leq 2\|x-s_n\|^2_H+2\|s_n-t_n\|^2_H<2\|x-s_n\|^2_H + \varepsilon^2/2 $$ Choose $M\geq N$ large enough so that $\{1,\ldots,N\}\subset\{\sigma(1),\ldots,\sigma(M)\}$. Then, for $n\geq M$ \begin{align} |t_n-s_n\|^2_H&=\|c_{\sigma(1)}f_{\sigma(1)}+\ldots +c_{\sigma(n)}f_{\sigma(n)}-c_1f_1-\ldots - c_nf_n\|^2_H\\ &\leq B^2\sum_{k>N}|c_k|^2<\varepsilon^2/4 \end{align} which follows again from \eqref{two} since all terms $c_jf_j$ with $1\leq j\leq N$ get cancelled.

Putting things together we get that for $n\geq M$, $\|t_n-x\|_H<\varepsilon$.