Question about Bessel sequence in a Hilbert space $\mathcal{H}$ I have a sequence in a Hilbert space $\mathcal{H}$, denoted by $\left\{f_k\right\}_{k=1}^{\infty}$. I would like to prove that, if there exists $B>0$ such that
$$
\left\|\sum_{k=1}^{\infty} c_k f_k\right\|^2 \leq B \sum_{k=1}^{\infty}\left|c_k\right|^2
$$
then $\left\{f_k\right\}_{k=1}^{\infty}$ is a Bessel sequence with bound $B$.
I tried to obtain a proof but I'm afraid I was wrong. Anyway it's this: what do you think?
Proof.
$$\left\|\sum_{k=1}^{\infty} c_k f_k\right\|=\sup_{\|g\|=1}\left|\langle \sum_{k=1}^{\infty} c_k f_k,g \rangle\right|\leq\sup_{\|g\|=1}\sum_{k=1}^{\infty} |c_k|\, \left|\langle  f_k,g \rangle\right|\leq \sqrt{\sum_{k=1}^{\infty} |c_k|^2}\, \sup_{\|g\|=1}\sqrt{\sum_{k=1}^{\infty} \left|\langle  f_k,g \rangle\right|^2}$$
Thus if in the latter I require that $
\left\|\sum_{k=1}^{\infty} c_k f_k\right\|^2 \leq B \sum_{k=1}^{\infty}\left|c_k\right|^2
$ then
$$\left\|\sum_{k=1}^{\infty} c_k f_k\right\|^2\leq \sum_{k=1}^{\infty} |c_k|^2\, \sup_{\|g\|=1} \sum_{k=1}^{\infty} \left|\langle  f_k,g \rangle\right|^2\leq B \sum_{k=1}^{\infty}\left|c_k\right|^2$$

This follows by duality.
Consider the operator $R:\ell_2(\mathbb{N})\rightarrow H$ as
$Rg=\sum_ng(n)h_n$. This operator is bounded by assumption:
$\|Rg\|_H\leq B^{1/2}\|g\|_{\ell_2}$.
The adjoint operator $R^*:H\rightarrow\ell_2$ is a bounded linear operator with  $\|R^*\|_{L(H,\ell_2)}=\|R\|_{L(\ell_2,H)}\leq B^{1/2}$ such that
\begin{align}
\langle Rg,h\rangle_H&=\langle \sum_ng(n)h_n, h\rangle_H=\sum_n\langle g(n)h_n,h\rangle_H\\
&=\sum_ng(n)\overline{\langle h,h_n\rangle_H}
=\langle g, R^*h\rangle_{\ell_2}
\end{align}
It follows that that the sequence $n\mapsto\langle h,h_n\rangle$ is in $\ell_2$ (and application of the uniform boundedness theorem for example) and that $(R^*h)(n)=\langle h,h_n\rangle$ for each $n\in\mathbb{N}$.
