Let $\left\{f_k\right\}_{k=1}^{\infty}$ be a sequence in $\mathcal{H}$ and $B>0$ be given. Then if $\left\{f_k\right\}_{k=1}^{\infty}$ is a Bessel sequence with Bessel bound $B$ 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 $\mathcal{H}$ and $\|T\| \leq \sqrt{B}$.
Don't focus on the part concerning that $T$ is a well-defined bounded operator (we can find it e.g. here), but consider instead the claim $\|T\| \leq \sqrt{B}$. I know that \begin{aligned} \left\| T\left\{c_k\right\}_{k=1}^{\infty} \right\| &=\sup _{\|g\|=1}\left|\left\langle T\left\{c_k\right\}_{k=1}^{\infty},g\right\rangle\right| \\ &=\sup _{\|g\|=1}\left|\left\langle\sum_{k=1}^\infty c_k f_k, g\right\rangle\right|\\ & \leq \sup _{\|g\|=1} \sum_{k=1}^\infty\left|c_k\left\langle f_k, g\right\rangle\right| \\ & \leq\left(\sum_{k=1}^\infty\left|c_k\right|^2\right)^{1 / 2} \sup _{\|g\|=1}\left(\sum_{k=1}^\infty\left|\left\langle f_k, g\right\rangle\right|^2\right)^{1 / 2} \\ & \leq \sqrt{B}\left(\sum_{k=1}^\infty\left|c_k\right|^2\right)^{1 / 2} . \end{aligned} The question is, why from this we can infer that $\|T\| \leq \sqrt{B}$?
