If a frame has upper frame bound $B$, why does the synthesis operator $T$ satisfy $\|T\|\le \sqrt B$?

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Let $\mathcal H$ be a Hilbert space. The following $$A\left\|f\right\|^{2}\leq \sum _{{k\in {\mathbb {N}}}}\left|\langle f,f_{k}\rangle \right|^{2}\leq B\left\|f\right\|^{2}$$ for each $f\in \mathcal H$, is the definition of frame $\{ f_k \}_{k=1}^\infty$ for $\mathcal H$ with bounds $A$ and $B$.

The following $$T:\ell^2(\mathbb N)\to \mathcal H, \ \ \ \ T\{c_k\}:=\sum_{k=1}^\infty c_k f_k\, .$$ is the pre-frame operator. My textbook claims (without proof!) that $\|T\|\leq \sqrt B$.

Can I have any suggestions please?

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Suppose $\{f_k\}_{k \in \mathbb{N}}$ is a frame for $\mathcal{H}$ with frame bounds $A,B > 0$. Then, in particular, it holds for all $f \in \mathcal{H}$ that $$ \sum_{k \in \mathbb{N}} |\langle f, f_k \rangle |^2 \leq B \|f\|^2, $$ which is equivalent to the map $C : \mathcal{H} \to \ell^2 (\mathbb{N}), \; f \mapsto (\langle f, f_k \rangle)_{k \in \mathbb{N}}$ being bounded and $\| C \|_{op} \leq B^{1/2}$. Since $\langle Cf, f \rangle = \langle f, Tf\rangle$ for all $f \in \mathcal{H}$, it follows that $T$ is the Hilbert adjoint of $C$. Thus $\|T\|_{op} = \|C\|_{op} \leq B^{1/2}$.