Let $\mathcal{H}$ be a Hilbert space and let $\mathcal{K}$ be a Hilbert space with an orthonormal basis $\{ e_i \}_{i \in I}$. Let $A$ be bounded linear operator from $\mathcal{H} \otimes \mathcal{K}$ to $\mathcal{H} \otimes \mathcal{K}$. How to show that $$\| A \|^2 \leq \sum_{i,j} \|(I_{\mathcal{H}} \otimes \left< e_i \right| ) A (I_{\mathcal{H}} \otimes \left| e_j \right>)\|^2,$$
where $\left< \cdot \right|$ and $\left| \cdot \right>$ are Dirac "bra" and "ket"?
Is it true that if $A$ is a bounded linear operator on a Hilbert space $\mathcal{H}$ then $\|A\| = \sup_{N}\|P_{N}A\|?$ The operator $P$ is a projection onto $N \subseteq \mathcal{H}$ and the supremum is taken over all finite-dimensional subspaces of $\mathcal{H}$.
Thank you for the help.
This is answer only to the second (simpler) part of your question.
Let $N\subset H$ be closed subspace, then $\Vert P_N\Vert\leq 1$ and $\Vert P_N A\Vert\leq \Vert P_N\Vert\Vert A\Vert=\Vert A\Vert$. Since $N$ is arbitrary we have $$ \sup\limits_{N\subset H}\Vert P_N A\Vert\leq\Vert A\Vert\tag{1} $$ Fix $\varepsilon>0$, then there exist $x\in H$ wth $\Vert x\Vert=1$, such that $\Vert Ax\Vert>\Vert A\Vert-\varepsilon$. Consider closed one-dimensional subspace $N_1=\mathrm{span}\{Ax\}$, then $$ \sup\limits_{N\subset H}\Vert P_N A\Vert\geq\Vert P_{N_1} A\Vert=\Vert P_{N_1} A\Vert\Vert x\Vert\geq\Vert P_{N_1} Ax\Vert=\Vert Ax\Vert>\Vert A\Vert-\varepsilon $$ Since $\varepsilon>0$ is arbitrary we have $$ \sup\limits_{N\subset H}\Vert P_N A\Vert\geq\Vert A\Vert\tag{2} $$ From $(1)$ and $(2)$ we get the desired equality.