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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.

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How $I_\mathcal{H}\otimes|e_j\rangle$ is defined? How it acts on vector $x\otimes y$? – Norbert Nov 27 '12 at 20:25
$(I_{\mathcal{H}} \otimes \left | e_j\right>) (x \otimes \lambda)= x \otimes \lambda e_j$, $x \in \mathcal{H}, \lambda \in \mathbb{C}$, $\left< e_i \right| y = \left<e_i, y\right>$, $y \in \mathcal{K}$. I mentioned that $\left< \cdot \right|, \left | \cdot\right>$ are Dirac "bra" and "ket". – Franz Nov 29 '12 at 9:51
if we try to work with simple tensors i.e. if we assume that $A$ can be decomposed as $A= A_{\mathcal{H}} \otimes A_\mathcal{K}$, and for simplicity we identify $x \otimes 1$ with $x$, $x \in \mathcal{K}$, then $$\| (I_{\mathcal{H}} \otimes \left< e_{i}\right|)A(I_{\mathcal{H}} \otimes \left| e_j\right>x\|^2 = \| A_{\mathcal{H}}x \otimes \left<A_{\mathcal{K}}e_j, e_i\right> \|^2 = |\left<A_{\mathcal{K}}e_j, e_i\right> |^2 \|A_{\mathcal{H}}\|^2 $$ – Franz Nov 29 '12 at 9:56
Above $x \in \mathcal{H}$ of course. Further we get $$\|A_{\mathcal{H}}x\|^2 \sum_{i \geq 1} \sum_{j \geq 1} |\left<A_{\mathcal{K}}e_j, e_i\right> |^2= \|A_{\mathcal{H}}\|^2 \sum_{i \geq 1} \| A_{\mathcal{K}}e_j\|^2 $$ so by continuity of $A$ and triangle inequality if we start from the left hand side i.e. $\|Ah\|^2, h = x \otimes y$, $y = \sum_{i \geq 0} \lambda_ie_i$ we got the inequality. Can we use some density arguments of the simple tensors in this case to justify the above reasoning? – Franz Nov 29 '12 at 10:10

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.

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