Let $(\mathcal{H}_1,\langle \cdot\mid \cdot\rangle_1), (\mathcal{H}_2,\langle \cdot\mid \cdot\rangle_2), \cdots, (\mathcal{H}_d,\langle \cdot\mid \cdot\rangle_d)$ be complex Hilbert spaces and let $\mathbb{H}=\oplus_{i=1}^d\mathcal{H}_k$. On $\mathbb{H}$, we have the following inner product $$\langle x,y\rangle=\sum_{k=1}^d\langle x_k\mid y_k\rangle_k,$$ for all $x=(x_1,\cdots,x_d)\in \mathbb{H}$ and $y=(y_1,\cdots,y_d)\in \mathbb{H}$.

Notice that every operator $\mathbb{T}\in \mathcal{B}(\oplus_{i=1}^d\mathcal{H}_k)$ can be represented as a $d\times d$ operator matrix (or partitioned operator) of the form $\mathbb{T}=[T_{ij}]_{i,j}$ with $T_{i,j}$ a bounded linear operator from $\mathcal{H}_j$ into $\mathcal{H}_i$. Note that for every vector $$ x=\begin{bmatrix}x_1 \\ x_2\\ \vdots \\ x_d \end{bmatrix}\in \displaystyle\bigoplus_{i=1}^d\mathcal{H}_k, $$ we have $$ \mathbb{T}x=\begin{bmatrix} \displaystyle\sum_{j=1}^dT_{1j}x_j \\ \displaystyle\sum_{j=1}^dT_{2j}x_j \\ \vdots \\ \displaystyle\sum_{j=1}^dT_{dj}x_j \end{bmatrix}\in \displaystyle\bigoplus_{i=1}^d\mathcal{H}_k, $$

Let $\mathbb{T}= (T_{ij})_{d \times d}$ be an operator matrix and $\tilde{\mathbb{T}} = (\|T_{ij} \|_{\mathcal{B}(\mathcal{H}_j,\mathcal{H}_i)})_{d\times d}$ its block-norm matrix. Why $$\|\mathbb{T}\|_{\mathcal{B}(\oplus_{i=1}^d\mathcal{H}_k)} \leq \| \tilde{\mathbb{T}} \|?$$

Attempt: Let $x=(x_1,\cdots,x_d)\in \oplus_{i=1}^d\mathcal{H}_k$. Then, $$ \|\mathbb{T}x\|^2=\sum_k\left\|\sum_jT_{kj}x_j\right\|_k^2\leq\sum_k\left(\sum_j\|T_{kj}\|_{\mathcal{B}(\mathcal{H}_j,\mathcal{H}_k)}\,\|x_j\|_j\right)^2 . $$

On the other hand, $$\| \tilde{\mathbb{T}} \|=\sup_{\|x\|_{\mathbb{R}^d}}\| \tilde{\mathbb{T}}x\|.$$ For all $x=(x_1,\cdots,x_d)\in \mathbb{R}^d$ we have $$ \|\tilde{\mathbb{T}}x\|^2=\sum_k\left|\sum_j\|T_{kj}\|x_j\right|^2. $$


You're almost there!

Given $x = (x_1,\dots,x_d) \in \Bbb H$, let $\tilde x$ denote $(\|x_1\|,\dots,\|x_d\|) \in \Bbb R^d$; note that $\|\tilde x\| = \|x\|$. It suffices to show that we always have $\|\Bbb Tx\| \leq \|\tilde {\Bbb T}\|\, \|x\|$. Indeed, we have $$ \|\mathbb{T}x\|^2=\sum_k\left\|\sum_jT_{kj}x_j\right\|_k^2\leq\sum_k\left(\sum_j\|T_{kj}\|_{\mathcal{B}(\mathcal{H}_j,\mathcal{H}_k)}\,\|x_j\|_j\right)^2\\ = \left\| \tilde {\Bbb T} \ \tilde x \right\|^2 \leq \|\tilde {\Bbb T}\|^2\|\tilde x\|^2 = \|\tilde {\Bbb T}\|^2 \| x\|^2. $$


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