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 $\mathbb{H}=\oplus_{i=1}^d\mathcal{H}_k$.
Let $\mathbb{T}= (T_{ij})_{d \times d}$ be an operator matrix on $\mathcal{B}(\oplus_{i=1}^d\mathcal{H}_k)$ 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
$$
\|\tilde{\mathbb{T}}x\|^2=\sum_k\left|\sum_j\|T_{kj}\|x_j\right|^2.
$$
Best Answer
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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 $$ \|\mathbb{T}x\|^2=\sum_k\left\|\sum_jT_{kj}x_j\right\|_k^2\leq\sum_k\left(\sum_j\|T_{kj}\|_{\mathcal{B}(\mathcal{H}_j,\mathcal{H}_k)}\,\|x_j\|_j\right)^2\\ = \left\| \tilde {\Bbb T} \ \tilde x \right\|^2 \leq \|\tilde {\Bbb T}\|^2\|\tilde x\|^2 = \|\tilde {\Bbb T}\|^2 \| x\|^2. $$