# Norm of a block of matrix operator

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 $$\|\tilde{\mathbb{T}}x\|^2=\sum_k\left|\sum_j\|T_{kj}\|x_j\right|^2.$$

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 $$\|\mathbb{T}x\|^2=\sum_k\left\|\sum_jT_{kj}x_j\right\|_k^2\leq\sum_k\left(\sum_j\|T_{kj}\|_{\mathcal{B}(\mathcal{H}_j,\mathcal{H}_k)}\,\|x_j\|_j\right)^2\\ = \left\| \tilde {\Bbb T} \ \tilde x \right\|^2 \leq \|\tilde {\Bbb T}\|^2\|\tilde x\|^2 = \|\tilde {\Bbb T}\|^2 \| x\|^2.$$