Let $\mathcal H_0$ and $\mathcal H_1$ be two Hilbert spaces. Then $\mathcal H_0 \oplus \mathcal H_1$ is also a Hilbert space endowed with the inner product $\langle x_0 \oplus x_1, y_0 \oplus y_1 \rangle_{\mathcal H_0 \oplus \mathcal H_1} = \langle x_0,y_0 \rangle_{\mathcal H_0} + \langle x_1,y_1 \rangle_{\mathcal H_1},$ $x_0,y_0 \in \mathcal H_0$ and $x_1,y_1 \in \mathcal H_1.$ Let $T_0 \in \mathrm B(\mathcal H_0)$ and $T_1 \in \mathrm B(\mathcal H_1).$ Define a function $T : \mathcal H_0 \oplus \mathcal H_1 \longrightarrow \mathcal H_0 \oplus \mathcal H_1$ defined by $$T(x_0 \oplus x_1) = T_0 (x_0) \oplus T_1 (x_1),\ x_0 \in \mathcal H_0, x_1 \in \mathcal H_1.$$ Then show that $T$ is a bounded linear operator and $\|T\| = \max \{\|T_0\|, \|T_1\| \}.$
I am able to prove linearity of $T$ and $\|T\| \leq \max \{\|T_0\|,\|T_1\| \}.$ But I am having hard time proving the equality. Would anybody please help me in this regard?
Thanks for your time.
EDIT $:$ I just come up with an argument which is as follows.
$$\begin{align*} \|T\| & = \sup\limits_{\substack {x_0 \oplus x_1 \in \mathcal H_0 \oplus \mathcal H_1} \\ \ \ \ \ \ \ \ {x_0 \oplus x_1 \neq 0}} \frac {\sqrt {\|T_0 (x_0)\|^2 + \|T_1 (x_1)\|^2}} {\sqrt {\|x_0\|^2 + \|x_1\|^2}} \\ & \geq \sup\limits_{\substack {x_0 \in \mathcal H_0} \\ \ \ {x_0 \neq 0}} \frac {\|T_0(x_0)\|} {\|x_0\|} \\ & = \|T_0\| \end{align*}$$ Similarly we can prove that $\|T\| \geq \|T_1\|.$ Hence $\|T\| \geq \max \{\|T_0\|,\|T_1\| \}.$