Why $\left\| \begin{pmatrix} 0 &A\\ B &0 \end{pmatrix}\right\|\geq\max\{\|A\|,\|B\|\}$? Let $E$ be a complex Hilbert space, with inner product $\langle\cdot\;, \;\cdot\rangle$ and the norm $\|\cdot\|$ and let $\mathcal{L}(E)$ the algebra of all operators on $E$.

Let $A,B\in\mathcal{L}(E)$. I want to prove that
  $$\left\| \begin{pmatrix}
0 &A\\
B &0
\end{pmatrix}\right\|=\max\{\|A\|,\|B\|\}.$$
  Here $\begin{pmatrix}
0 &A\\
B &0
\end{pmatrix}\in \mathcal{L}(E\oplus E)$.

It can be verified that 
$$\left\| \begin{pmatrix}
0 &A\\
B &0
\end{pmatrix}\right\|\leq\max\{\|A\|,\|B\|\}.$$

I'm facing difficulties to prove the converse inequality.

I see this result in a paper

 A: We have
$\left\{\left(\matrix{0\\ Y}\right) | \;\;\lVert Y\rVert^2=1\right\}\subseteq \left\{\left(\matrix{X\\ Y}\right) | \;\;\lVert X\rVert^2+\lVert Y\rVert^2=1\right\}$
So, by taking the sup on a bigger set,
$\sup_{\lVert X\rVert^2+\lVert Y\rVert^2=1}\left\lVert \pmatrix{0 & A\\B & 0}\pmatrix{ X \\ Y}\right\rVert\geq \sup_{\lVert Y\rVert^2=1}\left\lVert\pmatrix{0 & A\\B & 0}\pmatrix{0\\ Y}\right\rVert$
However, 
$\pmatrix{0 & A\\B & 0}\pmatrix{0\\ Y}=\pmatrix{AY \\ 0}$
So $\sup_{\lVert Y\rVert^2=1}\left\lVert\pmatrix{0 & A\\B & 0}\pmatrix{0\\ Y}\right\rVert=\sup_{\lVert Y\rVert^2=1}\left\lVert\pmatrix{AY \\ 0}\right\rVert$
$=\sup_{\lVert Y\rVert^2=1}\left\lVert AY \right\rVert$
$=\left\lVert A \right\rVert$
Hence
$\left\lVert \pmatrix{0 & A\\B & 0}\right\rVert\geq \left\lVert A \right\rVert$ 
idem for B.
A: Let $X=\begin{pmatrix}0&A\\B&0\end{pmatrix}$. By the $C^\ast$-identity we have
$$
\lVert X\rVert^2=\lVert X^\ast X\rVert=\left\lVert\begin{pmatrix}B^\ast B&0\\0&A^\ast A\end{pmatrix}\right\rVert.
$$
If $\xi_n\in E$ with $\|\xi_n\|=1$ and $\|A^\ast A\xi_n\|\to\|A\|^2$, then $\|(0,\xi_n)^T\|=1$ and
$$
\left\lVert X^\ast X\begin{pmatrix}0\\ \xi_n\end{pmatrix}\right\rVert=\left\lVert \begin{pmatrix}0\\ A^\ast A\xi_n\end{pmatrix}\right\rVert=\|A^\ast A\xi_n\|\to \|A\|^2.
$$
Thus $\|X\|^2\geq \|A\|^2$. The inequality $\|X\|^2\geq \|B\|^2$ can of course be proven similarly. Therefore $\|X\|\geq\max\{\|A\|,\|B\|\}$.
A: The formula holds for $B(X),$ the bounded operators on a normed space $X.$ No $C^*$-structure is needed.
Let  $X\times X$ be equipped with the norm $\|x\|=\|(x_1,x_2)\|=\sqrt{\|x_1\|^2+\|x_2\|^2}.$ Consider  operators  $T$ on $X\times X$ given by
$$Tx=\begin{pmatrix}
0 &A\\
B & 0\end{pmatrix}\begin{pmatrix} x_1 \\x_2\end{pmatrix}=\begin{pmatrix} Ax_2 \\ Bx_1\end{pmatrix}$$
where $A,B\in B(X)$ and $x_1,x_2\in X.$ Then
$$ \displaylines{\|Tx\|^2=\|Ax_2\|^2+\|Bx_1\|^2 \le \|A\|^2\|x_2\|^2+\|B\|^2\|x_1\|^2 \\ \le \max\{\|A\|^2,\|B\|^2\}\,\|x\|^2}$$
Therefore
$$\|T\|\le \max\{\|A\|,\|B\|\}$$
On the other hand
$$T\begin{pmatrix}x_1 \\ 0\end{pmatrix}= \begin{pmatrix} 0\\Bx_1\end{pmatrix},\qquad T\begin{pmatrix}0 \\ x_2\end{pmatrix}= \begin{pmatrix} Ax_2\\0\end{pmatrix}$$
Therefore $$\|T\|\ge \|B\|, \qquad \|T\|\ge \|A\|$$
Remark The same formula is valid for any reasonable norm on the product space $X\times X,$ like for example $\ell^p$-norm.
