Bound of direct sum of operators Let $X$ be a Banach space and $U,V$ complementary subspaces. Let $A: U \to U, B: V \to V$ be continuous and let $T(x) = (A \oplus B)(u \oplus v)= A(u) \oplus B(v)$. 
Does it hold that $\|T\| \le \|A \| + \|B\|$?  
 A: Not necessarily. $\lVert T\rVert$ can be bounded by $C(\lVert A\rVert + \lVert B\rVert)$, but the constant cannot always be chosen as $1$, it depends on how $U$ and $V$ lie relative to each other.
We can create a situation where the constant must be large even in $\mathbb{R}^2$ with the Euclidean norm. Let $U = \mathbb{R}\cdot (1,0)$ and $A = \operatorname{id}_U$, and $V = \mathbb{R}\cdot (\cos\varphi,\sin\varphi)$ with $B = -\operatorname{id}_V$ where $0 < \varphi < \pi$. Then
$$\begin{align}
(0,1) &= -\cot\varphi\cdot (1,0) + \frac{1}{\sin\varphi}(\cos\varphi,\sin\varphi)\\
&\mapsto -\cot\varphi\cdot (1,0) - \frac{1}{\sin\varphi}(\cos\varphi,\sin\varphi)\\
&= \left(-2\cot\varphi,-1\right)
\end{align}$$
and $\lVert T\rVert \geqslant \sqrt{1+4\cot^2\varphi}$ becomes arbitrarily large for $\varphi\to 0$ (or $\varphi\to \pi$).
A: Yet another proof: 
To show that $T$ is continuous if $A,B$ are one can use that a map $T$ is closed if given $x_n \to x \in X$ then $Tx_n \to Tx$.
Note that since $X = U \oplus V$ we have $x_n = u_n \oplus v_n$. Let $u = \lim u_n$ and $v= \lim v_n$.  Since $A,B$ are continuous, $Au_n \to Au$ and $Bv_n \to Bv$ and since $U,V$ are closed it follows that $Au \in U$ and $Bv \in V$. Hence 
$$ Au_n \oplus Bv_n \to Au \oplus Bv$$ which is the same as 
$$ T(x_n) \to Tx$$
hence $T$ is continuous.
