Bounded function is linear and continuous? Let A and B be topological spaces,$\quad \pi : \mathcal{B}(A,Y) \longrightarrow \mathcal{B}(B,Y)$ with $B\subseteq A, \quad \pi(f) := f|_B$.
Let $Y \neq0$ be a normed space.
Show that $\pi$ is a linear, continuous function with Norm $1$.
$\mathcal{B}(A,Y) =$ {$f:A\longrightarrow Y | f \quad \text{bounded}$}.
I'm not really sure why the function is linear or even continuous. I hope someone can give me an explanation here because i am totally clueless.
Thanks in advance.
 A: First you need to notice that you can equip $\mathcal{B}(A,Y)$ and $\mathcal{B}(B,Y)$ with norms. We define the norms
\begin{align}
\Vert \cdot \Vert_A &: \mathcal{B}(A,Y) \to [0, \infty), f \mapsto \sup_{x \in A} \Vert f(x) \Vert  \text{,}  \\
\Vert \cdot \Vert_B &: \mathcal{B}(B,Y) \to [0, \infty), f \mapsto \sup_{x \in B} \Vert f(x) \Vert  \text{.}
\end{align}
Now we have some structure to work with (for calculation of the norm of $\pi$ and continuity). Let us first check that $\pi$ is linear. Let $f, g \in \mathcal{B}(A,Y)$ and $\lambda \in \mathbb R$. Then we have
\begin{align*}\pi(f+g) &= (f+g)|_B = f|_B + g|_B = \pi(f) + \pi(g), \\
\pi(\lambda f) &= (\lambda f)|_B = \lambda f|_B = \lambda \pi(f).
\end{align*}
Thus $\pi$ is a linear function. Now let us figure out the continuity. We have 
\begin{align*} \Vert \pi(f) \Vert_B = \Vert f|_B \Vert_B = \sup_{x \in B} \Vert f|_B(x) \Vert = \sup_{x \in B} \Vert f(x) \Vert \leq \sup_{x \in A} \Vert f(x) \Vert = \Vert f \Vert_A.
\end{align*}
Hence $\pi$ is continuous because it is bounded. Further the calculation shows that $\Vert \pi \Vert \leq 1$. Now let $f(x) = y$ be a constant function for one $y \in Y$ and all $x \in A$. Then we have $f \in \mathcal{B}(A,Y)$ and 
\begin{align*} \Vert \pi(f) \Vert_B = \sup_{x \in B} \Vert f|_B(x) \Vert = \Vert y \Vert = \sup_{x \in A} \Vert f(x) \Vert = \Vert f \Vert_A.
\end{align*}
Hence we even have $\Vert \pi \Vert = 1$. That was what was to be shown. I hope I could help you :)
A: $\pi$ is clearly linear. So it is continuous iff it is bounded. But $||\pi(f)|| = ||f\mid_B|| = \sup_{||x|| = 1, x \in B} |f(x)| \le \sup_{||x||=1, x \in A} |f(x)| = ||f||$, so $\pi$ is bounded with norm $\le 1$. To see that $\pi$ has norm $1$, note that we can make equality in the above calculations hold if we take $f$ to have maximum value on the unit ball of $A$ at a point of $B$, which we can do (you should try to prove this).
