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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.

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    $\begingroup$ What are A and B? Also, it should be really really clear that $\pi$ is linear.. $\endgroup$ Jul 15 '17 at 17:46
  • $\begingroup$ @mathworker21 A and B are topological spaces. I will edit that thanks. But i dont really see that its clear that $\pi$ is linear. EDIT: ok i see now why $\pi$ is linear. i checked it the wrong way.. $\endgroup$ Jul 15 '17 at 17:50
  • $\begingroup$ Do you understand all the definitions involved? Like how B(A,Y) is a vector space? $\endgroup$ Jul 15 '17 at 17:54
  • $\begingroup$ @mathworker21Yes i see now why $\pi$ is a linear function. Am i able to use the operator norm here and show it is bounded? Therefore i can show its continous? $\endgroup$ Jul 15 '17 at 17:57
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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 :)

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  • $\begingroup$ Wow thanks for your effort. I was confused in the beginning using the definition of $\pi$, but got it now. Thanks again. $\endgroup$ Jul 15 '17 at 18:11
  • $\begingroup$ You're welcome :) $\endgroup$
    – Yaddle
    Jul 15 '17 at 18:13
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$\pi$ 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).

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  • $\begingroup$ Yes i got it now i was way to confused in the beginning. Thank you $\endgroup$ Jul 15 '17 at 18:05
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    $\begingroup$ No problem. I just wanted to make sure it was clear to you that$\pi$ is linear. I feel that any functional analysis book or whatever would be really hard to read otherwise.. $\endgroup$ Jul 15 '17 at 18:10

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