Let $(E,\|\cdot\|)$ be a normed $\mathbb{K}$-vector-space and $(\overline{E},\iota)$ a completion of $E$. Show, that
1) For every normed space $F$ is $\iota^\ast\colon L(\overline{E}, F)\to L(E, F), T\mapsto T\circ \iota$ an isometry.
2) Is $F$ a banach space, then is $\iota^\ast\colon L(\overline{E}, F)\to L(E, F)$ and isometric isomorphism.
3) In 2) it is important, that $F$ is complete.
I want to show these statements. I have to show, that $\iota^\ast\colon L(\overline{E}, F)\to L(E, F)$ is an isometry. Therefore:
$\|\iota^\ast(f)\|_E=\|f\|_{\overline{E}}$. Since $(\overline{E},\iota)$ is a completion of $E$ we know, that $\iota: E\to \overline{E}$ is an isometry.
It is $\|\iota^\ast(f)\|_E=\|f\circ \iota\|_E$. How can I proceed from here? I could stipulate like this, but do not think, that it helps to show equality:
$\|f\circ \iota\|_E\leq \|f\|_{op}\|\iota(f')\|_E$, for $f'\in L(E,F)$
Since $\iota$ is an isometry, then we get: $\|f\|_{op}\|\iota(f')\|_E=\|f\|_{op}\|f'\|_{E'}$
at 2):
Here I need to show, that $\iota^\ast$ is surjective, since from 1) we get that it is injective.
at 3):
I do not now how to give a counterexample.
Can you give me some hints on how to tackle this problem? Thanks in advance.
I gave a bounty worth 100 reputations, for a full answer.
For the clarity I’ll explicitily assume that you mean the following. Both $F$ and $E$ are normed spaces over the field $\Bbb K$, where $\Bbb K=\Bbb R$ or $\Bbb K=\Bbb C$, $\iota:E\to\overline{E}$ is an isometry, and for given normed spaces $E’,E’’$ over the field $\Bbb K$, $L(E’,E’’)$ denotes the space of continuous linear operators from $E’$ to $E’’$ endowed with the norm $\|T\|_{L(E’,E’’)}=\sup\{\|T(x)\|_{E’’}: x\in E’, \|x\|_{E’}=1\}$, for each $T\in L(E’,E’’)$. Now it remains to use Daniel Fischer’s comment.
Put $B=\{x\in E, \|x\|_{E}=1\}$ and $\overline{B}=\{x\in \overline{E}, \|x\|_{\overline{E}}=1\}$. Since $\iota$ is an isometry, $\iota(B)\subset \overline{B}$. Moreover, since $\iota$ is an isometric embedding of the metric $E$ into its completion, $\iota(B)$ is dense in $\overline{B}$. Indeed, let $x\in \overline{B}$ be an arbitrary point. Since $\iota(E)$ is dense in $\overline{E}$, there exists a sequence $\{x_n\}$ of points of $E\setminus\{0\}$ such that a sequence $\{\iota(x_n)\}$ converges to the point $x$. Then a sequence $\{y_n\}$, where $y_n=x_n/\|x_n\|_E$ for each $n$, belongs to $B$ and a sequence $\{\iota(y_n)\}$ converges to the point $x$, because of continuity of the map $\Bbb K\times E\to E$, $(\lambda, y)\mapsto \lambda y$ at a point $(1,x)$ and the convergence of the sequence $\{\|x_n\|_E\}$ to $1$.
1) Let $T\in L(\overline{E},F)$ be an arbitrary operator. Then $$\|T\|_{L(\overline{E},F)}=\sup\{\|T(y)\|_{F}: y\in \overline{B}\}=\sup\{\|T(y)\|_{F}: y\in \iota(B)\}= \sup\{\|T(\iota(x))\|_{F}: x\in B\}= \sup\{\|(T\iota)(x)\|_{F}: x\in B\}=$$ $$\sup\{\|\iota^*(T)(x)\|_{F}: x\in B\}= \|\iota^*(T)\|_{L(E,F)}.$$
The second equality holds because the map $T$ is continuous and $\iota(B)$ is dense in $\overline{B}$.
2) 3) To be continued.