6
$\begingroup$

Let $(V, \| \cdot \|)$ be a normed vector space and $W$ be a linear subspace.

Prove that $T: V^*/W^{\perp} \to W^*, \ T(x^* + W^{\perp}) = y^*$ where $y^*(x) = x^*(x)$ for all $x \in W$, is an isometric isomorphism.

$\perp$ denotes the annihilator, and $*$ the dual. There was a hint included that said

"First show that $W^{\perp}$ is a closed linear subspace of $V^*$. Prove that $T$ is a well-defined linear operator. To show that $T$ is an isometric isomorphism apply the Hahn-Banach theorem."

I'm stuck at the last part. Let $y^* \in W^*$ then from Hahn-Banach we have that $\exists \ x^* \in V^*$ s.t $x^* = y^*$ on $W$, and $\|x^*\|_{V^*} = \|y^* \|_{W^*}$. But how can I arrive at $\|T(x^* + W^{\perp})\|_{W^*} = \|x^*+ W^{\perp}\|_{V^*/W^{\perp}}$?

$\endgroup$
  • $\begingroup$ How is the norm induced on the quotient space? $\endgroup$ – Berci Apr 10 '17 at 20:45
  • $\begingroup$ I have no clue. @Berci $\endgroup$ – Olba12 Apr 10 '17 at 20:54
  • $\begingroup$ That was a bad response from me. But I think this is very difficult. And I dont know if that was meant as a question or a hint. @Berci $\endgroup$ – Olba12 Apr 10 '17 at 21:15
  • $\begingroup$ @Berci Norm should be the usual norm on quotients in normed spaces. $\endgroup$ – el_tenedor Apr 10 '17 at 21:28
3
$\begingroup$

The norm on the quotient should be defined as usual: $$ \|x^* + W^\perp\| := \inf\{ \|x^* + y^*\| \colon y^* \in W^\perp\}. $$

We will first show that $\|T(x^* + W^\perp)\| \leq \|x^* + W^\perp\|$.

For all $y \in W^\perp$ we calculate:

$$ \|T(x^* + W^\perp)\| = \sup_{x \in W_1} |x^*(x)| = \sup_{x \in W_1}|x^*(x) + y^*(x)| \leq \sup_{x \in V_1}|x^*(x) + y^*(x)| = \|x^* + y^*\|, $$ where $W_1$ and $V_1$ denote the unit balls in $W$ and $V$ respectively.

This proves the first claim by passing to the infimum.

Now we show $\|T(x^* + W^\perp)\| \geq \|x^* + W^\perp\|$.

Let $T(x^* + W^\perp) \in W^*$. By Hahn-Banach, there exists some extension $v^* \in V^*$ with $$x^*(z) = T(x^* + W^\perp)(z) = v^*(z)$$ for all $z \in W$ and $$\|T(x^* + W^\perp)\| = \|v^*\| \tag{1}.$$ Per constructionem, we have $$v^* + W^\perp = x^* + W^\perp \tag{2}$$ since for all $z \in W$ we have $(v^* - x^*)(z) = v^*(z) - x^*(z) = 0$, i.e. $(v^* - x^*) \in W^\perp$. In particular, (2) gives $$ \|v^* + W^\perp\| = \|x^* + W^\perp \|. \tag{3} $$

This gives $$ \|x^* + w^\perp\| \overset{(3)}{=} \|v^* + W^\perp\| \leq \|v^*\| \overset{(1)}{=} \|T(x^* + W^\perp)\|, $$ where the second inequality holds since the quotient mapping $x^* \mapsto x^* + W^\perp$ is always a contraction.

$\endgroup$
  • $\begingroup$ I don't understand $\|T(x^* + W^\perp)\| = \sup_{x \in W_1} |x^*(x)| = \sup_{x \in W_1}|x^*(x) + y^*(x)| = \|x^* + y^*\|.$ nor "Per constructionem, we have ..." In the second. Are you interested in providing some details? :) $\endgroup$ – Olba12 Apr 11 '17 at 1:51
  • 1
    $\begingroup$ @Olba12 I have updated my post. The first chain of inequalities does only use the definition of the operator norm and the fact that $y \in W^\perp$. $\endgroup$ – el_tenedor Apr 11 '17 at 7:22
  • $\begingroup$ Can I think of (2) as the same equivalence class, just different representation? Like $[2]_5 = [7]_5$? $\endgroup$ – Olba12 Apr 11 '17 at 9:26
  • 2
    $\begingroup$ @Olba12 This is correct, in fact (2) is just a different notation for $[v^*] = [x^*]$, which is oftenly used for quotients of vectorspaces. Have a look at quotients of vector spaces, maybe this will make thinks clearer. $\endgroup$ – el_tenedor Apr 11 '17 at 9:44
  • 1
    $\begingroup$ @Olba12: You're welcome! $\endgroup$ – el_tenedor Apr 11 '17 at 9:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.