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 $\|T(x^* + W^{\perp})\|_{W^*} = \|x^*+ W^{\perp}\|_{V^*/W^{\perp}}$?

  • $\begingroup$ How is the norm induced on the quotient space? $\endgroup$
    – Berci
    Apr 10, 2017 at 20:45
  • $\begingroup$ I have no clue. @Berci $\endgroup$
    – Olba12
    Apr 10, 2017 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, 2017 at 21:15
  • $\begingroup$ @Berci Norm should be the usual norm on quotients in normed spaces. $\endgroup$
    – el_tenedor
    Apr 10, 2017 at 21:28

1 Answer 1


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.

  • $\begingroup$ 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, 2017 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, 2017 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, 2017 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, 2017 at 9:44
  • 1
    $\begingroup$ @Olba12: You're welcome! $\endgroup$
    – el_tenedor
    Apr 11, 2017 at 9:47

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