TITLE: Prove that $T(x^* + W^{\perp}) = y^*$ is an isometric isomorphism from $V^*/W^\perp$ to $W^*$
QUESTION [9 upvotes]: 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 $\|T(x^* + W^{\perp})\|_{W^*} = \|x^*+ W^{\perp}\|_{V^*/W^{\perp}}$?

REPLY [6 votes]: 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.