This is a well-known result in functional analysis, in terms of dual of quotient spaces and annihilators of subspaces. Let me formulate the problem first and a new attempt to prove it:
Let $W$ closed vector subspace of $V$, $V/W$ the quotient space, and $W^\perp$ the annihilator of $W$.
(1) Show that the map $S$ taking an element $f\in W^\perp$ into the linear functional $S(f)$ defined by
$$S(f)([x])=f(x),\forall[x]\in V/M$$
is well-defined, maps $W^\perp$ into $(V/W)^*$ and induces an isometry between $W^\perp$ and $(V/W)^*$.
(2) Show that the map $T$ taking an element $[f]\in V^*/W^\perp$to the linear functional $T([f])$ defined by $$T([f])(x)=f(x),\forall[f]\in V^*/W^\perp\text{ and }x\in W$$
is well-defined, maps $V^*/W^\perp$ into $W^*$ and induces an isometry between $ V^*/W^\perp$ and $W^*$.
Many texts, like in Brezis', directly proves the existence of isometric isomorphisms in both cases. That is, prove both $S$ and $T$ are bijective, linear, and are isometries. After reading the proofs in Brezis' book, I'm thinking of constructing inverse maps of $S$ and $T$, another way to prove both results and here are my thoughts so far:
For (1), I consider $g\in (V/W)^*$ and define the map $T(g)=g\circ Q$, where $Q:V\to V/W$. I wish to show that $T:(V/W)^*\to W^\perp$ (this is not the same as $T$ in (2)), continuous, $T$ is inverse of $S$ with both norm equal to $1$.
For (2), to construct the inverse map, I'm thinking of using Hahn-Banach theorem to extend an element $g\in W^*$ to an element $f\in V^*$ with the same norm. Then show that $[f]=[\tilde{f}]$ for any two such extensions, then the map $g\to [f]$ is well-defined.
Up to now, I don't know how to continue with these forumulations. I can prove that $S$ and $T$ are well-defined, but having problem showing the rest using this method. Can anyone help me finish the argument? Thank you.
Best Answer
I figured this out:
Take $v_1-W,v_2-W\in V/W$, this implies $v_1-v_2\in W$. Since $f\in W^\perp$, $f(v_1-v_2)=0$, $f(v_1)=f(v_2)$ by linearity. Hence $S(f)$ is well-defined.
By the hint, we show that $T$ maps $V/W$ into $W^\perp$, and is linear, bijective, and is an isometry: It is clear that $T$ is linear. For $x\in V$, $$T(g)(x)|=|g(x-W)|\leq\|g\|\|x-W\|\leq\|g\|\|x\|$$
Hence $T(g)\in V^*$. If we take $x\in W$, $T(g)(x)=g(W)=0$, so $T(g)\in W^\perp$. So $T$ maps $(V/W)^*$ into $W^\perp$. By the above argument, it is clear that $T$ is linear and injective. To show that it is surjective, let $h\in W^\perp$ and define the map $g:V/W\to\mathbb{C};g(x-W)=h(x),\forall x\in V$. $g$ is well-defined, since $W\subset\ker h$. $g$ is actually linear, and $\forall x\in V$ and $\forall y\in W$, $$|g(x-W)|=|h(x)|=|h(x-y)|\leq\|h\|\|x-y\|$$
which implies $g\in(V/W)^*$. Hence, $T(g)=h$, $T$ is surjective.
And recall that in Exercise 2, we prove that $\|Q\|_{op}=1$, so we have $\|T(g)\|=\|g\circ Q\|\leq\|g\|$. To prove the converse, let $\{x_n-W\}\subset V/W$ be a sequence s.t. $\|x_n-W\|<1$ and $|g(x_n-W)|\to\|g\|$. Then, $\forall n$, $\exists y_n\in W$ s.t. $\|x_n-y_n\|<1$. Hence, $$\|T(g)\|\ge\|T(g)(x_n-y_n)\|=|g(x_n-y_n)|\to\|g\|$$
Hence, $\|T(g)\|=\|g\|$. Then we have shown the above three results, and it remains to check that $S$ is the inverse of $T$. $$\forall[x]\in V/W,T\circ S(f)([x])=T(f)(x)=f([x])\Rightarrow T\circ S=\text{id}_{W^\perp}$$ $$\forall x\in V,S\circ T(g)(x)=S(g)([x])=g(x)\Rightarrow S\circ T=\text{id}_{(V/W)^*}$$
Since the inverse of an isometry is again an isometry, we conclude that $S$ is an isometry.
To show $T$ is well-defined, take $[f],[f']\in V^*/W^\perp$, then $\forall x\in W$, $[f-f'](x)=0$, which implies $f-f'\in W^\perp$. $0=(f-f')(x)\Leftrightarrow f(x)=f'(x)\Leftrightarrow T[f](x)=T[f'](x)$.
Consider the map $S:W^*\to V^*/W^\perp;S(w^*)=v^*-W^\perp$. First, we show it is well defined, independent of choices of Hahn-Banach extension. Suppose $w^*$ has Hahn-Banach extensions $v_1^*$ and $v_2^*$, then $w^*=v_1^*|_W=v_2^*|_W$. This implies $(v_1^*-v_2^*)|_W=0\Rightarrow v_1^*-v_2^*\in W^\perp\Rightarrow v_1^*-W^\perp=v_2^*-W^\perp$.
Obviously, $S$ is linear. $S$ is injective: We can rewrite $S(v^*|_W)=v^*-W^\perp$, and $v^*|_W=0$ if and only if $v^*\in W^\perp$. By analytic Hahn-Banach theorem, $\forall w^*\in W^*$, $\exists v^*\in V^*$ s.t. $v^*|_W=w^*$, which implies $S$ is surjective. Furthermore, the theorem implies that $\|v^*\|=\|w^*\|$. Use this result to prove that $S$ is an isometry: $\|S(w^*)\|\leq\|v^*\|=\|w^*\|$. To prove the converse, as noticed when proving injectivity, $w^*=(v^*-W^\perp)|_W$, hence $\| w^{*}\|\leq\| v^*-w^{*'}\|,\forall w^{*'}\in W^\perp$. Take infimum over all $w^{*'}\in W^\perp$, $\|w^*\|\leq\|v^*-W^\perp\|=\|S(w^*)\|$. Hence $\|S(w^*)\|=\|w^*\|$.
It remains to check that $T$ is the inverse of $S$: $$\forall w^*\in W, T\circ S(w^*)(x)=T([v^*])(x)=v^*(x)=w^*(x)\Rightarrow T\circ S=\text{id}_{W^*}$$
where in the last equality is consequence of Hahn-Banach extension. $$\forall[f]\in V^*/W^\perp, S\circ T([f])(x)=S(f)(x)=[f](x)\Rightarrow S\circ T=\text{id}_{V^*/W^\perp}$$
Note that in the last equality, there is actually another extension $[\tilde{f}]$, as given in the hint. But we have shown that $S$ is well-defined, so it is independent of choices of extensions. Since the inverse of an isometry is again an isometry, we conclude that $T$ is an isometry.