Isometry between $X^\ast/M^\perp$ and $M^\ast$.

Question

Let $X$ be a normed linear space, $M \subset X$ be a subspace, $M^\perp = \{x^\ast \in X^\ast \mid x^\ast\big|_M = 0\}$ be the annihilator of $M$, $X^\ast$ the topological dual of $X$, and let's define $\Phi: X^\ast/M^\perp \to M^*$ by $\Phi(x^\ast+M^\perp) = x^\ast\big|_M$.

I have already proved that $M^\perp$ is closed, and that $\Phi$ is well-defined, linear, and surjective.

I finally want to prove that $\Phi$ is an isometry. On one hand, we take $y^\ast \in M^\perp$ and now: $$\begin{align} \|\Phi(x^\ast+M^\perp)\| &= \|x^\ast\big|_M\| = \|x^\ast\big|_M+y^\ast\big|_M\| \\ &=\|(x^\ast+y^\ast)\big|_M\| \leq \|x^\ast+y^\ast\|,\end{align}$$and taking the infimum on $y^\ast$ gives $\|\Phi(x^\ast+M^\perp)\| \leq \|x^\ast+M^\perp\|$.

I'm having trouble on the other inequality. One first attempt was to fix $m \in M$, take any $y^\ast \in M^\perp$ and do: $$\|(x^\ast+y^\ast)(m)\| \leq \|x^\ast(m)\|+\|y^\ast(m)\| = \|x^\ast(m)\| \leq \|x^\ast\big|_M\|\|m\|,$$and this gives $\|x^\ast\big|_M + y^\ast\big|_M\| \leq \|x^\ast\big|_M\|$ instead of the $\|x^\ast + y^\ast\|\leq \|x^\ast\big|_M\|$ that I wanted. Can someone help?

Answer 1

Edit: Figured it out.

So $\Phi: X^{\ast}/M^{\perp}\to M^{\ast}$ is defined as before, and you have shown that:

(1) $M^{\perp}$ is closed.

(2) $\Phi$ is continuous and surjective.

(3) For $x^{\ast}+M^{\perp}\in X^{\ast}/M^{\perp}$, we have $||\Phi(x^{\ast}+M^{\perp})||_{M^{\ast}}\leq ||x^{\ast}+M^{\perp}||_{X^{\ast}/M^{\perp}}$

Now we can proceed. Since $\Phi(x^{\ast}+M^{\perp})\in M^{\ast}$, Hahn-Banach tells us there is a $\phi_x\in X^{\ast}$ with the property that $||\phi_x||_{X^{\ast}}=||\Phi(x^{\ast}+M^{\perp})||_{M^{\ast}}$. Now consider $\phi_x+M^{\perp}\in X^{\ast}/M^{\perp}$ (if you like, use the natural projection).

We have that $\phi_x+M^{\perp}=x^{\ast}+M^{\perp}$ because, by construction of $\phi_x$, $\phi_x=x^{\ast}$ on $M$, since $\phi_x$ is the extension of $x^{\ast}|_{M}$. Thus $\phi_x-x^{\ast}$ annihilates $M$, so $\phi_x-x^{\ast}\in M^{\perp}$ and it holds.

Now we have that $$||x^{\ast}+M^{\perp}||_{X^{\ast}/M^{\perp}}=\inf_{y^{\ast}\in M^{\perp}} ||\phi_x+y^{\ast}||_{X^{\ast}}\leq ||\phi_x||_{X^{\ast}}=||\Phi(x^{\ast}+M^{\perp})||_{M^{\ast}}$$

Combining with $(3)$ gives the result, as all the inequalities become equalities.