$\|f\vert_M\|=\|f+M^0\|$, Exercise from Folland's functional analysis chapter I have a question on functional analysis, on something rather simple, but I'm stuck for good. Here it is.
If $X$ is a Banach space and $M<X$ a closed subspace, define $M^0=\{f\in X^*: f\vert_M=0\}$. It is easy to check that $M^0$ is a closed subspace of $X^*$. I need to show that for $f\in X^*$ it is $$\| f\vert_M\|=\|f+M^0\|=\text{(by def.)}\inf_{g\in M^0}\|f-g\|$$
I've been able to prove that $\|f\vert_M\|\leq\|f+M^0\|$ but the other inequality is driving me crazy. I've been trying to use the Hahn- Banach theorem to extend functionals that are in M$^0$, but I get nothing useful. Any hints or ideas?
 A: Okay, I think I got it: If $f\in X^*$, it is indeed true that $\|f\vert_M\|\leq\|f+M^0\|$. For the other inequality, by the Hahn- Banach theorem, $f\vert_M$ extends to a linear functional $F\in X^*$ s.t. $\|F\|_{X^*}=\|f\vert_M\|=\|f\|_{M^*}$. But $f+M^0=F+M^0$, since both $f$ and $F$ extend $f\vert_{M}$. So in order to prove that $\|f+M^0\|\leq\|f\vert_{M}\|$ it suffices to prove that $\|F+M^0\|\leq\|f\vert_M\|$. But it is $$\|F+M^0\|=\inf_{g\in M^0}\|F-g\|=\inf_{g\in M^0}\sup_{\|x\|=1}|F(x)-g(x)|\leq\sup_{\|x\|=1}|F(x)|=\|F\|_{X^*}=\|f\vert_M\|$$, since $0\in M^0$.
A: The result follows from this fact: $M^*\cong X^*/M^{\perp}$. Indeed, define $\phi:M^*\to X^*/M^{\perp}$ by $\phi (f)=F+M^{\perp}$ where $F$ is a Hahn-Banach extension of $f$. If $F_1$ is another extension of $f$ then $F-F_1\in M^{\perp}$ so $\phi$ is well-defined. $\phi$ is linear and surjective. 
It is clear that $\|\phi(f)\|\le \|F\|,$ by definition of the quotient norm.
Now, $\|f\|\le \|F\|,$ and this is so for $any$ extension of $f$. But, for all $g\in M^{\perp},\ F+g$ also extends $f$  so $\|f\|\le \|F+g\|$ and so, by definition of the quotient norm, we have $\|f\|\le \|F\|_{X/M^{\perp}}=\|\phi(f)\|$ and this is so for any extension of $f$.
To finish, we use a corollary of the Hahn-Banach Theorem, to find an extension $F_1$ of $f$ such that $\|f\|=\|F_1\|.$  Then, $\|f\|\le \|F_1\|_{X/M^{\perp}}=\|\phi(f)\|\le \|F_1\|=\|f\|\Rightarrow \|\phi(f)\|=\|f\|$.
