I have a closed subspace $Y$ of a Banach space $X$ and a map $T: X'/Y^{\circ} \to Y'$ given by $[f] \to f|_y$. The norm in $X'/Y^\circ$ is given by $\|[f]\| = \inf \{ \|f-h\| : h \in Y^\circ \}$. I'm trying to show this map is an isometry, i.e. $\inf \{ \|f-h\| : h \in Y^\circ \} = \inf \{M > 0 : |f(y)| \le M \|y\| \}$ I've shown it is well-defined, linear and I've shown one side of the isometry inequality:
If $h \in Y^{\circ}$ then $|f(y)| = |(f-h)(y)| \le \|f-h\|\|y\|$. Hence $||f|_y \le \|[f]\|$. Now for the other inequality I want to show that if I take an $M$ in the set $\{M > 0 : |f(y)| \le M \|y\| \}$ then this is the "best" bound for some particular $\|f-h\|$ and then we would have $\inf \{ \|f-h\| : h \in Y^{\circ} \} \ge \inf \{M > 0 : |f(y)| \le M \|y\| \}$.
To do this I can see for a general $h \in Y^{\circ}$ we have $|f(y)-h(y)| \le M\|y\|$ and so $\|f-h\| \le M$. Now I suppose $K$ is a bound for $f-h$ and hence $K \ge \frac{|f(x)-h(x)|}{\|x\|}$ and now I'd like to show $K \ge M$ which would then give $\|f-h\| = M$ but I'm unsure how to pick $h$ to achieve this!
Apologies for the lengthy explanation, I hope this is okay to follow.