TITLE: Find a norm in the dual space
QUESTION [5 upvotes]: Let $X$ be a normed space and $Y$ a linear subspace of $X$. We define
  $$Y^{\perp}=\{f\in X^*: f(y)=0, \; \forall y\in Y\}$$
  and 
  $$\|f\|_Y=\sup\{|f(y)|: y\in Y, \; \|y\|=1\}.$$
  Prove that 
  $$\|f\|_Y=\inf\{\|f-g\|: g\in Y^{\perp}\}$$

First of all, given $f\in X^*\backslash Y^\perp$, then for any $g\in Y^\perp$ we get that 
\begin{equation*}
\begin{split}
\|f-g\|&=\sup\{|f(x)-g(x)|: x\in X, \; \|x\|=1\}\\
&\geq\sup\{|f(y)+g(y)|: y\in Y, \; \|y\|=1\} \\
&= \sup\{|f(y)|: y\in Y, \; \|y\|=1\}\\
&=\|f\|_Y
\end{split}
\end{equation*}
As $g\in Y^\perp$ was arbitrary, we find that 
$$\|f\|_Y\leq\inf\{\|f-g\|: g\in Y^{\perp}\}$$
How can I prove the other inequality?

REPLY [4 votes]: By the Hahn-Banach theorem, there is an extension of $f|_Y$ ($f$ restricted to $Y$), call it $h$, such that $\|h\|=\|f|_Y\|=\|f\|_Y$ and $h|_Y=f|_Y$. So $h-f=g\in Y^\perp$. Hence $\|f\|_Y=\|f+g\|$ and the infimum is attained.