The Question
The problem is from "Methods of Modern Mathematical Physics", Volume I, and it's closely related to the Hahn-Banach theorem. It asks to prove the following:
Corollary 3. Let Z be a subspace of a normed linear space X and suppose that y is an element of X whose distance from Z is d. Then there exists a $\Lambda\in X^*$ so that $||\Lambda||\le1$, $\Lambda(y)=d$, and $\Lambda(z)=0$ for all $z\in Z$.
To me this theorem seems like a sort of a normed space version of the Banach separation theorem for locally convex spaces. A proof of the latter can be found easily, but I couldn't make a connection between that argument and my problem. I gave it a go on my own, but I couldn't find a definitive proof.
My Understanding
Here is what I've come up with:
Let $Y$ be the subspace generated by $y$. We can define a bounded functional $\lambda:Y\longrightarrow\mathbb{C}$ by
\begin{equation}
\lambda(\alpha y) = \alpha d, \quad \forall\alpha\in\mathbb{C}.
\end{equation}
The Hahn-Banach theorem ensures there is a bounded linear functional $\Lambda$ defined on all $X$ which extends $\lambda$ and such that $||\Lambda||_{X^*}=||\lambda||_{Y^*}$ (Corollary 1 of Theorem III.6 in Reed-Simon).
However, there are two difficulties: for starters, I can't convince myself that $||\lambda||_{Y^*}\le1$: in fact
\begin{equation}
\frac{|\lambda(\alpha y)|}{||\alpha y||} = \frac{|\alpha|\cdot|\lambda(y)|}{|\alpha|\cdot||y||} = \frac{d}{||y||} \le \frac{||y-z||}{||y||} \le \frac{||y||+||z||}{||y||} = 1 + \frac{||z||}{||y||},
\end{equation}
where $z\in Z$. More importantly, I can't find a reason why $\Lambda$ should vanish on $Z$.
For completeness, here is the statement of the theorem I used:
Corollary 1. Let X be a normed linear space, Y a subspace of X, and $\lambda$ an element of $Y^*$. Then there exists a $\Lambda\in X^*$ extending $\lambda$ and satisfying $||\Lambda||_{X^*}=||\lambda||_{Y^*}$.
Best Answer
Instead, try the following: consider the subspace $E= Z+ Y$, and define $\lambda:E\to \Bbb{C}$ by $\lambda(z+ty)=td$. So, by construction, $\lambda$ vanishes on $Z$. Now, verify $\lambda$ is bounded by $1$ on $E$ and use Hahn-Banach to extend $\lambda$ to $X$. In fact, if $Z$ is closed and $y\notin Z$ then then you can even ensure the extension has norm exactly $1$.