Using Hahn-Banach for existence of a functional Let $A\subset X$ (A is closed and linear subspace). And there is a point $x_0\in X$ such that $\inf_{y\in A}\|y-x_0\|= \epsilon$ ($\epsilon>0$). I want to prove that there is a functional $f\in X^*$ such that $f(x_0)=1$ and $\|f\|=1/\epsilon$.
I think I have to use Hahn-Banach for this. I'm going to re-write vector $x\in A_1=span(A,x_0)$ as $x=y+\lambda x_0$ ($y\in A, \lambda\in R$). I'm not sure how to go about using Hahn-Banach to prove the existence of the functional and that $f(x_0)=1$.
 A: Since $A$ is a closed subspace of $X$, then it exists $y \in A$ such that : $\inf\{\|y-x_0\| : x_0 \in X\} = d(x_0,Y)= \varepsilon$.
Let $I = A \bigcup\langle x_0\rangle = \{ y + \lambda x_0 : y \in A, \lambda \in \mathbb R\}$. Consider the functional $f:I \to \mathbb R$ such that :
$$f(y+\lambda x_0) =  \lambda d(x_0,Y)=\lambda \varepsilon$$
Note that for $y=0$ and $\lambda =1$, it is : 
$$f(x_0) = d(x_0,Y)=\varepsilon$$
For $\lambda =0$ it is : $f(y) = 0$.
Now, note that :
$$|f(y+\lambda x_0)| = |\lambda|d(x_0,Y) \leq |\lambda|\left\|x_0 +\frac{1}{\lambda}y \right\| = \|y + \lambda x_0\|$$
Using that and a small other argument, you can prove that $\|f\| = 1$. I'll leave this up to you.
After you are done with that, just use the Hahn-Banach Theorem to extend that function to the whole $X$.
Now, consider the functional : 
$$g = \frac{1}{\varepsilon}f$$
Then, you can clearly see that :
$$g(x_0) = \frac{1}{\varepsilon}f(x_0) = 1 $$
$$g(y) = \frac{1}{\varepsilon} f(y) = 0$$
$$\|g\| = \left\| \frac{1}{\varepsilon}f\right\|= \frac{1}{\varepsilon}$$
The exercise is complete.
