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 $$f(x_0)=1$$.

• $x_o \in X$ and $y \in A$ I assume ? – Rebellos Feb 25 at 21:25
• yes, edited the question – Jack Feb 25 at 21:30
• no, want to show f(x_0)=1 and also f(y)=0. I think for y it would be obvious after showing it for x_0. – Jack Feb 25 at 21:38
• I suppose that the infimum may be equal to $\varepsilon$. – Rebellos Feb 25 at 22:20

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.