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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 $f(x_0)=1$.

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    $\begingroup$ $x_o \in X$ and $y \in A$ I assume ? $\endgroup$ – Rebellos Feb 25 at 21:25
  • $\begingroup$ yes, edited the question $\endgroup$ – Jack Feb 25 at 21:30
  • $\begingroup$ 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. $\endgroup$ – Jack Feb 25 at 21:38
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    $\begingroup$ I suppose that the infimum may be equal to $\varepsilon$. $\endgroup$ – Rebellos Feb 25 at 22:20
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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.

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