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What is the strongest form of the Hahn-Banach separation theorem for Hilbert spaces? Could you please provide a reference?

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  • $\begingroup$ Hi Tom! Do you have suggestions of what kinds of things a HB separation theorem on Hilbert space would accomplish that can't be done in Banach space? $\endgroup$ Feb 13, 2014 at 16:54
  • $\begingroup$ Hi @ABlumenthal! Orthogonality of the separating hyperplanes to the distance-minimizing vector, relationship between the metric distance and the marginal difference between the hyperplanes, etc. $\endgroup$ Feb 13, 2014 at 17:23
  • $\begingroup$ Riesz representation. $\endgroup$
    – Michael
    Feb 13, 2014 at 18:05
  • $\begingroup$ Sorry, @Michael, that is not helpful. Could you expand on what you mean? The Riesz representation theorem is not a form of the Hahn-Banach separation theorem, though it is useful in stating it. $\endgroup$ Feb 13, 2014 at 19:31
  • $\begingroup$ I guess the proof of the usual separation theorem would be easier because you don't have to appeal to Hahn-Banach. If $A$ is a nonempty closed convex set in a Hilbert space and $x \in X$, then there exists a unique closest element of $A$ to $x$, call it $a$. I think the linear functional $\ell(y)=(y,x-a)$ separates $x$ from $A$. $\endgroup$ Sep 27, 2018 at 17:38

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This doesn't answer the question of giving the strongest Hahn-Banach separation theorem for Hilbert spaces. But it does address a closely related question of how to easily get separation in Hilbert spaces, without needing Hahn-Banach. So maybe the method suggests how to get a really stronger separation theorem in Hilbert spaces.

If $A$ is a nonempty closed convex set in a Hilbert space and $x \in X$, then there exists a unique closest element of $A$ to $x$, call it $a$. The linear functional $\ell(y)=(y,x-a)$ separates $x$ from $A$. I think you can use this idea to separate $A$ from other sets besides singeltons $\{x\}$.

Edit: Let me explain how I came up with this. I drew a picture in $\mathbb{R}^2$ with a convex set $A$ and a point $x$ not in $A$. I drew the shortest line from $A$ to $x$. That's the vector $x-a$. Then I drew a second line orthogonal to the first that separates $A$ from $x$. The second line is $\{y: y \cdot (x-a) = c\}$ for some $c$. In other words, the second line is $\ell(y)=c$. One reason Hilbert spaces are great is that you can do exactly same thing.

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