fa.functional analysis - A counter example to Hahn-Banach separation theorem of convex sets.

The Hahn-Banach theorem for a locally convex space X says that for any disjoint pair of convex sets A, B with A closed and B compact, there is a linear functional $lin X^*$ separating A and B. So, it would be nice to have a counterexample where both A and B are closed, but not compact. As no-one has posted such an example, I'll do that now, where the space X is a separable Hilbert space. In fact, as with fedjas example, there will be no separating linear functionals at all, not even noncontinuous ones.

Take μ to be the Lebesgue measure on the unit interval [0,1] and X = L²(μ). Then let,

• A be the set of f ∈ L²(μ) with f ≥ 1 almost everywhere.


• B be the one dimensional subspace of f ∈ L²(μ) of the form f(x) = λx for real λ.

These can't be separated by a linear function $lcolon Xtomathbb{R}$. A similar argument to fedja's can be used here, although it necessarily makes use of the topology. Suppose that $l(f)ge l(g)$ for all f in A and g in B. Then $l$ is nonnegative on the set A-B of f ∈ L² satisfying $f(x)ge 1-lambda x$ for some λ. For any $fin L^2$ and for each $ninmathbb{N}$, choose $lambda_n$ large enough that $Vert(1-lambda_nx+vert fvert)_+Vert_2le 4^{-n}$ and set $g=sum_n 2^n(1-lambda_n x+vert fvert)_+in L^2$. This satisfies $pm f+2^{-n}gge1-lambda_nx$, so $pm l(f)+2^{-n}l(g)ge 0$ and, therefore, $l$ vanishes everywhere.

If you prefer, you can create a similar example in $ell^2$ by letting $A={xinell^2colon x_nge n^{-1}}$ and B be the one dimensional subspace of $xinell^2$ with $x_n=lambda n^{-2}$ for real λ.

Note: A and B here are necessarily both unbounded sets, otherwise one would be weakly compact and the Hahn-Banach theorem would apply.