# Wildly discontinuous linear functionals

Let $$X$$ be a Banach space, $$H\subseteq X$$ be a dense hyperplane, and $$f$$ be a continuous linear functional defined on $$H$$. Then $$f$$ is uniformly continuous and hence it admits a unique continuous extension to $$X$$.

However, let us instead choose a discontinuous linear functional $$g$$ extending $$f$$ to the whole of $$X$$.

One cannot say that $$g$$ is too bad since, after all, its restriction to a big subspace, namely $$H$$, is continuous.

Does every Banach space admit a linear functional which is either zero or discontinuous when restricted to every infinite dimensional subspace?

The kernel of such a functional will then have a dense intersection with every infinite dimensional subspace.

• For any linear functional $f$ on an infinite-dimensional Banach space, the kernel of $f$ is a subspace with codimension 1, hence also infinite dimensional, and the restriction of $f$ to its kernel is 0 which is certainly continuous... did you have something else in mind? – Nate Eldredge Sep 30 at 20:59
• @Nate, thanks! That does highlight a big flaw in my thinking. Would it still be a trivial question if we restrict to infinite dimensional subspaces not contained in the kernel? – Black Sep 30 at 21:16
• moreover, there aren't that many ways of extending f from H to X. All extensions are proportional. – Pietro Majer Sep 30 at 21:16
• @Pietro, that is true, but these are precisely the functionals I would not consider wildly discontinuous. – Black Sep 30 at 21:17
• @Pietro, there is only one continuous extension, but infinitely many linear ones, so surely one will be discontinuous (I take back my "that is true" of 2 comments ago). – Black Sep 30 at 21:24

No non zero linear functional has the property you ask for. Suppose $$F$$ is a non zero linear functional. Choose $$x$$ s.t. $$F(x)=1$$. Let $$G$$ be a continuous linear functional s.t. $$G(x)=1$$. Let $$Y$$ be the intersection of the kernels of $$F$$ and $$G$$, so that $$Y$$ has codimension $$2$$. Then $$F$$ is continuous on the linear span of $$Y$$ and $$x$$ since $$F$$ agrees with $$G$$ on this codimension one subspace.
• Excellent! Thank you. I guess this also shows that there is no subspace $H$ whose intersection with any closed hyperplane is dense. If there was one, you could put $H$ inside some hyperplane, see it as the kernel of a functional $F$ and then use the same argument. – Black Oct 1 at 0:33