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.

  • 6
    $\begingroup$ 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? $\endgroup$ – Nate Eldredge Sep 30 at 20:59
  • $\begingroup$ @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? $\endgroup$ – Black Sep 30 at 21:16
  • $\begingroup$ moreover, there aren't that many ways of extending f from H to X. All extensions are proportional. $\endgroup$ – Pietro Majer Sep 30 at 21:16
  • $\begingroup$ @Pietro, that is true, but these are precisely the functionals I would not consider wildly discontinuous. $\endgroup$ – Black Sep 30 at 21:17
  • $\begingroup$ @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). $\endgroup$ – 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.

| cite | improve this answer | |
  • $\begingroup$ 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. $\endgroup$ – Black Oct 1 at 0:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.