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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.

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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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