4

Let $(X, \tau)$ be a Hausdorff topological vector space, $\mathbb{K}$ the scalar field with its usual topology and $\Lambda : X \to \mathbb{K}$ a linear functional.

There is this general criterion concerning the continuity of $\Lambda$, which you can find a concise proof of in Rudin's Functional Analysis as the Theorem $1.18$ for example:

The assertions below are equivalent:

  1. $\Lambda$ is continuous;
  2. $\ker \Lambda$ is closed;
  3. $\ker \Lambda$ is not dense;
  4. $\Lambda$ is bounded on at least one neighbourhood of $0$.

The equivalence "1. $\Leftrightarrow$ 2." can also be shown directly in normed spaces due to continuity and boundedness coinciding in metrisable TVSs and hence in normed spaces.
This in particular has been done several times on this site: see A second proof of the fact that a linear functional $f$ on a normed vector space $\mathcal{X}$ is bounded iff $f^{-1}(0)$ is closed., or The kernel of a continuous linear operator is a closed subspace?, and so on...

My question is then simple:

Do we have that $\Lambda$ is sequentially continuous iff $\ker \Lambda$ is sequentially closed?

The "only if" direction is fairly easy to prove: consider a sequence $(x_n)_n$ in $\ker \Lambda$ converging to $x \in X$.
By sequential continuity, $\Lambda(x_n) = 0 \to \Lambda(x)$, yet $0 \to 0$, hence by uniqueness of the limit in $\mathbb{K}$ we get $\Lambda(x) = 0$, and $x$ belongs in $\ker \Lambda$.

Thus, what I have trouble with is the "if" direction. I have a suspicion that it's just not a true claim due to the lack of MSE questions about functionals with sequentially closed kernels (if I type "sequentially closed kernels" in the search bar, there are only ten results, with seemingly only one mentioning those concepts in the same sentence, namely When is a sequentially closed subspace the kernel of a sequentially continuous linear functional? but it's in the opposite direction from my current question), though the lack of questions is of course not actual evidence of the claim being true nor of it being false. I'm probably just missing something simple to be fair.

What I've tried is first starting from "$\Lambda$ not sequentially continuous" and attempting to prove that $\ker \Lambda$ is not sequentially closed.
Since sequential continuity is the same as sequential continuity at $0$ thanks to linearity, there exists a sequence $(x_n)_n$ in $X$ converging to $0$ and such that $(\Lambda(x_n))_n$ does not converge to $\Lambda(0) = 0$, and then, taking a $y \in X$ such that $\Lambda(y) = 1$, we can consider the sequence $(x'_n)_n := (x_n - \Lambda(x_n) y)_n$ to have a sequence belonging to $\ker \Lambda$... but then what?
If $(\Lambda(x_n))_n$ was known to have a finite non-zero limit point $\alpha$, I could exhibit a subsequence $(x'_{\varphi(n)})_n$ converging to $-\alpha y$ which would not belong in $\ker \Lambda$, hence the latter would not be sequentially closed and we would be done, however what if all we have is $|\Lambda(x_n)| \to \infty$ or the only finite limit point is $0$?

Feel free to edit or re-tag if/when adequate.

Bruno B
  • 5,027

1 Answers1

2

If $(\Lambda(x_n))$ is bounded, then you're done, since it must either converge to $0$ or have some finite nonzero accumulation point. If $(\Lambda(x_n))$ is unbounded, passing to a subsequence we may assume $|\Lambda(x_n)|\to \infty$. Now let $y_n=x_n/\Lambda(x_n)$. Since $|\Lambda(x_n)|\to\infty$ and $x_n\to 0$, $y_n\to 0$ as well. But $\Lambda(y_n)=1$ for all $n$, so now your argument applied to $(y_n)$ yields a contradiction.

Eric Wofsey
  • 330,363