6
$\begingroup$

It is well known continuous linear functionals are (Borel) measurable. I have read, as a remark, the converse is also true for separable Banach spaces, but I could not find any references.

$\endgroup$
  • $\begingroup$ You are talking about linear functionals? $\endgroup$ – Michael Greinecker Dec 6 '16 at 14:12
  • $\begingroup$ @MichaelGreinecker Yes. Sorry for being imprecise. $\endgroup$ – newbie Dec 6 '16 at 14:15
  • 1
    $\begingroup$ Your question provides rather little information.. Is it Lusin's theorem what you are looking for? $\endgroup$ – Hannes Dec 6 '16 at 14:15
  • 1
    $\begingroup$ @Hannes I am asking if all measurable linear functionals on separable Banach spaces are continuous. It is not the exact statement of Lusin. $\endgroup$ – newbie Dec 6 '16 at 14:17
9
$\begingroup$

If "measurable" means "Borel" or even "Baire measurable", then this is true. It's a special case of a more general result that any Borel homomorphism of Polish groups is continuous. See for instance Kechris, Classical Descriptive Set Theory, Theorem 9.10.

Also, you can drop the word "separable". Suppose $X$ is an arbitrary Banach space and $f : X \to \mathbb{R}$ is a Borel linear functional. To show $f$ is continuous, it suffices to show that for any sequence $x_n \to 0$, we have $f(x_n) \to 0$. But if we let $X_0$ be the closed linear span of $\{x_n\}$, then $X_0$ is a separable Banach space and the restriction of $f$ to $X_0$ is Borel. So $f$ is continuous on $X_0$, meaning $f(x_n) \to 0$ as desired.

If "measurable" means "measurable with respect to a particular Borel measure $\mu$", then this is not true. It could be that $X$ contains a dense subspace $X_0$ of full measure. In that case, we can use Zorn's lemma to choose a linear functional $f$ which is 0 on $X_0$ and nonzero elsewhere. Since $f = 0$ $\mu$-almost everywhere, it is $\mu$-measurable, but not continuous.

One can also come up with examples where $f$ is $\mu$-measurable and discontinuous, and every linear functional which is $\mu$-a.e. equal to $f$ is also discontinuous. This happens, in particular, for Gaussian measures $\mu$; see Bogachev, Gaussian Measures, Theorem 3.7.6, for an equivalent construction. (Of course, as a consequence of measurability, there will be a Borel function $g$ with $f=g$ a.e.; but then $g$ will not be linear!)

$\endgroup$
3
$\begingroup$

You do not even need seperability. Proposition 1.2.29 in the book Barrelled Locally Convex Spaces of Bonet and Perez Carreras says that every Borel measurable linear map from a Baire locally convex space to a locally convex space is continuous. This result is due to Laurent Schwartz and it is related to his more famous Borelian graph theorem (linear maps from a separable Banach space into Souslin locally convex spaces with borelian graph are continuous).

$\endgroup$

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.