Consider the following fragment from Takesaki's second volume of "Theory of operator algebras" (lemma 2.4, chapter VI "Left Hilbert algebras").

enter image description here

In another post, it was explained that we should interpret the integral $$\int_{\mathbb{R}} e^{-rt^2}x(t)dt$$ as a Gelfand-Pettis integral.

Why does this Gelfand-Pettis integral exist? Is there a general existence result that garantuees the existence of this integral as en element of the compact subset $K$?

The closed I could find was theorem 3.27 in Rudin's book "Functional analysis" but there Rudin works with probability measures (or more generally, finite measures) on compact spaces, whereas here we are dealing with Lebesgue measure on $\mathbb{R}$.

Thanks in advance for your help!

  • 2
    $\begingroup$ You can find a complete description (by double-duality) of this integral in Bourbaki Integration Chapter III § 3.1. $\endgroup$ Mar 21 at 5:47
  • $\begingroup$ @DuchampGérardH.E. I quickly skimmed this section in Bourbaki's book but did not see anything in this section outside the case that the integrand function is compactly supported. Can you be a little more specific which result in this section is relevant? Thanks! $\endgroup$
    – Andromeda
    Mar 21 at 7:17
  • 1
    $\begingroup$ Okay I will send you some more elaborated route ASAP (when back, I'm travelling). $\endgroup$ Mar 21 at 7:57
  • $\begingroup$ @DuchampGérardH.E. Thanks! Safe travels. $\endgroup$
    – Andromeda
    Mar 21 at 8:08
  • $\begingroup$ Does [ this ](:en.wikipedia.org/wiki/Pettis_integral) help? $\endgroup$ Mar 21 at 17:10

1 Answer 1


You can reduce your integral to the case treated by Rudin. Let $\beta\mathbb R$ be the Stone-Cech compactification of $\mathbb R$. By the universal property, $x$ extends to a continuous map $\tilde x\colon\beta\mathbb R\to K$.

Define a probability measure on $B(\beta\mathbb R)$ by $\mu(A)=\sqrt{\frac{r}{\pi}}\int_{A\cap \mathbb R}e^{-rt^2}\,dt$. Note that since the embedding $\mathbb R\to \beta\mathbb R$ is a homeomorphism onto its image, the trace $\sigma$-algebra of $B(\beta\mathbb R)$ on $\mathbb R$ coincides with $B(\mathbb R)$.

By Rudin's Theorem 3.27, the Pettis integral $\int_{\beta\mathbb R}\tilde x\,d\mu$ exists and belongs to $K$. Moreover, approximation by step functions one can show that $\int_{\beta \mathbb R}\omega(\tilde x)\,d\mu=\sqrt{\frac{r}{\pi}}\int_{\mathbb R}\omega(x(t))e^{-rt^2}\,dt$ for $\omega\in E^\ast$. Thus $\int_{\beta\mathbb R}\tilde x\,d\mu=\sqrt{\frac{r}{\pi}}\int_{\mathbb R}x(t)e^{-rt^2}\,dt$ as Pettis integrals.

  • $\begingroup$ I did not know this route. Will study it ! $\endgroup$ Mar 22 at 12:19
  • $\begingroup$ @MaoWao Wonderful answer! Thanks for sharing! $\endgroup$
    – Andromeda
    Mar 22 at 16:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.