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Given the following proposition :

$\textbf{Proposition :}$ $X$ Hilbert space with countable basis, $\mu$ Radon measure on $X$, $\phi \in L_{\mu}^{1}(X)$ and $f \in \left\lbrace f : X \longmapsto \overline{\mathbb{R}} : f = \sup\limits_{i \in \mathbb{N}}f_{i} \hspace{0.1cm} \mbox{with $f_{i}$ affine} \right\rbrace - \left\lbrace -\infty,+\infty\right\rbrace$. Then $f(\frac{1}{\mu(X)}\int \phi(x)d\mu(x)) \leq \frac{1}{\mu(X)} \int f\phi(x)d \mu(x)$

During the proof I have the necessity of proving the following : I'd like to prove that given $h$ affine function, $h = \langle w,z \rangle + \beta $ then it holds $h (\frac{1}{\mu(X)} \int \phi(x)d\mu(x)) = \langle w, \frac{1}{\mu(X)} \int \phi(x)d\mu(x) \rangle +\beta = \frac{1}{\mu(X)} \int \langle w,\phi(x) \rangle + \beta$

In particular I don't understand how the last equality follows from the linearity and continuity of the scalar product. More generally it would be useful knowing if the following fact holds :

If $\psi$ is linear, $\psi(\int \phi d\mu) = \int \psi(\phi) d\mu ? $

The notation $\frac{1}{\mu(X)} \int$ is quite ugly but I don't how to write the symbol \fint on MSE. Any help or solution would be appreciated.

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1 Answer 1

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If $\phi$ is an $X$ valued $L^{1}$ function then $\int \phi d\mu$ is also equal to the Pettis integral of $\phi$ and $f(\int \phi d\mu)=\int f\circ \phi d\mu$ for any continuous linear function $f$. Ref: Vector Measures by Diestel and Uhl.

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  • $\begingroup$ Could you please give me a reference of the fact that if $\phi$ is an $X$ valued $L^{1}$ then $\int \phi d\mu$ is also equal to the Pettis integral ? $\endgroup$ Sep 28, 2020 at 11:50
  • $\begingroup$ @jacopoburelli See Theorem 6 (Hille), p.47 in Diestel and Uhl's book. $\endgroup$ Sep 28, 2020 at 12:00

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