# Undertanding Jensen's inequality proof on countable basis Hilbert with Radon measure

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.

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.
• 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 ? Sep 28, 2020 at 11:50