A question on linearity of inner product The linearity of inner product on $(X,\langle.,.\rangle)$ is usually written as 
$$\langle x+\alpha y,z\rangle = \langle x,z\rangle + \alpha\langle y,z\rangle,\qquad \forall (\alpha,x,y,z)\in R\times X^3$$
Let $\xi$ be $X$-valued random element. Can we say that $E\langle \xi,x\rangle=\langle E\xi,x\rangle$, where $x\in X$ is not random. I see that it is true when for example $X=L^2([0,1],\mathcal{B}[0,1],\lambda)$ with $\langle f,g\rangle = \int_0^1f(x)g(x)dx$, or when $X$ is Euclidean space with usual inner product. Can we say this for arbitrary inner product? 
 A: The point in doing so in general is, that $x^* = \left<\cdot, x\right>\colon X \to \mathbb R$ is a continuous linear functional. Let $\xi \colon \Omega \to X$ be a random variable, we will show that $\def\E{\mathbb E}\E\def\<#1>{\left<#1\right>}x^*(\xi) = x^*(\E \xi)$ holds if $\E\|\xi\|$ exists. 
We start with a characteristic $\xi = \chi_{A}\cdot x$ for some measurable $A \subseteq \Omega$ and some $x \in X$. Then $\def\P{\mathbb P}$
$$ \E\bigl(x^*(\xi)\bigr) = \E\chi_Ax^*(x) = x^*(\E\chi_Ax) = x^*(\E\xi) $$
By linearity of $x^*$ this holds true for every simple $\xi$, that is finitely valued $\xi$. Now let $\xi  \in L^1(\Omega, X)$ be arbitrary, then there is a sequence $\xi_n$ of simple functions such that $\|\xi_n - \xi\|_1 \to 0$. We have 
\begin{align*}
  \int_\Omega \def\abs#1{\left|#1\right|}\abs{x^*\xi - x^*\xi_n}\, d\P 
   &\le \int_\Omega \|x^*\|\abs{\xi - \xi_n}\, d\P\\
   &= \|x^*\|\|\xi_n - \xi\|_1 \to 0
\end{align*}
That is $x^*(\xi_n) \to x^*(\xi)$ in $L^1(\Omega)$, hence $\E x^*(\xi_n)\to \E x^*(\xi)$. On the other hand $\xi_n \to \xi$ in $L^1(\Omega)$ gives $\E\xi_n \to \E\xi$, hence as $x^*$ is continuous $x^*(\E\xi_n) \to x^*(\E\xi)$. So we have
$$ x^*(\E\xi) = \lim_n x^*(\E\xi_n) = \lim_n \E x^*(\xi_n) = \E x^*(\xi). $$
