Exchangeability of inner product with the integral
Let $(X,\mathcal{E})$ be a measure space. Let $\mu : \mathcal{E} \to \mathbb{R}^+$ be a positive measure. If $f: X \to \mathbb{R}^m$ is measurable and $z \in \mathbb{R}^m$, is it true that: $\int_B \langle f, z \rangle d\mu = \langle \int_B f d\mu, z \rangle$?
I could prove this for simple functions $f$ but not for general functions.
Any help is much appreciated.
Thanks, Phanindra
Solution 1:
Writing $f=(f_1,\ldots,f_m)$ and $z=(z_1,\ldots,z_m)$, both sides are $\displaystyle\sum\limits_{k=1}^mz_k\int_Bf_k\text{d}\mu$, so, yes, they are equal as soon as everything written makes sense, that is, as soon as every $f_k$ is integrable on $B$.
On the LHS, this follows from the definition of the scalar product $\langle\ ,\ \rangle$ and the linearity of the integral. On the RHS, this follows from the definition of the integral of a vector-valued function and the definition of the scalar product $\langle\ ,\ \rangle$.
Solution 2:
Have you any mind for the case that $f$ and $z$ depend on both $B$ and $X$ (I mean to the elements of these spaces)?
If not can we put conditions on $z$? for example $z$ becomes sufficiently small.
As I deal with this problem, the answer of my first question is NO, for the second question, I write: We define $s := \max_{\xi \in B} \{z (\xi , x)\}$, by Fubini's theorem we have $\int_{B}\int_{D} g (z - s) dx d\mu(\xi) = \int_{B\times D} g (z - s) dx d\mu(\xi)$ $\leq \|(z - s)\| \|g\|$ $\leq 2 \|s\| \|g\|$ as $\|z\| \rightarrow 0$ thus $\|s\| \rightarrow 0$, now one can write $\int_{B \times X} g z dx d\mu(\xi) = \int_{B \times X} g s dx d\mu(\xi) $ $= \int_{X} s \int_{B} g d\mu(\xi) dx = \langle E[g] ,s \rangle,$ also noting that
$$\langle E[g] ,E[z] - s \rangle \leq \int_{D}\int_{B} (z - s) d\mu(\xi)dx \int_{D} \int_{B} g d\mu(\xi)dx $$
Am I right? In advance thanks for your contribution.
And one more, norms belongs to proper spaces and are not same.
If you have any question I am eager to here.