Demonstration that the $\langle f(x)\rangle$ of an odd function $f(x)=-f(-x)$ of position $x$ in a symmetric potential well $V(x)=V(-x)$ is null Consider a potential infinite well, which borders are $x=-a$ and $x=a$. I pretend to demonstrate that the expected value of a odd function $f(x)$, i.e., $\langle f(x)\rangle$, is null. 
We have the following wave function: $$\psi(x,t)=\sum_n A_n u_n(x) e^{-\frac{iE_n}{\hbar}t}$$
I know that for a symmetric potential infinite well, the eigenfunctions $u(x)$ of $H$ are real, and have a defined parity. So,
$$\langle f(x)\rangle=\int_{-a}^{a}\text{dx } \psi^*(x)f(x)\psi(x)=\int_{-a}^{a}\text{dx } \left(\sum_n A_n u_n(x)\right)f(x)\left(\sum_n A_n u_n(x)\right)$$
$\langle f(x)\rangle$ will be null if only $u_n(x)$ and $u_m(x)f(x),$ where ($n\neq m)$ are orthogonal. 
Is it true that if $u_n(x)$ and $u_m(x)$ are orthogonal (which is true), then $u_n(x)$ and $u_m(x)f(x)$ will be also orthogonal?
 A: 
Is it true that if $u_n(x)$ and $u_m(x)$ are orthogonal (which is true), then $u_n(x)$ and $u_m(x)f(x)$ will be also orthogonal?

No.  The simplest example of this is the case $f(x) = u_n(x)/u_m(x)$ for whatever $n$ and $m$ you're considering.
More broadly, the result you're trying to prove is false.  Consider the infinite square well between $\pm \pi/2$, with the wavefunction $\psi(x) = \cos(x) + \sin(2x)$:

The expectation value of $f(x) = x$ is quite obviously not zero, even though $f(x)$ is an odd function, and in fact works out to be $\langle x \rangle = 16/9\pi.$
Now, if $\psi$ happens to be an eigenstate of parity, then I think your result holds:  If $P \psi(x) = \pm \psi(x)$ (where $P$ is the parity operator), then $\psi^*(x) \psi(x)$ is obviously an even function, which means that $\psi^*(x) \psi(x) f(x)$ is an odd function, and we get zero when we integrate it over an even interval.  What's more, if the potential is itself symmetric, then $[H, P] = 0$ and all of the energy eigenstates $u_n(x)$ are (or can be chosen to be) parity eigenstates.  But if you're combining states of different parity (which you seem to want to do), then all bets are off.
