Let $f$ be a nonnegative function defined on $[-a,a]$ for some $a>0.$ Let $f_{o}$ and $f_{e}$ be its odd and even parts defined as $f_{e}=(f(x)+f(-x))/2$ and $f_{o}=(f(x)-f(-x))/2$ so that $f=f_{o}+f_{e}$. Let $y$ be a point in $[-a,a]$ which achieves the mean value of $f,$ i.e., $f(y)=\frac{1}{2a} \int_{-a}^a f(x) \,dx.$ $y$ exists by the mean value theorem. However note that $\int_{-a}^a f(x) \,dx=\int_{-a}^a f_e(x) \,dx+\int_{-a}^a f_o(x) \,dx$ and that $\int_{-a}^a f_o(x) \,dx=0.$ Thus we also have $f(y)=\frac{1}{2a}\int_{-a}^a f_e(x) \,dx.$ This implies that the mean value for the even function is also achieved at $y$ however for $f_e(x)$ it is also achieved at $-y$ by symmetry.
My question is, under which conditions on $f$ [the obvious one being that $f$ is itself symmetric] can we say that $f$ will achieve its mean value at symmetric points? My guess is that this is quite hard.
A more interesting question is the following. Let $y(0)$ be defined by the same relation as before, i.e., $f(y(0))=\frac{1}{2a} \int_{-a}^a f(x) \,dx.$ and define a more general quantity $y(d)$ by $f(y(d))=\frac{1}{2a} \int_{-a+d}^{a+d} f(x) \,dx.$ where $d\in [-a,a]$ which means we're considering sliding intervals $I_d=[-a+d,a+d ]$.
Is it true that there exists a value of $d \in [-a,a]$ so that $y(d)=0$?