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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 $y(d)=0$?

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You have the even/odd definitions swapped. Also, not sure what you mean by $f_e$ also achieves it at $-y$. –  anon Jan 18 '12 at 5:57

1 Answer 1

You're right re: odd/even. $ f$ achieves its mean at say +y and -y iff y achieves it and $ f(y)=f(-y).$ I now realise my first question was trivial. As for the second since $f$ is continuous the points achieving the mean in each interval should describe smooth continuous curves as the interval moves to the right and since 0 is the only point common to all intervals $I_d$ as $d$ ranges from -1 to 1 each curve should cross 0 somewhere. This means if we focus on one of the curves we track the continuous curve $\{(y(d),d): -1 \leq d \leq 1\}$ in the parallelogram with corners $(-2,-1),(0,-1),(2,1),(0,1)$ if we let $a=1$ wlog.

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