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$f:R\to R$ is a continuous function such that for every $r>0$ and $x\in R$.

If $\frac{1}{2r}∫^{x+r}_{x-r}f(t)dt=f(x)$, show that there exist constants $a,b$ such that $f(x)=ax+b$.

I tried to do this by fixing x at $x_0$ and r at $r_0$ and tried the contrapositive approach but got stuck. Can someone guide me in the right in how to start on this problem...and what ideas would be relevant in trying to prove this.

I understand what is happening. The mean value of the function is basically the value of the fucntion at the center of the interval. I just don't know how to start on the proof.

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Does the equation hold for all $r$ or for some fixed $r>0$? –  gt6989b Dec 13 '12 at 22:14
edited. Its for every r>0. –  user52932 Dec 13 '12 at 22:18
Just a clarification note. It must hold for all $x\in \mathbb{R}$ as well, otherwise $f(x) = x^3$ around $x=0$ is a counterexample. –  gt6989b Dec 13 '12 at 22:49

1 Answer 1

Here is perhaps a thought of a direction.

Fix some $x \in \mathbb{R}$ and define $g(t) = f(t) - f(x)$. Then,

$\frac{1}{2r} \int_{x-r}^{x+r} g(t) dt = \frac{1}{2r} \int_{x-r}^{x+r} [f(t) - f(x)] dt = \frac{1}{2r} \int_{x-r}^{x+r} f(t) dt - f(x) = 0, $

which happens $\forall r > 0$. Thus, the graph of $g$ is anti-symmetric around the origin, so the graph of $f$ is anti-symmetric around every real point.

This last point implies that $f$ must be linear. Geometrically it cannot happen any other way, but I'm not sure how to prove it formally yet. Will edit this answer if I think of something better.

EDIT 1 This is inspired by comments from @DavidMitra below.

Note that we can differentiate in $r$ the relationship $2rf(x) = \int_{x-r}^{x+r} f(t)dt$ to get $2f(x) = f(x+r) - f(x-r)*(-1) = f(x+r) + f(x-r)$ and again to get $f'(x+r) = f'(x-r)$, which must hold for all $x,r>0$, so $f'$ is constant, so $f$ is linear.

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If you differentiate both sides of the defining equation for $f$, you obtain that the expression $f(x+r)-f(x-r)\over 2r$ is constant with respect to $r$ with value $f'(x)$. So any secant line to the graph of $f$ "centered" at $x$ has slope $f'(x)$. Is this what you mean by "anti-symmetric" about every point? –  David Mitra Dec 13 '12 at 23:12
@DavidMitra I think I'm saying something equivalent, or maybe stronger. If you rotate the graph of $f$ over $(-\infty,x)$ by 180 degrees around $(x, f(x))$, you end up with the graph of $f$ over $(x,\infty)$. –  gt6989b Dec 13 '12 at 23:15
@DavidMitra It's a nice idea to use Leibniz's Rule, but we cannot assume $f$ is differentiable, only integrable... –  gt6989b Dec 13 '12 at 23:16
$f$ is continuous, so both $g(x)=\int_0^{x+1} f(t)\,dt$ and $h(x)=\int_0^{x-1} f(t)\,dt$ are differentiable. So $f=g-h$ is. Or, am I missing something? –  David Mitra Dec 13 '12 at 23:31
@DavidMitra You are right. Sorry for not thinking about this before. This helped figure out the problem altogether - just differentiated wrt $r$... –  gt6989b Dec 13 '12 at 23:37

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