2
$\begingroup$

$f:R\to R$ 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 direction...as 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.

  • 0
    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.2012-12-13

1 Answers 1

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.

  • 0
    I think a more relevant question is that how do we know that$f$is differentiable wrt$r$at f(x+r) and f(x-r)? Also if f' is constant wrt r, don't we get a linear equation for r? .2012-12-14