I'm currently studying the theory of PDEs and, in particular, harmonic functions. I've been given this question:
Show that if $u:(a,b) \rightarrow \mathbb{R}$ is continuous, and satisfies the following:
$$u(x_0) = {1\over2}(u(x_0 - r) + u(x_0 + r))$$ $\forall$ intervals $[x_0 - r,\space x_0 + r] \subseteq (a,b)$. Then u is harmonic (and therefore affine since u is $1$D). I've been given the hint; divide the interval $[x_0 - r,\space x_0 + r]$ in half again and again, however I can't see how this would help prove that the function is harmonic. All I seem to get after a long time splitting that interval is that:
$$ u(x_0) = {1\over2}(u(x_0-{2n-1\over2n}r) + u(x_0-{2n-3\over2n}r) + ... +u(x_0+{2n-1\over2n}r))$$
Would anyone please be able to either enlighten me on how the hint is linked to the question or if there is an easier way to go about showing this?
If you do the subdivision correctly, in the limit you end out getting that for any $x$ and any $r > 0$ you have $$f(x) = {1 \over 2r}\int_{x - r}^{x + r} f(y)\,dy$$ You can even set $r = x$ so that the right-hand side becomes continuously differentiable by the fundamental theorem of calculus. This shows $f(x)$ is also continuously differentiable. You can then differentiate the expression $f(x) = 1/2(f(x + r) + f(x- r))$ to show $f'(x)$ is harmonic, then you can iterate the above to show $f(x)$ is actually $C^{\infty}$.
The goal becomes to show that $f''(x) = 0$ for all $x$. The idea for this is that if $f''(x_0)$ were positive or negative, $f(x)$ would either be concave or convex near $x_0$, so that $f(x_0) = 1/2(f(x_0 + r) + f(x_0 - r))$ would not be satisfied for small values of $r$.
Note that if you already know $f(x)$ is $C^2$, then you just have to do the last paragraph. But since you only were given $f(x)$ is continuous you have to do the above as well.