Prove the set of points where $f$ is differentiable is dense.

Let $I\subset \Bbb{R}$ be an open interval and consider a continuous function $f:I\to \Bbb{R}$ satisfying, for all $x\in I$ $$\displaystyle \lim_{h\to 0} \frac{f(x+h)+f(x-h)-2f(x)}{h}=0$$ Prove that the set of points at which $f$ is differentiable is dense in $I$.

I know that $$\displaystyle \lim_{h\to 0} \frac{f(x+h)+f(x-h)-2f(x)}{h^2}=f''(x)$$ when $f''(x)$ exists, but it seems to be no use here.

Solution 1:

To show the set of points on which $f$ is differentiable is dense, we need to show an arbitrary open subinterval of $I$ contains a point where $f$ is differentiable. Restrict our attention to any open interval in $I$, say $(a,b)$.

We know that $f$, being continuous, attains a maximum on $[a,b]$. Hence, it either attains a maximum in $(a,b)$, say at $x_0$, or at one of the endpoints (so $f(a)$ or $f(b)$ are greater than $f(x)$ for $x \in (a,b)$). Suppose the mxamimum is at $x_0$, contained in $(a,b)$. If we look at your limit at this maximum, we see that it is non-positive ($\leq 0$), and can be $0$ only if $f$ is "constant" near $x_0$, implying it is differentiable. More rigorously:

$\forall \epsilon>0 \ \ \exists \ \ \delta>0$ such that $2f(x_0)-f(x_0+h)-f(x_0-h)<\epsilon |h|$ whenever $|h|<\delta$. This trivially gives $f(x_0)-f(x_0+h)<\epsilon |h|$ (as the other remaining part is non-negative, $x_0$ being a maximum) whenever $|h|<\delta$, implying $f$ differentiable at $x_0$ with derivative $0$.

If on the other hand, our maximum is at one of the end points, then do the following. Try and sub-restrict $f$ to a smaller interval $(c,d) \in (a,b)$. If, for all such restrictions, the maximum is at one of the end-points, then $f$ is monotone on $(a,b)$ and differentiable almost everywhere on $(a,b)$ (Monotone+continuous but not differentiable). If not, then the above argument applies, and we are done.