If $f'$ is differentiable at $a$ then $f'$ is continuous at $(a-\delta,a+\delta)$

Is there a counterexample?

Proposition: Let $f:\mathbb{R} \longrightarrow \mathbb{R}$ be a differentiable function such that $f':\mathbb{R} \longrightarrow \mathbb{R}$ is differentiable at $a\in\mathbb{R}$ (possibly differentiable at a single point). Then $f':\mathbb{R} \longrightarrow \mathbb{R}$ is continuous at an interval $(a-\delta,a+\delta)$ for some $\delta>0$.

Any hints would be appreciated.

Solution 1:

The answer is: "Yes, there is a counterexample". However, I'm no longer absolutely convinced my proposed example in the comments "works". Here's (a sketch of) an example about which I feel confident. No claim of simplicity is made; after so many false starts, I'm in an overkill mood.

Let $s(x) = x^2 \sin(1/x)$, extended by continuity.

Let $\psi:\mathbf{R} \to \mathbf{R}$ be a smooth bump function supported in $[-1, 1]$ and equal to $1$ in some neighborhood of $0$.

Let $\phi(x) = \psi(x) s'(x)$. The important points are: (i) $\phi$ is the derivative of some function $\Phi$ on $\mathbf{R}$; (ii) $\Phi' = \phi$ is discontinuous at $0$ and continuous everywhere else; (iii) For convenience, and without loss of generality (by tweaking the bump function, for example), we may assume $\Phi$ itself is supported in $[-1, 1]$.

Following Ted Shifrin's construction, we'll first construct $g = f'$. The idea is to sum suitably shifted and scaled $\Phi$s in such a way that (i) each point of $\mathbf{R}$ lies in the support of at most one summand (so $g$ is differentiable away from accumulation points of endpoints of the supports); (ii) the discontinuities of the summands accumulate at $0$ (so that $g'$ is discontinuous somewhere in every neighborhood of $0$); (iii) $|g(x)| \leq Cx^2$ for some $C > 0$ (so that $g$ is differentiable at $0$, the only accumulation point of the supports of the summands).

That the properties of the preceding paragraph can be achieved is probably safe to leave as an exercise. One strategy is, for each positive integer $n$, to scale and shift a copy of $\Phi$ to make the support $[1/(2n+1), 1/2n]$ and the maximum height $1/n^2$.

The desired counterexample is $f(x) = \int_0^x g$.

As G. H. Hardy apocryphally said, "Yes, it's trivial."

Solution 2:

The issue is that a function $g$ can be continuous at $a$ without being continuous on any interval around $a$. So you want such a function $g$ that is integrable, so that you can then consider $f(x) = \int_a^x g(t)\,dt$.

I would suggest something like (taking $a=0$) $$g(t) = \begin{cases} 1/[1/t], & 0<t\le 1 \\ 0, &\text{otherwise}\,. \end{cases}$$ Now set $f(x) = \int_0^x g(t)\,dt$.

It seems the hypotheses in the problem may have changed, or I missed something. So we want such a function $g$ that is differentiable at $0$ without being continuous on an interval around $0$. How can we modify this $g$? (HINT: We have a sequence of points $x_n\to 0$ with $g(x_n)=x_n$. Can we make it $g(x_n)=x_n^2$ instead?)