12
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @TonyK: In the OP's defense, this is a tricky question to state (I for one mis-construed it more than once, not due to fault of the OP), and the edits do not seem to have changed the mathematical meaning. As I read it: "Let $f$ be differentiable on $\mathbf{R}$. If $f'$ itself is differentiable at $a$, does there exist a $\delta>0$ such that $f'$ is continuous on $(a-\delta, a+\delta)$?" $\endgroup$ – Andrew D. Hwang Dec 24 '13 at 20:08
  • $\begingroup$ Ah, I see. I misconstrued it too! I have deleted my erroneous comment. $\endgroup$ – TonyK Dec 24 '13 at 20:12
1
$\begingroup$

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."

$\endgroup$
1
$\begingroup$

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?)

$\endgroup$
  • 1
    $\begingroup$ Because of the jump discontinuities, $f$ isn't differentiable everywhere...? But this detail can be fixed by taking $\phi(t) = t^2\sin(\frac{1}{t})$ and $g(t) = \sum_{n=1}^\infty \frac{1}{n^2} \phi'(t - \frac{1}{n})$. Since $\phi'$ is bounded the series for $g$ converges uniformly, so the integral, $f(x)$, can be differentiated termwise; $g = f'$ is differentiable at $0$ but discontinuous at $\frac{1}{n}$ for every $n > 0$. $\endgroup$ – Andrew D. Hwang Dec 24 '13 at 19:38
  • $\begingroup$ Yes, fair enough, I was violating $f'$ continuous on $(0,\delta)$ by having $f'$ undefined at $x=1/n$. That disregarded the hypothesis that $f$ be everywhere differentiable. My apologies. The question was one step more subtle than I my approach. ... Fundamental fact for everyone: If $f$ is differentiable, $f'$ can never have jump discontinuities. (Darboux's Theorem) $\endgroup$ – Ted Shifrin Dec 24 '13 at 19:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.