Let $u \in C^0([-1,1])$ such that $u(0)=0$. Suppose that $u$ satisfies the following property:

For every $\{t_k\}\subset \mathbb{R}$ such that $t_k \to 0$, there exist a real number $\alpha$ (depending on the sequence $\{t_k\}$), and a subsequence $\{t_{k_n}\}$, such that, if we set $u_k(x)={u(t_k x) \over t_k}$, we have that $u_{k_n}(x) \to \alpha x$ uniformly on $[-1,1]$.

My question is the following: Is the function $u$ necessarily differentiable at the origin? Even if $\alpha$ depends on the sequence $\{t_k\}$, I can't imagine how such a counterexample should behave.

  • 2
    $\begingroup$ What if we choose $t_k=1$? $\endgroup$ Apr 9 '19 at 16:05
  • $\begingroup$ sorry, I forgot to add $t_k \to 0$ $\endgroup$ Apr 9 '19 at 17:12
  • $\begingroup$ I believe $f(x)=x\cos(\log\log(1/x))$ is a counterexample. $\endgroup$ Apr 10 '19 at 5:31
  • $\begingroup$ @AnthonyQuas how can you prove it? I think that yours is not enough. Take for instance some very fast-converging to 0 sequence as $t_k = e^{-e^k}$ and you will see the graph oscillating between $x$ and $-x$, which I think is enough to prove that yours does not satisfy the hypotheses. If I'm wrong, please explain. $\endgroup$ Apr 10 '19 at 11:05
  • $\begingroup$ It’s true that there is this oscillation, but it takes place on a different scale, so that the place where it looks like $-x$ is a tiny part in the centre. This does not affect the uniform convergence. $\endgroup$ Apr 10 '19 at 14:36

Set $f(x)=x\cos(\log(\log(1/|x|)))$ (and 0 at 0).

Let $\epsilon>0$. Now if $0<t<\epsilon^{1/\epsilon}$, consider $\Delta_t(x):=|f(tx)/t - x\cos(\log\log(1/t)|$.

On $[-\epsilon,\epsilon]$, this is bounded above by $2\epsilon$.

If $\epsilon<|x|\le 1$, we have \begin{align*} \Delta_t(x)&=|x[\cos(\log\log(1/(t|x|))-\cos(\log\log(1/t))]|\\ &\le \log\log(1/t|x|)-\log\log 1/t\\ &= \log(\log(1/t)+\log(1/|x|))-\log\log(1/t)\\ &\le \log(1/|x|)/\log(1/t)<\epsilon, \end{align*} where I used the concavity of $\log$ in the inequality before last.

  • $\begingroup$ Thank you. Now this is clear. $\endgroup$ Apr 10 '19 at 18:13

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.