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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.

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    $\begingroup$ 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
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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.

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

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