Differentiability of the blow-up of a function

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.

• What if we choose $t_k=1$? Apr 9 '19 at 16:05
• sorry, I forgot to add $t_k \to 0$ Apr 9 '19 at 17:12
• I believe $f(x)=x\cos(\log\log(1/x))$ is a counterexample. Apr 10 '19 at 5:31
• @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. Apr 10 '19 at 11:05
• 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. Apr 10 '19 at 14:36

Set $$f(x)=x\cos(\log(\log(1/|x|)))$$ (and 0 at 0).
Let $$\epsilon>0$$. Now if $$0, consider $$\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.