Differentiability of the blow-up of a function

Question

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.

Answer 1

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.