Is the limit of the scaling for Lipschitz functions unique? Let $ B(0,1) $ be a ball with center $ 0 $ and radius $ 1 $. If $ u $ is a $ C^{0,1}(\overline{B(0,1)}) $ function such that $ u(0)=0 $ and $ [u]_{C^{0,1}(B(0,1))}\leq M $ with $ M>0 $. Let $ u^r(x)=r^{-1}u(rx) $, with $ 0<r\leq 1 $ it can be find that $ [u^r]_{C^{0,1}B(0,1)}\leq M $ and $ u^r(0)=0 $. By using Ascoli-Arzela lemma, we can choose a subsequence $ \{r_{j}\} $ such that $ r_j\to 0 $ when $ j\to\infty $ and $ u^{r_j}\to u^0\in C^{0,1}(\overline{B(0,1)}) $ uniformly in $ C^0(\overline{B(0,1)}) $. I want to ask is the limit $ u^0 $ unique here?
 A: The answer is negative in general. The limit of the subsequences needs not be unique.
Indeed, a standard topological argument shows that if such limit $u^0$ were unique, then $u^r\to u^0$. So I am going to disprove the latter statement.
Let $f=f(t)$ be one of these examples of a Lipschitz continuous function of $t\in [0, 1]$ such that $f(0)=0$ but the limit
$$
\lim_{t\to 0^+}\frac{f(t)}{t}
$$
does not exist. Define $u(x)=f(\lvert x\rvert)$. This function is clearly Lipschitz continuous for $x\in B(0, 1)$ (see below), but
$$
\frac{u(rx)}{r}=\frac{f(r\lvert x \rvert)}{r}, $$
hence the limit as $r\to 0$ does not exist.
Proof that $u$ is Lipschitz continuous. For $x, y\in B(0, 1)$,
$$
\lvert u(x)-u(y)\rvert=\lvert f(\lvert x\rvert) -f(\lvert y \rvert)\rvert \le C\lvert \lvert x \rvert - \lvert y \rvert \rvert \le C\lvert x- y\rvert,$$
where $C$ denotes any Lipschitz constant of $f$. $\Box$

On the other hand, as discussed in comments, the limit does exist if $u$ is differentiable at $0$. This is not the case of the example above.
