Smoothness of quotient of Holder continuous functions, provided the decay

Let $$I=(-1,1)$$ and $$u \in \text{Lips}(I)$$, the space of Lipschitz continuous functions in I. Suppose that $$|u(x)| \le C |x|^\alpha$$ for $$x \in I$$, for some $$\alpha \in (1,2)$$. I would like to estimate the Holder continuity of $$v(x)= \begin{cases} |x|^{-\beta}\cdot u(x) &\text{ for } x \neq 0 \\ 0& \text{ for } x =0 \end{cases}$$ for some $$1<\beta<\alpha$$ (so that $$|x|^{\beta}$$ is also in $$\text{Lips}(I)$$) in the whole interval $$I$$.

It is clear that $$v$$ is in $$\text{Lips}(I \setminus (-\varepsilon,\varepsilon))$$ (and therefore $$C^{\alpha-\beta}(I\setminus (-\varepsilon,\varepsilon))$$) for any $$\varepsilon \in (0,1)$$. and We also know that $$v$$ is "Holder continuous at $$0$$", that is, $$|v(x)-v(0)| \le C |x|^{\alpha-\beta}$$. Is it true that $$v$$ is Holder of order $$\alpha - \beta$$? The problem, of course, is to check if the Holder condition holds over sequences $$\{x_n\}_n$$ and $$\{y_n\}_n$$ such that $$|x_n-y_n|\ll |y_n| \longrightarrow 0$$.

I managed to prove that $$v$$ is in $$C^{1-\beta/\alpha}(I)$$. Without further hypothesis, is this the best Holder smoothness $$v$$ can have in $$I$$?

The choices of $$\alpha$$ and $$\beta$$ seemed to be a convenient regime to start, but the question could be done in a more general setting.

EDIT Let me put my proof of the Holder regularity $$1- \beta/\alpha$$. Assum w.l.o.g that $$|x|<|y|$$. We have from hypothesis that $$|u(x)-u(y)| \le L |x-y|$$ and for some constant $$L>0$$ and that $$|u(x)-u(y)| \le C (|x|^\alpha+|y|^\alpha) \le 2C |y|^\alpha.$$ thefore, writing $$|u(x)-u(y)|=|u(x)-u(y)|^{\delta}|u(x)-u(y)|^{1-\delta}$$, we have $$\tag{1} |u(x)-u(y)| \le C^\prime |x-y|^{(1-\delta)}|y|^{\delta\alpha}.$$

Now, we write $$\left | \frac{u(x)}{|x|^\beta}-\frac{u(y)}{|y|^\beta} \right| = \left | \frac{u(x)}{|x|^\beta} \frac{u(x)-u(y)}{|y|^\beta}-\frac{u(x)-u(y)}{|y|^\beta} \right|.$$

Using the triangular inequality and $$(1)$$ and the fact that $$|x|^\alpha$$ is Lipchitz, we have \begin{align*} \left | \frac{u(x)}{|x|^\beta}-\frac{u(y)}{|y|^\beta} \right| & \le C^{\prime \prime} \left [ |x|^{\alpha-\beta} |x-y| +|y|^{\delta\alpha-\beta}|x-y|^{1-\delta} \right] \end{align*} We then choose $$\delta = \beta/\alpha$$, and we get that the RHS is bounded by $$C^{\prime \prime}|x-y|^{1-\beta/\alpha}$$.

No it's not, in fact you found the sharp exponents! The example showcasing sharpness is a function which is 1-Lipschitz, has slope almost always -1 or 1, and oscillates between $$|x|^{\alpha}$$ and $$-|x|^{\alpha}$$. I don't think writing it explicitly will help much, so here's a picture instead:

Notice the dimensions in purple (which you can do at any crossing point $$x_0$$, of which there is infinitely many. Let's see what happens to them when we multiply by $$|x|^{-\beta}$$. Again, I think a picture does a better job:

That means that for points $$x_0$$ arbitrarily close to zero, we can find a point $$x'$$ (the point where $$u$$ meets with $$|x|^{\alpha}$$) such that $$|x'-x_0|\sim x_0^\alpha$$ and $$|u(x')-u(x_0)|\sim x_0^{\alpha-\beta}$$. From the first of those equalities we get that $$x_0\sim |x'-x_0|^{1/\alpha}$$. Plugging that into the second equation, we obtain the bound that shows sharpness:

$$|u(x')-u(x_0)|\sim |x'-x_0|^{1-\beta/\alpha}$$

and in particular, $$u(x)$$ is not $$C^\lambda$$ for any $$\lambda>1-\beta/\alpha$$.