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 $C^{\prime \prime}|x-y|^{1-\beta/\alpha}$.


1 Answer 1


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:

Description of the counterexample

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:

enter image description here

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 $\lambda>1-\beta/\alpha$.


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