Is the set of exponents for which a monotone function is Hölder-continuous closed in $(0,\infty)$?

Fix an a bounded (non-degenerate) interval $I\subset\mathbb R$. For $F:I\to\mathbb R$ and $\alpha>0$, a function is said to be Hölder-continuous of exponent $\mathbf{\alpha}$ if $$\sup_{\substack{x,y\in I\\x\neq y}}\frac{|F(x)-F(y)|}{|x-y|^{\alpha}}<\infty.$$

It is well-known (and quite easy to prove) that if $0<\alpha<\beta$ and $F$ is Hölder-continuous of order $\beta$, then it is also Hölder-continuous of order $\alpha$. This implies that the set of exponents for which a given function is Hölder-continuous is one of the following forms:

• empty,
• $(0,\infty)$,
• $(0, \alpha^*)$ for some $\alpha^*>0$, or
• $(0,\alpha^*]$.

The question is whether this set of exponents is closed or not. To exclude pathologically oscillating functions, suppose that $F$ is non-decreasing and is Hölder-continuous for $\alpha\in(0,\alpha^*)$. Is it also Hölder-continuous of order $\alpha^*$?

• I am a bit rusty on Real Analysis, so this might sound naive, but for $\alpha >0$ does it not follow that $f$ is a constant? So in that sense $a^*$ must be $<1$? Sep 27 '16 at 8:19
• @MathematicianByMistake It would follow only for $\alpha>1$. But you are right in that the question is interesting only for $\alpha^*\leq1$. Sep 27 '16 at 8:21
• Yes, I meant $\alpha >1$ but wrote $0$ instead..I guess I am rusty in typing as well :-) Sep 27 '16 at 8:25

Choose your favourite $\alpha \in (0,1]$, let $I = \bigl[0,\exp\bigl(-\frac{1}{\alpha}\bigr)\bigr]$, and

$$f(x) = \begin{cases} x^{\alpha}\log \frac{1}{x} &, x > 0 \\ \quad 0 &, x = 0.\end{cases}$$

Then $f$ is not $\alpha$-Hölder continuous, since

$$\frac{f(x) - f(0)}{x^{\alpha}} = \log \tfrac{1}{x} \xrightarrow{x \downarrow 0} +\infty.$$

Also, $f$ is strictly increasing on $I$ since for $x > 0$ we have $f'(x) = -x^{\alpha-1}(1+\alpha\log x)$, which is positive for $0 < x < \exp\bigl(-\frac{1}{\alpha}\bigr)$.

It remains to see that $f$ is $\beta$-Hölder continuous for $\beta \in (0,\alpha)$. Choose an arbitrary such $\beta$ and let $\gamma = \beta^{-1}$. Consider the function $g(x) = f(x)^{\gamma}$. Clearly $g$ is continuously differentiable on $I \setminus \{0\}$ with

$$g'(x) = \gamma x^{\alpha\gamma - 1}\bigl(\log \tfrac{1}{x}\bigr)^{\gamma-1}\bigl(\alpha \log \tfrac{1}{x} - 1\bigr)$$

there. We have $\lim\limits_{x \downarrow 0} g'(x) = 0$ since $\alpha\gamma > 1$, hence $g$ is also differentiable at $0$ with $g'(0) = 0$. Thus $g$ is continuously differentiable on $I$, and by the compactness of $I$ (and monotonicity of $g$) we have $0 \leqslant g'(x) \leqslant K$ for some $K \in (0,+\infty)$ on $I$. By the mean value theorem, we have

$$\frac{\lvert f(y) - f(x)\rvert}{\lvert y-x\rvert^{\beta}} = \biggl\lvert \frac{g(y) - g(x)}{y-x}\biggr\rvert^{\beta} \leqslant K^{\beta}$$

for $x,y\in I,\, x\neq y$, i.e. $f$ is $\beta$-Hölder continuous with constant $K^{\beta}$ on $I$.

• Thank you for this clever example! Sep 29 '16 at 5:03
• Just one remark: You may want to write $\log (1/x)$ instead of $-\log(x)$. This is because when you write, for example, $-(\log x)^{\gamma}$, you may be taking a fractional power of a negative number, technically speaking. Writing $[\log(1/x)]^{\gamma}$ instead circumvents this annoying technicality. Sep 29 '16 at 5:09
• Good point, thanks. Sep 29 '16 at 8:32