(listed items are just the definitions, you can skip to "Clearly" if you are familiar with them)
- Let $E \subset \mathbb{R}^N$ and let $f \colon E \to \mathbb{R}.$ The modulus of continuity of $f$ is the increasing function $\omega_f \colon [0,\infty) \to [0,\infty)$ defined by $$\omega_f(t) := \sup\{|f(x) – f(y)| : x,y \in E, \|x – y\| \le t\}.$$The function $f$ is Dini's continuous if $$\int_0^1\frac{\omega_f(t)}{t}\ dt < \infty$$
- Let $E \subset \mathbb{R}^N$ and let $f \colon E \to \mathbb{R}.$ The function $f$ is Holder continuous with exponent $\alpha \in (0,1)$ if $$|f|_{C^{0,\alpha}(E)} := \sup\Big\{\frac{|f(x) – f(y)|}{\|x – y\|^{\alpha}} : x,y \in E, x \neq y\Big\} < \infty.$$
Clearly if $f$ is Holder continuous with exponent $\alpha$, then $\omega_f(t) \le Lt^{\alpha}$ for some $L > 0$ and hence $f$ is Dini's continuous. Do you know a counterexample to the other inclusion? (namely, I am looking for a Dini's continuous function such that it is not Holder continuous for any $\alpha$).
[Motivation for the question: I know that the Fourier series of a Holder continuous function converges uniformly to the function. I also know that the Fourier series of a continuous function can be very bad (it can diverge on a dense subset). Today I read that the Fourier series of a Dini's continuous function converges uniformly, so, as it is natural, I want to know what functions I was missing before.]
Best Answer
Here is an example: $N=1$, $E=[0,e^{-3})$ and $$f(x)=\left\{\begin{array}{ccc}(\log x)^{-2}&,& x\in(0,e^{-3})\\ 0&,&x=0\end{array}\right..$$ For every $\alpha>0$, $$\lim_{x\to 0^+}\frac{|f(x)-f(0)|}{x^\alpha}=\infty,$$ so $f$ is not $\alpha$-Hölder continuous.
Now let us show that $f$ is Dini continuous. To begin with, note that $$f'(x)=-\frac{2}{x(\log x)^3}>0,\quad x\in(0,e^{-3}),$$ and $$f''(x)=\frac{2}{x^2(\log x)^3}+\frac{6}{x^2(\log x)^4}<0,\quad x\in (0,e^{-3}).$$ Therefore, $f$ is increasing on $[0,e^{-3})$ and $f'$ is decreasing on $(0,e^{-3})$. As a result, for every $t\in (0,e^{-3})$, when $0\le x<y\le x+t<e^{-3}$, $$0\le f(y)-f(x)\le f(x+t)-f(x)=\int_0^tf'(x+s)ds \le \int_0^tf'(s)ds=f(t).$$ It follows that $$\omega_f(t)\le f(t),\ \forall t\in (0,e^{-3})\Longrightarrow \int_0^{e^{-3}}\frac{\omega_f(t)}{t}dt\le \int_0^{e^{-3}}\frac{f(t)}{t}dt<\infty.$$