Dini's continuity vs Holder continuity

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 $$\omega_f(t)\le f(t),\ \forall t\in (0,e^{-3})\Longrightarrow \int_0^{e^{-3}}\frac{\omega_f(t)}{t}dt\le \int_0^{e^{-3}}\frac{f(t)}{t}dt<\infty.$$