Let $F$ be a non-negative continuous function. Is there someone who explain why from mentioned condition we have a sequence with those conditions?
From the condition $$\liminf_{\xi\to0}\frac{f(\xi)}{\xi^p}=0$$ there is a sequence $\{\theta_n\}\subseteq ]0,+\infty[$ such that $\lim_{n\to\infty} \theta_n=0$ and $$\lim_{n\to\infty} \frac{\sup_{|\xi|<\gamma_n} F(\xi)}{\theta_n^p}=0.$$ Indeed, one has $$\lim_{n\to\infty} \frac{\sup_{|\xi|<\theta_n} F(\xi)}{\theta_n^p}=\lim_{n\to\infty} \frac{F(\xi_n)}{\xi_{\theta_n}^p} \cdot \frac{\xi_{\theta_n}^p}{\theta_n^p} = 0,$$ where $F(\xi_n)=\sup_{|\xi|<\theta_n}F(\xi)$.
I know that $\lim\inf_{x\to 0^+}f(x)=\lim_{x\to 0^+}(\inf_{0<y<x}f(y))$.
Let $F(x)=x\sin^2{1\over x}.$ Then $$\liminf_{x\to 0}{F(x)\over |x|}=0,\qquad \sup_{|t|\le x}{F(t)\over |t|}=1$$