Show that $F(x)=o(x^{1/q} )$, where $x \to +\infty$, whith $q=\frac{p-1}{p}.$

Question

Let $f \in \mathcal{L}^p$, with $1<p<+\infty$. For all $x\geqslant 0$, we define $\displaystyle{F(x)= \int_0^x f(t)dt}$.

1. Show that $F$ is uniformely continuous on $\mathbb{R}$.
2. Show that $F(x)=o(x^{1/q} )$, where $x \to +\infty$, whith $q=\frac{p}{1-p}.$

My attempt :

1. let $0\leqslant x \leqslant y,$ \begin{align*}|F(y)-F(x)|&=\left|\int_x^yf(t)dt\right|\\&\leqslant\int_{\mathbb{R}}1_{[x,y]}(t)|f(t)|dt\\& \leqslant (y-x)^{1/q} \left(\int_{\mathbb{R}} |f(t)|^p dt \right)^{1/p} \text{(by Hölder's inequality)} \\&\leqslant(y-x)^{1/q}\; ||f||p.\end{align*}
2. Let $a>0$ (fixed) and $x>a$, $|F(x)-F(a)|\leqslant (x-a)^{1/q}\; ||f||p$, then

$|F(x)|\leqslant |F(a)|+(x-a)^{1/q}\; ||f||p$ $\implies $ $\dfrac{|F(x)|}{x^{1/q}}\leqslant \dfrac{|F(a)|}{x^{1/q}}+(1-\frac{a}{x})^{1/q}\; ||f||p$.

I got stuck here, any help is highly appreciated.

Answer 1

I notice that $\displaystyle{ \left(\int_a^x|f(t)|^pdt \right)^{1/p} \xrightarrow{\color{red}{a} \to +\infty}0}$ $\implies$ $\displaystyle{ \left(\int_a^x|f(t)|^pdt \right)^{1/p} \xrightarrow{x \to +\infty}0}$, so

for all $\varepsilon>0$, $\exists$ $ a_{\varepsilon}$ : $\forall x\geqslant a_{\varepsilon}$, $\displaystyle{ \left(\int_a^x|f(t)|^pdt \right)^{1/p}} \leqslant \varepsilon/2.$

$ \dfrac{|F(a)|}{x^{1/q}} \xrightarrow{x \to +\infty}{}0 $ $\implies$ $\exists$ $x_{\varepsilon} >0$ : $\forall x> x_{\varepsilon}$, $\dfrac{|F(a)|}{x^{1/q}} \leqslant \varepsilon/2.$

For $\displaystyle {m=max(x_{\varepsilon} ,a_{\varepsilon})}$, we have $\dfrac{|F(x)|}{x^{1/q}} \leqslant \varepsilon/2+\varepsilon/2=\varepsilon, $ for all $x\geqslant m$.

what do you think?