$1<p<\infty$, $f\in L^{p}(0,\infty)$, $p^{-1}+q^{-1}=1$, define $$F(x)=\int_{0}^{x}f(t)dt,$$ then I need to show that $\frac{F(x)}{x^{\frac{1}{q}}}\rightarrow 0$ as $x\rightarrow 0$ and $x\rightarrow \infty $.
We can write $F(x)=\int_{(0,\infty))}f{1}_{[0,x]}$, and then since $f\in L^p$ and indicator function is in $L^q$, applying Holder gives us: $\frac{F(x)}{x^{\frac{1}{q}}}\leq ||f||_p$. I do not know how to go from here. The hint says I need to use some kind of density argument after this, i.e may be first we need to show that the result holds on a dense subset of $L^p$ and then go from there, honestly I do not quite understand the hint. Thank you for your help.
After having done the substitution $t:=sx$ we get $$F(x)=x\int_0^1f(xt)dt,$$ and using Hölder's inequality, $$|F(x)|\leqslant x\left(\int_0^1|f(xt)|^pdt\right)^{\frac 1p},$$ which gives, doing back the substitution, $$\frac{|F(x)|}{x^{\frac 1q}}\leqslant x^{1-\frac 1q}\left(\int_0^x\frac 1x |f(s)|^pds\right)^{\frac 1p}= \left(\int_0^x|f(t)|^pdt\right)^{\frac 1p}.$$ We can conclude from this as $|f|^p$ is integrable.
An other solution would be the following: consider the map $T\colon L^p(0,\infty) \to L^\infty(0,+\infty)$ given by $T(f)(x):=\frac{f(x)}{x^{\frac 1q}}$. An use of Hölder's inequality gives that $\lVert T\rVert\leqslant 1$. By a $2\varepsilon$-argument, it's enough to show the result when $f$ is a simple function (that is, a linear combination of characteristic function of finite measure).