How to show that a limit vanishes Let $1<p<\infty$, $f \in L^p([0,\infty))$. Why is it then true that
$$\lim_{x\to\infty} \frac{1}{x^{1-\frac{1}{p}}} \left( \int_0^x f(t)dt\right) =0?$$
I know that from Hölder's inequality, we can get that $||f||_{L^1([0,\infty))} \leq ||f||_{L^p([0,\infty))} x^{1-\frac{1}{p}}, \; \forall x >0$, and I also think that this inequality must be strict, since equality would imply that $f$ is constant, which would contradict the fact that $f \in L^p$ on a domain with infinite measure. 
That's only a necessary condition though, and I'm not really sure how to proceed any further.
 A: An idea is to split the integral from $0$ to $x$ into two parts, a first one whose contribution will disappear when $x\to+\infty$ and a second one to which one applies the obvious argument which failed when applied to the whole thing. 
More specifically, for every positive $u$, pick $x(u)$ such that 
$$
\int_{x(u)}^{+\infty}|f(t)|^p\mathrm{d}t\le u.
$$
For every $x\ge x(u)$, decompose the integral of $f$ from $0$ to $x$ into the integral from $0$ to $x(u)$, whose value is $C(u)$, say, and the integral from $x(u)$ to $x$, to which one can apply Hölder's inequality. All this yields
$$
\left|\int_0^xf(t)\mathrm{d}t\right|\le C(u)+(x-x(u))^{1-1/p}u^{1/p}\le C(u)+x^{1-1/p}u^{1/p}.
$$
When $x\to+\infty$, this shows that the limsup of the absolute value of the ratio you are interested in is at most $u^{1/p}$. Since $u$ can be any positive real number, you are done.
A: First, we can find continuous monotonic $g(x)>0$ such that $\lim_{x\rightarrow \infty} g(x)=\infty$ and $f\cdot g \in L^p[0,\infty)$.  This simply says that there is no "fastest growing convergent" function.  That is, for any $f\in L^p[0,\infty)$ we can find a function which grows slightly faster but still converges in $L^p$.
Let  $q>1$ satisfy $\frac{1}{p}+\frac{1}{q}=1.$  Then by Hölders inequality
$$\int_{0}^{x}|f(t)|dt\leq\left(\int_{0}^{x}\frac{1}{g(t)}dt\right)^{\frac{1}{q}}\left(\int_{0}^{x}|f(t)g(t)|^{p}dt\right)^{\frac{1}{p}}\leq C_{1}\left(\int_{0}^{x}\frac{1}{g(t)}dt\right)^{\frac{1}{q}}$$ where $C_{1}$ is the norm of $fg$ in $L^{p}[0,\infty)$.  
The result then follows since $$\lim_{x\rightarrow \infty} \frac{1}{x^{1-p}}\left(\int_{0}^{x}\frac{1}{g(t)}dt\right)^{1-\frac{1}{p}}= 0.$$ 
Hope that helps,
