# 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.

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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.

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I like this approach! – Eric Naslund Jul 20 '11 at 19:44
@Eric: And I like yours... :-) To which, considering that this kind of stuff is not as widely known as it should, you could add the (simple) argument which shows that your $g$ indeed exists. – Did Jul 20 '11 at 19:53
Nice approach, and thanks! – user1736 Jul 20 '11 at 21:40

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,

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Cute argument. I like it. Maybe it would be nice to add why such a $g$ exists for completeness. – Jonas Teuwen Jul 20 '11 at 20:06
Thanks for the help! Sorry, I had a couple of followup questions that I was hoping you could clarify: 1)If you're applying Holder, should the integrand of your first expression be $\frac{1}{g(t)^q}$ instead of just $\frac{1}{g(t)}$ ? And 2) Could also explain why the limit of the integral at the end goes to 0? Are you assuming that you can find $g$ satisfying all the properties above and with $\frac{1}{g} \in L^q$? I didn't catch that in your argument, if that is the case. – user1736 Jul 20 '11 at 21:39
One does not need $1/g$ to be in $L^q$, simply that the integral of $1/g^q$ up to $x$ is $o(x)$, which holds as soon as $1/g(x)=o(1)$. – Did Jul 20 '11 at 21:47
@Didier: Oh, I think I can see how that would be true. Thanks again for your assistance. – user1736 Jul 20 '11 at 22:22
@user For every nonnegative integer $n$, pick a nonnegative $x_n$ such that the integral of $|f(x)|^p$ on $x\ge x_n$ is at most $2^{-n}$. Assume for simplicity that $(x_n)$ is nondecreasing and that $x_n\to+\infty$. Then define $g(x)$ as the maximal $n$ such that $x\ge x_n$. – Did Jul 21 '11 at 16:47