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I've encountered the following problem while preparing for a qualifying exam.

Let $f \in L^p [0,\infty)$. Show that for $1 < p < \infty$ we have $$\lim_{x \rightarrow \infty} \frac{1}{x^{1 - 1/p}} \int\limits_0^x f(t) dt = 0.$$

At first, I thought this would be an immediate application of Holder's inequality. However, Holder only gives us $$\left|\frac{1}{x^{1 - 1/p}} \int\limits_0^x f(t) dt\right| \leq ||f||_p.$$

Does anyone have any hints/suggestions on how to proceed?

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    The space of functions with compact support is dense in $L^p[0,\infty)$, and $\{T_x : x \in (0,\infty)\}$ is an equicontinuous family. – 2017-01-05
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    [Similar](http://math.stackexchange.com/questions/876346/limit-of-integral-of-lp-functions?rq=1) – 2017-01-05
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    Thanks, I believe I figured it out – 2017-01-05
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    @DanielFischer Do we need that translations are an equicontinuous family? Here is my proof. Let $g \in C_c [0, \infty )$. Then $g = 0$ outside some ball $B_R \subset (0 , \infty)$ and so if $x > R$, then $x^{1/p - 1} \int_0^x g = x^{1/p - 1} \int_{B_R} g \rightarrow 0$ as $x \rightarrow \infty$. Let $f \in L^p [0,\infty)$ and $\epsilon > 0$. There exists a $g \in C_c [0,\infty)$ such that $||f - g||_p < \epsilon$. Thus, $$\lim_{x \rightarrow \infty} \left| x^{1/p - 1} \int_0^x f \right| = \lim_{x\rightarrow \infty} \left| x^{1/p - 1} \int_0^x (f-g) \right| \leq ||f - g||_p < \epsilon$$ – 2017-01-06
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    By $T_x$ I meant the map $f \mapsto x^{1/p-1}\int_0^x f(t)\,dt$, not the translations. And your argument is the one I had in mind. – 2017-01-06

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Don't give up on Holder: Let $f\in L^p[0,\infty).$ Let $\epsilon>0.$ Because $|f|^p \in L^1[0,\infty),$ there exists $x_0>0$ such that $\int_{x_0}^\infty |f|^p < \epsilon.$ Thus

$$ \int_0^x |f| = \int_0^{x_0}|f| + \int_{x_0}^x |f|$$ $$ \le \int_0^{x_0}|f| + \left(\int_{x_0}^x |f|^p \right)^{1/p}(x-x_0)^{1-1/p} \le \int_0^{x_0}|f| + \epsilon^{1/p}(x-x_0)^{1-1/p} .$$

Now divide by $x^{1-1/p}$ to get

$$\limsup_{x\to \infty}\frac{1}{x^{1-1/p}}\int_0^x |f| \le 0 + \epsilon^{1/p}.$$

Since $\epsilon$ is arbitray, the $\limsup$ is $0,$ giving the result.

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    I am sorry, but I don't quite see how this gives us the result. Can you provide a little bit more of a hint? – 2017-01-06
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    If $g\in L^1,$ then $\int_0^x |g| \to 0$ as $x\to 0.$ This follows from the DCT. – 2017-01-06
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    Yes, that is relatively easy to see. However, i am not concerned with the limit as x goes to zero. Unless your suggesting that we can somehow use this information to extract info about the limit as x goes to infinity, in which case, I don't exactly see how this can be done. – 2017-01-06
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    Sorry, I misread the problem and thought $x\to 0^.$ Will rethink. – 2017-01-06
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    I have a new answer. – 2017-01-06