# Integral of $L^2(\mathbb R^+)$ function is $o(\sqrt x)$

I stumbled upon an exercise which goes as follows :

Let $$f \in L^2(\mathbb R^+)$$, show that $$\int_0^x f(t)\text{d}t = o\left(\sqrt x\right)$$.

1. Cauchy-Schwarz inequality gives the fact that $$\int_0^x f = O\left(\sqrt x\right)$$
2. For decreasing functions, the results holds since $$xf(x)^2 \leq \int_x^{2x} f^2 = o(1)$$.
3. Simple limit examples such as $$f(t) = \dfrac 1 {\sqrt t\ln(t)}$$ or $$f(t) = \displaystyle\sum_{k} \dfrac{1_{[k,k+1]}}{\sqrt k}$$ advocate for the result.

However I cannot produce a proof of that result nor find any counterexample, I don't even know what to believe. Given the source of the problem, a proof, if there is, should be elementary.

Happy Hollidays

• May I ask what the source is? Dec 25, 2021 at 0:29
• @CalvinKhor : compendium of exercises for french "classe préparatoire" math students (~20 yo). Some statements are false, which makes me cautious. Link (in french) madore.org/~david/math/kholles/annee97_spe/ex97_e24.dvi Dec 25, 2021 at 0:34
• I haven’t worked the details, but there might be a way to use decreasing rearrangements cf en.m.wikipedia.org/wiki/Symmetric_decreasing_rearrangement Dec 25, 2021 at 0:43
• Note that for $f$ in $L^2 \cap L^p$ with $1<p<2$ we get improved asymptotics of $\int_0^x f(t) dt$ as $x\to \infty$, which satisfies the claim. Now we use the fact that $L^2 \cap L^p$ is dense in $L^2$. Dec 25, 2021 at 0:51

To elaborate on my comment. Fix an arbitrary $$f\in L^2$$ and let $$g\in L^2 \cap L^p$$ for some $$1. A simple triangle inequality followed by Hölder's inequality gives
$$\lvert \int_0^x f(t)dt \rvert \leq \lvert \lvert f-g\rvert \rvert_{L^2}\sqrt{x} + \lvert \lvert g \rvert \rvert_{L^p} x^{1-1/p}.$$ This implies that $$\limsup_{x \to \infty} \frac{1}{\sqrt{x}} \lvert \int_0^x f(t) dt \rvert \leq \lvert \lvert f-g\rvert \rvert_{L^2} \qquad \forall g \in L^2 \cap L^p.$$ Using the fact that $$L^2 \cap L^p$$ is dense in $$L^2$$, we can take the infimum of all such $$g$$'s, thus proving the claim.
For any $$0 < a < x$$, using the Cauchy-Schwartz inequality one has $$\bigg|\int_0^x f(t) dt\bigg| \leq \int_0^a |f(t)| dt + \int_a^x |f(t)| dt$$ $$\leq \int_0^a |f(t)| dt + \bigg(\int_a^x|f(t)|^2 dt\bigg)^{1/2} (x - a)^{1 \over 2}$$ Given any $$\epsilon > 0$$ one can choose $$a$$ so that $$\bigg(\int_a^{\infty}|f(t)|^2 dt\bigg)^{1/2} < {\epsilon \over 2}$$ Thus for such $$a$$ one has $$\bigg|\int_0^x f(t) dt\bigg| \leq \int_0^a |f(t)| dt + {\epsilon\over 2}x^{1 \over 2}$$ So if $$x$$ is large enough one has the desired result that $$\bigg|\int_0^x f(t) dt\bigg| \leq \epsilon x^{1 \over 2}$$
• We're taking $x \to \infty$, not $x \to 0$. Dec 25, 2021 at 2:15
• Ok I changed to the $x \rightarrow \infty$ case. Dec 25, 2021 at 3:48
• (+1) This is an explicit version of the other answer by putting $\require{unicode}g=f \unicode{x1D7D9}_{[0,a]}$ and sending $a\to\infty$ Dec 25, 2021 at 13:04
• @CalvinKhor Well, there is one distinction... we're not using $L^p$ for $p < 2$ or using Holder's inequality... instead we just use that $f {\mathbb 1}_{[0,a]}$ converge to $f$ in $L^2$. Dec 25, 2021 at 19:49