6

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

B. Pillet
  • 98
  • May I ask what the source is? – Calvin Khor Dec 25 '21 at 00:29
  • 1
    @CalvinKhor : compendium of exercises for french "classe préparatoire" math students (~20 yo). Some statements are false, which makes me cautious. Link (in french) http://www.madore.org/~david/math/kholles/annee97_spe/ex97_e24.dvi – B. Pillet Dec 25 '21 at 00:34
  • 1
    I haven’t worked the details, but there might be a way to use decreasing rearrangements cf https://en.m.wikipedia.org/wiki/Symmetric_decreasing_rearrangement – Calvin Khor Dec 25 '21 at 00:43
  • 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 $L^2$. – TheOscillator Dec 25 '21 at 00:51

2 Answers2

3

To elaborate on my comment. Fix an arbitrary $f\in L^2$ and let $g\in L^2 \cap L^p$ for some $1<p<2$. 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.

TheOscillator
  • 2,899
3

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 $$\bigg|\int_0^x f(t) dt\bigg| \leq \epsilon x^{1 \over 2}$$

Zarrax
  • 44,950