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

Solution 1:

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.

Solution 2:

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