I am trying to prove an inequality that is

$\left\| \frac{1}{x}\int_{0}^{x} f(s) ds \right\|_2 \leq 2\|f\|_2=2\left( \int_{0}^{\infty} |f(s)|^2ds \right)^\frac{1}{2}$

where $f\in L^2([0,\infty))$ and $\|\cdot\|_2$ is $L^2([0,\infty))$ norm.

I have used several ways but those ways fail to prove the inequality. Especially, if we use Holder's inequality for functions $\chi_{[0,x]}$ and $f$, we get infinity... I have no idea.... this inequality might be helpful but I am not sure.

$|\int_{0}^{x} f(s) ds|^2\leq 2\sqrt{x}\int_{0}^{x} \sqrt{s} |f(s)|^2 ds$

thanks in advance!


We can use the other inequality to get $$\left\lVert x \mapsto \frac{1}{x} \int_0^x f(s) \, \mathrm{d} s \right \rVert_2^2 \leq \int \limits_0^\infty \frac{2 \sqrt{x}}{x^2} \int \limits_0^x \sqrt{s} |f(s)|^2 \, \mathrm{d} s \, \mathrm{d} x = 2 \int \limits_0^\infty \frac{1}{x^{3/2}} \int \limits_0^x \sqrt{s} |f(s)|^2 \, \mathrm{d} s \, \mathrm{d} x \, .$$ Changing the order of integration (Tonelli's theorem) we find $$ \left\lVert x \mapsto \frac{1}{x} \int_0^x f(s) \, \mathrm{d} s \right \rVert_2^2 \leq 2 \int \limits_0^\infty \sqrt{s} |f(s)|^2 \int \limits_s^\infty \frac{1}{x^{3/2}} \, \mathrm{d} x \, \mathrm{d} s = 4 \int \limits_0^\infty |f(s)|^2 \, \mathrm{d} s = 4 \lVert f \rVert_2^2$$ as desired.

  • $\begingroup$ Could you give me more detail on the equality of $2\int_{0}^{\infty}\frac{1}{x^{3/2}}\int_{0}^{x} \sqrt{s}|f(s)|^2 ds dx=2\int_{0}^{\infty} \sqrt{s}|f(s)|^2 \int_{s}^{\infty} \frac{1}{x^{3/2}}dx ds$? $\endgroup$ – Lev Ban Aug 8 '18 at 23:49
  • $\begingroup$ @LevBan Sure! We want to integrate the function $(x,s) \mapsto 2 x^{-3/2} \sqrt{s} |f(s)|^2$ over the part of $[0,\infty)^2$ in which $s$ is smaller than $x$ . We can write this region in two different ways: $$\{(x,s) \in \mathbb{R}^2 \colon 0 \leq x < \infty , 0 \leq s \leq x \} = \{(x,s) \in \mathbb{R}^2 \colon 0 \leq s < \infty , s \leq x \leq \infty \} $$ If we integrate with respect to $s$ first, we can find the limits of integration from the first representation. After switching the order of integration we see that $x$ must run from $s$ to $\infty$ according to the second version. $\endgroup$ – ComplexYetTrivial Aug 9 '18 at 0:00

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