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Consider the Lebesgue measure on Borel sets of $(0,\infty)$. Prove that for every $f\in L^2(0,\infty)$,

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

My trial:

Naturally, we can think of Holder's inequality

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

And it is totally different from the desired result. I have no idea how to put $|\chi_{[0,x)}|$ into the integral of $f$ anyway.

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  • $\begingroup$ since you want a $\sqrt{s}|f(s)|^2$ in the right side, you may try the decomposition $f(s) = \frac{1}{\sqrt[4]{s}}\sqrt[4]{s}f(s)$.and then use Holder $\endgroup$ – Veridian Dynamics Aug 8 '18 at 20:12
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What about the Cauchy-Schwarz Inequality? Note that $$\left|\int_0^x\,f(s)\,\text{d}s\right|^2\leq\left|\int_0^x\,\frac{1}{\sqrt[4]{s}}\,\Big(\sqrt[4]{s}\,\big|f(s)\big|\Big)\,\text{d}s\right|^2\leq \left(\int_0^x\,\frac{1}{\sqrt{s}}\,\text{d}s\right) \,\left(\int_0^x\,\sqrt{s}\,\big|f(s)\big|^2\,\text{d}s\right)\,.$$ The equality holds iff there exists a constant $c\in\mathbb{C}$ such that $f(s)=\dfrac{c}{\sqrt{s}}$ for almost every $s\in[0,x]$.

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  • $\begingroup$ Three identical solutions at roughly the same time. $\endgroup$ – Batominovski Aug 8 '18 at 20:18
  • $\begingroup$ Can we do like $f(s)=\frac{1}{\sqrt[4]{s}} (\sqrt[4]{s})f(s)$ on $[0,x]$? What happens when $s=0?$ $\endgroup$ – Lev Ban Aug 8 '18 at 20:23
  • $\begingroup$ You can think of $\frac{1}{\sqrt[4]{s}}$ as some $g(s)$ which is $0$ at $s=0$, and equals $\frac{1}{\sqrt[4]{s}}$ for $s>0$. Or you can think of the integral as the integration on $(0,x)$, or on $(0,x]$. $\endgroup$ – Batominovski Aug 8 '18 at 20:24
  • $\begingroup$ Okay that makes sense. Thanks $\endgroup$ – Lev Ban Aug 8 '18 at 20:25
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Write $f(s) = \frac{1}{\sqrt[4]{s}}\sqrt[4]{s}f(s)$ and use Holder: $$\left|\int_0^xf(s)ds\right|^2 \leq \int_0^xs^{-1/2}ds\int_0^1s^{1/2}f(s)ds =2\sqrt{x}\int_0^1\sqrt{s}f(s)ds$$

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