Real Analysis: Hardy Littlewood

nateHI
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Homework Statement


Establish the Inequality ##f^*(x)\ge \frac{c}{|x|ln\frac{1}{x}}## for
##f(x)=\frac{1}{|x|(ln\frac{1}{x})^2}## if ##|x|\le 1/2## and 0 otherwise

Homework Equations


##f^*(x)=\sup_{x\in B} \frac{1}{m(B)} \int_B|f(y)|dy \quad x\in \mathbb{R}^d##

The Attempt at a Solution


Disregard, I figured it out.
 
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What was the sticking point that you overcame?
 
I was stuck on the first step. I was able to work in reverse from the solution but felt like I was missing a key idea doing it that way. Namely,

##\sup_{x\in B} \frac{1}{m(B)} \int_B \frac{1}{|x|(ln\frac{1}{x})^2}\ge \frac{1}{2|x|}\int_{-|x|}^{|x|} \frac{1}{|x|(ln\frac{1}{x})^2}##

from there you just work out the integral.
 
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Likes jedishrfu
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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