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Isn't it just a convolution operator with $h(x) = \chi_{[0, 1]}(x)\frac{1}{\sqrt{x}}$? So, by the Young's convolution inequality, $f*h$ is in $L^p([0, 1])$ for all $p < \infty$ (in particular for $p = …

Yes, there is a bound for this ratio -- it is always between $\frac{1}{2}$ and $1$. Upper bound is obvious since $|X+Y| \le |X| + |Y|$ so let us prove the lower bound. Let $X$ be positive with probab …

This is false: Let the probability distribution be $P(X = 0, Y = 0) = P(X = 1, Y = 1) = \frac{1}{2}$ and let $A = \{ 0\}$, $B = \{ 0, 1\}$. Then the left-hand side of your inequality is $1$ while th …

No, this equality does not hold in general. What is true is that rhs is at most twice as large as lhs and this is sharp. For the sake of brevity I will only show an example where lhs is $1$ and rhs …

Moreover, it is known that Sobolev's and Morrey's inequalities are essentially sharp (in our case, we can not have a uniform estimate with a better modulus of continuity, say). … So informally we are asking if there're inequalities which work for Lipschitz functions and fail for $\frac{1}{2}$-Hölder, and for me it seems natural that it is a real possibility, since the Lipschitz …

Your conjecture is indeed correct. The proof is based on the following simple yet powerful trick I learned many years ago: $|z| = \sup_{|v| = 1} \Re (zv)$. Therefore it is enough to only bound from ab …

First of all such a Csiszar–Kullback-Pinsker inequality or whatever cannot possibly be true since $x^2$ explodes faster than $x\log x$ so you can make a local adjustment so that the right-hand side is …

I will show that $a \ge \sqrt{\pi}$ which I assume is what you wanted to ask (and it complements counterexample of Iosif Pinelis which appeared while I was writing this answer, showing that this is sh …