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In the rewritten form it doesn't require any ingenuity at all. For brevity, introduce the notation $$ (Th)(s)=\frac 1\pi\int_0^s\frac{h(s-u)h(u)}{\sqrt{u(s-u)}}\,du=\frac 1\pi\int_0^1\frac{h(s(1-v))h( …

Certainly. All you need is $EX^2=+\infty$. Then the characteristic function $f_X(t)$ satisfies $\lim_{t\to 0}\frac{1-|f(t)|}{t^2}=+\infty$, so for every finite interval $I\subset \mathbb R$, we have $ …

The situation is governed by the following instance of a very general comparison principle (so general that I wouldn't even try to formulate it in full). Let $f:[0,+\infty)\to[0,+\infty)$ be a decreas …

It looks like you care only about the order of magnitude (i.e., an answer up to a constant factor), in which case it is fairly easy. First, ignore all coefficients. Setting them to $1$ just changes th …

I would suspect that non-classical arguments are needed to do so. All you need to know is that $t\mapsto \frac 1{1+t}$ is a decreasing function, so for $0<x\le x'$ we have $\frac{\log(1+x')}{\log(1+x) …

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Here is a sketch. Feel free to ask for clarifications if my writing gets too terse or confusing in places :-). First recall the Bernstein (a.k.a. Hoeffding, Chernov, etc.) bound. If $Y_m$ are mean $0 …

Looks correct. Split $H=H_n\oplus H^n$ where $H_n=span(e_1,\dots,e_n)$. Let $P_n$ and $P^n$ be the corresponding projections. The conditions are $P_nA_kP_n\to P_nAP_n$, $\operatorname{Tr}(P^nA_kP^n)\l …

Take $T_1(y)=y-y^2$ with $y\in[0,1]$ and $T_2(x,y)=e^{iy}x$, $x\in\mathbb C, |x|\le 1$. Now take $x_0=1$, $y_0=1/2$, say. Then all assumptions hold, but $y_n\approx c/n$, so the rotations in the itera …

OK, let's try to get a few terms. We obviously have total mass $1$, so let's get some idea of how it is distributed. Clearly, what we are really interested in is $U(h)=\int_{0}^{1-h}|x^2-h^2|^\gamma$. …

Canceling the innermost operations, you see that the sequence is increasing. On the other hand, the related sequence in which you replace the tower of $3$'s by $10$ times that tower is decreasing for …