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I think it is easier if you rewrite your inequality as \begin{align} & P( \bigcap_{i=1}^{s-1}A_i ) - P ( \bigcap_{i=1}^{s-1}A_i \cap A_s) \geq P(\bigcap_{i=1}^{s-1}A_i)(1-P(A_s)) \\ & \iff P(A_s …

In the paper "Hilbert's inequality", Montgomery and Vaughan proved that a generalization of the discrete Hilbert transform is bounded in $\ell^2$. The inequality reads as follows $$ \Big| \sum_{k\neq …

Suppose you have a power series with coefficients $a_n$ $$ f(z):= \sum_{k=1}^\infty a_k z^k .$$ Then the coefficients of $f^2$ are exactly $c_n$. Also if we denote by $\odot$ the Hadamard multiplicati …

You can probably prove that $$ \Big( \int_\mathbb{A_r} |f(z)|^2 \frac{dxdy}{\pi(1-r^2)} \Big)^{1/2} \leq \int_{ \mathbb{T}} |f(e^{i\theta})| \frac{d\theta}{2\pi}+\int_{ \mathbb{T_r}} |f(re^{i\theta}) …

So first of all, what is claimed in Love's paper is slightly different, It says that for a sufficient large choice of the parameter $c>0$ the function $$ g(x): = \sum_{n=0}^\infty c^{-n/p}\cos(c^n \pi …

It should be true for $p>1$. For a function $\varphi$ in the Schwartz class it holds that \begin{equation} D_1D_2 \varphi(x) = -R_1 R_2 \Delta \varphi(x), \end{equation} where $R_1, R_2$ are the Riesz …

I think similar questions always translate in the $L^p$ boundedness of a Fourier multiplier. In this case you want a Fourier multiplier which "exchanges the operator $D_1D_2D_3$ with the operator $D_1 …