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Nov 25, 2022 at 16:00 history edited Alexandre Eremenko CC BY-SA 4.0
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Nov 11, 2022 at 1:09 comment added Alexandre Eremenko Such function $G$ has a Cauchy integral representation, and its real part is represented by Poisson formula. Then one see by a direct computation that $A$ is the jump of derivative of $G$ at $0$.
Nov 11, 2022 at 1:08 comment added Alexandre Eremenko @Giorgio Metafune: Thanks for pointing the misprint, I corrected. My method of deriving the formula for $A$ was to introduce one half of the Fourier transform $F(z)=\int_0^\infty e^{-izt}f(t)dt$. It is analytic in the lower half-plane. Then, since $f$ is even $\hat{f}(x)=F(x)+F(-x)$ by a change of the variable in the integral. So if we define $G(z)=F(z)$ in the lower half-plane and $G(z)=-F(-z)$ in the upper half-plane, then the resulting function $G$ will be analytic in $C\backslash[-1,1]$ tend to $0$ at infinity and on $(-1,1)$ it will have a jump $\hat{f}$.
Nov 11, 2022 at 1:02 history edited Alexandre Eremenko CC BY-SA 4.0
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Nov 10, 2022 at 15:59 comment added Giorgio Metafune It is a very nice way to obtain to obtain this kind of estimates. I see that the formula for A is true by solving the Dirichlet problem in the half plane with boundary data $\hat f$, using the Poisson kernel and the Fourier transform. I wonder if there is a simpler way. By the way, you forgot a "hat before $f$, in Cramesr's estimate.
Nov 9, 2022 at 23:50 history edited Alexandre Eremenko CC BY-SA 4.0
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Nov 9, 2022 at 23:35 history edited Alexandre Eremenko CC BY-SA 4.0
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Nov 9, 2022 at 23:35 comment added Alexandre Eremenko @Giorgio Metafune: Thanks! I corrected.
Nov 9, 2022 at 23:13 comment added Giorgio Metafune A guess that in the formula for $A$ a minus sign is missing before the limit.
Nov 9, 2022 at 19:03 history answered Alexandre Eremenko CC BY-SA 4.0