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I've been reading Imayoshi & Taniguchi's text "An Introduction to Teichmuller Spaces" and while going through the proof of Lemma 4.20 realized that I'm not comfortable integrating functions over the complex plane (i.e. over an area form, not a contour integral.)

In the proof they state that for $1<p<2,$ the function $$\frac{\zeta}{z(z-\zeta)}\in L^p(\mathbb{C})$$ where $z$ is our variable and $\zeta=a+ib$ is a fixed complex number. I want to verify this claim, my attempt is as follows: $$\int \int_{\mathbb{C}}\left\lvert \frac{\zeta}{z(z-\zeta)} \right\rvert^p dxdy =\lvert\zeta\rvert^p \int\int_{\mathbb{C}} \frac{dxdy}{\lvert x+iy\rvert^p \lvert x+iy-(a+ib)\rvert^p}$$ $$=\lvert\zeta\rvert^p\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \frac{dxdy}{(x^2+y^2)^{p/2} ((x-a)^2+(y-b)^2)^{p/2}}.$$

My puestions are: 1) Is what I've written above correct so far? 2) Any ideas on how to proceed or a better way to justify that this function is in fact in $L^p(\mathbb{C})$?

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  • $\begingroup$ You have $p$ in the title, but then it disappears yet $q$ appears, but then you talk about $p$ again? Please clear this up. $\endgroup$ – zhw. Jan 19 '18 at 18:28
  • $\begingroup$ I edited the sentence some and just used $p$ in the question title because of the convention to refer to these spaces as "L-p" spaces. $\endgroup$ – ThePiper Jan 19 '18 at 18:43
  • $\begingroup$ Then it should be $p$ the whole way through, the other index should be dropped as it adds nothing. $\endgroup$ – zhw. Jan 19 '18 at 19:07
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Really, if all you're doing is integrating the modulus of a function, then you just need to check an integral over $\mathbb R^2$ written in a funny way. First prove that $1/z \in L^q_{loc}$ for $q<2$. This follows from the fact that $1/|x|\in L^q_{loc}(\mathbb R^n)$ for $q<n$, which can be seen by using radial coordinates around $0$. This implies that $\frac1{z(z-\zeta)} \in L^q_{loc}$. To finish, notice that outside of a ball $B$ so large that it contains both $0$ and $\zeta$, you should be able to prove you have sufficient decay at infinity for the function to be integrable.

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  • $\begingroup$ Thanks sifuuuuuu $\endgroup$ – Chee Han Jan 20 '18 at 15:32
  • $\begingroup$ +1 And sorry, I edited the question to use $L^p$ as in my comment $\endgroup$ – zhw. Jan 20 '18 at 21:16
  • $\begingroup$ @zhw. I did see actually! but it was too late and i didnt want to bump this question further by editing my answer $\endgroup$ – Calvin Khor Jan 20 '18 at 21:58

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