Conplex/real Integration and poles of function So I am working on the following problem: 
Let $\Delta $ be the unit disk centered at origin, and $f$ is holomorphic on $\Delta-\{0\}$. If $$\int_\Delta|f|dxdy<\infty$$ show that $f$ has at most a degree 1 pole at 0.
I start with writting the integral in polar coordinate: $$\int_0^1\int_0^{2\pi} r|f(re^{i\theta})|d\theta dr$$
The inner integral is finite for various value of $r$. So the integral $$\int_{|z|=1}fdz\leq \int_{|z|=1}|f||dz|<\infty$$ So from here where do I go, my goal is show that $zf(z)$ is a holormophic function on $\Delta$. I thought of the function $$F(z)=\int_{|\xi|=1}\frac{\xi f(\xi)}{\xi-z}d\xi"="2\pi izf(z) $$
I put a quotation mark because I want some this equality to come out. This equality does not follow unless $F(z)$ is holormophic and finite. But how can we show that based on what we have?
Thanks
 A: From the previous post we know that $$|\int_{|z|=1}fdz|\leq \int_{|z|=1}|f||dz|$$ The right hand is exactly the integral $$\int_{|z|=1}|f|\cdot |z|dz=\int_0^{2\pi} r|f(re^{i\theta}|d\theta$$, where $r=1$. So $\int_{|z|=1}zf(z)$ is finite. Assume the integral is zero, if not u can subtract the constant on both side. By Morera Thorem, $zf(z)$ is holmorphic on $\Delta$. 
A: Suppose $f(z)$ has a pole of order $n \ge 2$ at the origin.  Then is it not the case that
$g(z) = z^n f(z) \tag{1}$
for some $g(z) \in H(\Delta)$?  It is indeed; see this widipedia entry. Furthermore, $g(0) \ne 0$, whence $\vert g(0) \vert > 0$.  Thus there is a real $\gamma > 0$ with $\vert g(0) \vert > \gamma$, and since $g(z)$, hence $\vert g(z) \vert: \Delta \to \Bbb R$ is continuous,  there exists real $\rho \in (0, 1)$, such that $\vert g(z) \vert > \gamma$ for $z \in \bar D (0, \rho)$, where $D(0, \rho)$ is the disk or radius $\rho$ centered at $0$, and $\bar D(0, \rho)$ is its closure.  In $\bar D(0, \rho) \setminus \{0\}$ we have, from (1),
$f(z) = z^{-n}g(z); \tag{2}$
now writing $z = re^{i \theta}$ so that $z^n = r^n e^{i n \theta}$ in $\Delta$ and $z^{-n} = r^{-n}e^{-i n \theta}$ in $\Delta \setminus \{0\}$, we see from (2) that we may write
$f(z) = r^{-n}e^{-in \theta}g(z) \tag{3}$
in $\bar D(0, \rho) \setminus \{0\}$, whence
$\vert f(z) \vert^2 = f(z) \bar f(z) = r^{-2n} g(z) \bar g(z) = r^{-2n} \vert g(z) \vert^2 \tag{4}$
or
$\vert f(z) \vert = r^{-n} \vert g(z) \vert, \tag{5}$
also holding in $\bar D(0, \rho) \setminus \{0\}$; and since $\vert g(z) \vert > \gamma$ on $\bar D(0, \rho)$, we see that
$\vert f(z) \vert > \gamma r^{-n} \tag{6}$
in $\bar D(0, \rho) \setminus \{0\}$.
Having prepared the estimate (6) for $f(z)$ on $\bar D(0, \rho) \setminus \{0\}$, we may proceed to evaluate the integral
$\int_{\bar D(0, \rho) \setminus \{0\}} \vert f(z) \vert dx dy \tag{7}$
by the usual means:  viz., we begin by picking real $\epsilon$ such that $0 < \epsilon < \rho$ and evaluate the integral over the annular region $A = \bar D(0, \rho) \setminus D(0, \epsilon)$; switching to polar coordinates, we have
$\int_A \vert f(z) \vert dx dy = \int_A \vert f(z) \vert r dr d\theta = \int_0^{2\pi} \int_\epsilon^\rho \vert f(z) \vert r dr d\theta > \gamma \int_0^{2\pi} \int_\epsilon^\rho r^{-n} r dr d\theta, \tag{8}$
where use has been made of (6).  Continuing in this direction,
$\gamma \int_0^{2\pi} \int_\epsilon^\rho r^{-n} r dr d\theta = 2 \pi \gamma \int_\epsilon^\rho r^{1 - n} dr; \tag{9}$
if $n = 2$, then we obtain
$\int_A \vert f(z) \vert dx dy > 2\pi \gamma \int_\epsilon^\rho r^{-1} dr = 2\pi \gamma (\ln r \mid_\epsilon^\rho = 2 \pi \gamma \ln(\dfrac{\rho}{\epsilon}), \tag{10}$
whereas if $n > 2$ we have
$\int_A \vert f(z) \vert dx dy > 2\pi \gamma \int_\epsilon^\rho r^{1 - n} dr = 2\pi \gamma (\dfrac{1}{2 - n} r^{2 - n}\mid_\epsilon^\rho$
$= 2 \pi \gamma \dfrac{1}{2 - n}(\rho^{2 - n} - \epsilon^{2 - n}); \tag{11}$
in either case we see that
$\int_A \vert f(z) \vert dx dy \to \infty \; \; \text{as} \;\; \epsilon \to 0. \tag{12}$
This shows that, with the conventional means of evaluation, $\int_{\bar D(0, \rho)} \vert f(z) \vert dx dy$ is unbounded, but since
$\int_\Delta \vert f(z) \vert dx dy \ge \int_{\bar D(0, \rho)} \vert f(z) \vert dx dy, \tag{13}$
$\int_\Delta \vert f(z) \vert dx dy$ is unbounded as well; thus
$\int_\Delta \vert f(z) \vert dx dy < \infty \tag{14}$
implies we must have $n < 2$; the pole of $f(z)$ at $0$ must be of order at most $1$.  QED.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
A: Assume that 
$$
f(z)=\sum_{n=-\infty}^\infty a_nz^n.
$$
If $a_{k}\ne 0$, for some $k>1$, then
$$
a_{-k}=\frac{1}{2\pi i}\int_{\lvert z\rvert=r}z^{k-1}f(z)\,dz,
$$
and hence
$$
\lvert a_{-k}\rvert\le  r^{k}
\int_{0}^{2\pi}\lvert\, f(r\mathrm{e}^{i\vartheta})\rvert\,d\vartheta 
$$
and thus
$$
\frac{\lvert a_{-k}\rvert}{r^k}\le
\int_{0}^{2\pi}\lvert\, f(r\mathrm{e}^{i\vartheta})\rvert\,d\vartheta.
$$
But
$$
\int_{\Delta}\lvert\,f\rvert\,dx\,dy=\int_0^{2\pi}\left(\int_0^1 \lvert f(r\mathrm{e}^{i\vartheta})\rvert \,r\,dr\right)d\vartheta\ge
\int_0^{1}
\frac{\lvert a_{-k}\rvert\,dr}{r^{k-1}}=\infty,
$$
as $\int_0^1 r^{-s}\,dr=\infty$ iff $s\ge-1$
