This looks almost too easy where did I go wrong? Complex Analysis.

[l'Hôpital]

Homework Statement

http://www.math.northwestern.edu/graduate/prelims/AnalysisPrelim2010FallFinalVersion.pdf

Problem 2 of Part III.

Homework Equations

Complex Analysis.

The Attempt at a Solution

So, I think my proof is wrong (since I never used the fact that it was $f^2$) as opposed to $f$. So, could you point out at where?

Since $f$ is meromorphic, it can only have poles as discontinuities. We'll argue by means of contradiction. Let $a$ be a pole of $f$. Since meromorphic functions have isolated zeroes, there exists a small disk with border $\gamma$ around $a$ such that $f$ has no zeroes, nor other singulaties. In particular, this implies $g := \frac{1}{f^2}$ is analytic on said disk. Applying Runge's theorem, we can obtain a sequence of polynomials $g_n$ uniformly converging to $g$. So, $$\int_{\gamma} g_n f^2 dz \rightarrow \int_{\gamma} gf^2 dz = length(\gamma)$$ But by the given,

$$\int_{\gamma} g_n f^2 dz = 0$$ for all $g_n$, which is a contradiction.

Am I just invoking a theorem too powerful?

Homework Statement

Homework Equations

The Attempt at a Solution

 

[mfb]

An integral over a closed loop with a constant function is 0, not the length of the loop. This does not lead to a contradiction.

 

[l'Hôpital]

Oh wow. Shameful. Shameful, shameful, shameful. Been doing too much real stuff lately haha. How about this?

By the integral condition, it suffices to only consider $f$ having simple poles as discontinuities. Wlog, we assume it has a pole at $z = 0$ and choose a contour around it which only has that as the only discontinuity in the region it defines. Thus, by the residue theorem and the integral condition, we have that $\lim_{ z \rightarrow 0} z f(z) = 0$. Thus, there exists some small c such that $|z| \leq c \rightarrow |z f(z)| < 1$. The function $g(z) = z f(z)$ has only 0 as a discontinuity in previously-mentioned region, and it's removable, so it's analytic on the domain defined by the contour. For $|z| < c$ we have by the maximum modulus principle that $|g|$ attains a max value at $|z| = c$, so in particular, $|f(z)| < 1/c$, for $|z| < c$, hence the discontinuity is removable, so not a pole.

 

[mfb]

Thus, by the residue theorem and the integral condition, we have that $\lim_{ z \rightarrow 0} z f(z) = 0$.

I don't understand that reasoning.

Thus, there exists some small c such that $|z| \leq c \rightarrow |z f(z)| < 1$. The function $g(z) = z f(z)$ has only 0 as a discontinuity in previously-mentioned region, and it's removable, so it's analytic on the domain defined by the contour. For $|z| < c$ we have by the maximum modulus principle that $|g|$ attains a max value at $|z| = c$, so in particular, $|f(z)| < 1/c$, for $|z| < c$, hence the discontinuity is removable, so not a pole.

I would simplify this to "no pole of any order could satisfy the lim condition, therefore f has not a pole". But you have to get the lim condition first.

 

[l'Hôpital]

Alright. So, by the integral condition and the residue theorem (in that order), we get the equalities

$$0 = \int zf^2 dz = Res(zf^2,0)$$

Since we are assuming it's a simple pole of $f$, it's an order 2 pole of $f^2$, hence simple pole of $zf^2$, so the residue boils down to the limit condition $\lim_{z \rightarrow z} z(z f^2 (z) ) = 0$ which is equivalent to the stated limit condition before by taking a square root.

We can assume it's a simple pole since if it was any higher, say $m$, we can simply replace $f$ with $z^{m-1} f$, which would then have a simple pole at $z= 0$ , then show it's not really a pole by the argument I produced earlier which means it couldn't have been a pole of said order for $f$.

I tried to reduce to the simple pole case since for higher order, the residue involves taking derivatives, and that looked scary.

 

[mfb]

We can assume it's a simple pole since if it was any higher, say $m$, we can simply replace $f$ with $z^{m-1} f$, which would then have a simple pole at $z= 0$ , then show it's not really a pole by the argument I produced earlier which means it couldn't have been a pole of said order for $f$.

Better start with modified polynomials directly, otherwise your assumption is not valid - your other cases do not look like the case you considered.