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Let $f(z)$ be an entire function, $R_n$ a sequence of positive real numbers tending to $\infty$ such that $f(z) \neq 0$ on $|z|=R_n$ and there exists $M>0$ such that $$\int_{|z|=R_n} \left|\frac{f'(z)}{f(z)}\right| ~dz<M$$ for all $n$. Show that $f$ is a polynomial.

What came to my mind is to consider that $f(z)=a_0+a_1z+\cdots \;\;\forall z\in\mathbb{C}$, and to try proving that $a_k=0$ from a certain $k$, maybe using the Cauchy formula for these coefficients, but I can't use the hypotesis on that bounded integral. Is observing that there is a logarithmic derivative of any use?

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An idea: try to show that $f^{-1}(a)$ is finite for every $a\in\mathbb C$, using the argument principle. –  user31373 Jul 5 '12 at 21:55
    
So indicating with $N$ the number of zeros inside $|z|<R_n$ by the argument principle I have $2\pi i N=\int_{|z|=R_n}\frac{f'}{f}$ which can be estimated by its modulus and eventually by M, and taking the limit i have that there are at most M zeros. If i wanted to do something analogous with $f(z)-a$ i need an estimate like $|f(z)-a|>|f(z)|$ on $|z|=R_n$.. –  balestrav Jul 5 '12 at 22:16
    
Right, we are stuck with $a=0$. Luckily, a better idea arrived meanwhile. –  user31373 Jul 5 '12 at 22:22

1 Answer 1

up vote 7 down vote accepted

As $\,\displaystyle{\frac{1}{2\pi i}\oint \frac{f'}{f}\,dz}\,$ is the number of roots inside the circle, the total number of roots is finite

(bounded by $\,\frac{M}{2\pi}\,$). Let $$g(z)=\frac{f(z)}{(z-a_1)\cdot\ldots\cdot(z-a_k)}$$, where $a_1,\dots,a_k$ are the roots of $f$. Then $g$ satisfies the same condition as $f$ (with a different $\tilde M$). As $g$ has no root, it is of the form $g(z)=\exp(h(z))$ for some entire function $h$. We thus have $\,\displaystyle{\oint |h'(z)|\,|dz|<\tilde M}\,$ for circles of radii $R_n\to\infty$. This implies $h'=0$, hence $f$ is a polynomial.

edit: why $h'=0$: if $$h'(z)=c_1 z^m+c_2 z^{m+1}+\dots$$ ($c_1\neq0$) then $$\oint \frac{h'(z)}{z^{m+1}}\,dz=2\pi i c_1$$, which certainly implies $\oint |h'(z)|\,|dz|\to\infty$.

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