My question is that: On $C-\{0\}$ the function is holomorphic and satisfies $|f(z)|\le C|z|^{3/2}$ (or $|f(z)|\le C|z|^{1/2}$). $C$ here is the constant. Show that $f=0$？
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For the case of $3/2$ take note that linear functions fill the bill. – ncmathsadist Jan 02 '13 at 00:25
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1Linear functions do not satisfy the inequality when $|z|<1$. – JSchlather Jan 02 '13 at 00:29
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Please see here for correct solutions to the case of $|z|^{\frac{3}{2}}$. – 23rd Apr 24 '13 at 10:53
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Let $f$ be holomorphic on $\mathbb C \setminus \{0\}$ and satisfy the inequality $|f(z)| \leq C|z|^{1/2}$. First note that the singularity at $0$ is removable and indeed $\lim_{z\rightarrow 0} f(z)=0$ from the inequality. So $f$ extends to an entire function $f$ such that $f(0)=0$. Note that $g(z)=f(z)/z$ is also an entire function. But we also have that
$$|f(z)/z| \leq \frac{C}{|z|^{1/2}},$$
in particular $g$ is bounded by $C$ outside the unit disk and since it is continuous it is bounded inside the unit disk. Thereby $G$ is a constant function and we must have that $g=0$.
JSchlather
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take g=f/z^(3/2), then g has a analytic continuation or has pole. in the neighbourhood of 0, a meromorphic function takes all values in a nhbd. of inifinity.so it cannot be bounded by any constant.but |g| < C by assumption. in the same way other one.
Koushik
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Thank you! So you can show that g is a constant a, $f=az^{1/2}$, but why a must be zero? – Siming HE Jan 02 '13 at 00:46
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i can show that g has a analytic continuation which implies f has a zero of order atleast 2 at zero. – Koushik Jan 02 '13 at 01:21
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You can not conclude that $f(z)/z^{\frac{3}{2}}$ is holomorphic on $\mathbb{C}\setminus{0}$ a priori, because when $f$ is not $0$, $f(z)/z^{\frac{3}{2}}$ is even not well defined on $\mathbb{C}\setminus{0}$. @SimingHE – 23rd Apr 24 '13 at 10:29