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Thread: Proofs about an analytic function

  1. #1
    MHF Contributor arbolis's Avatar
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    Proofs about an analytic function

    Let $\displaystyle f(z)$ be an analytic function in the open disk $\displaystyle D(0,R)$ which satisfies $\displaystyle |f(z)| \leq |z|^n$ for some $\displaystyle n \in \mathbb{N}^*$ and for all $\displaystyle z \in D(0,R)$.
    1)Prove that $\displaystyle f$ has a zero in $\displaystyle z=0$ or order $\displaystyle \geq n$.
    2)Prove that $\displaystyle \frac{f(z)}{z^n}$ has a removable singularity at $\displaystyle z=0$.
    3)Assume that $\displaystyle f$ is entire and that $\displaystyle |f(z)| \leq |z|^n$ for some $\displaystyle n \in \mathbb{N}^*$ and for all $\displaystyle z \in \mathbb{C}$. Show that $\displaystyle f(z)=c z^n$ with $\displaystyle c \in \mathbb{C}$.
    Attempt:1) Let $\displaystyle g(z)=|z|^n$. Then we have $\displaystyle g(0)=0$. Since $\displaystyle |f(z)| \leq g(z)$, it follows that $\displaystyle f(0)=0$. Does this also mean that $\displaystyle |f^{(n)}(z)|\leq n! |z|$? If so, then I think I've proven part 1).
    Part 2): I think they want me to prove that $\displaystyle \lim _{z \to 0} z\frac{f(z)}{z^n}=l$ where $\displaystyle l \in \mathbb{C}$.
    Using the fact that $\displaystyle \frac{|f(z)|}{|z|^n} \leq 1$, I get that $\displaystyle -z \leq \frac{f(z)}{z^{n-1}} \leq z$ which tend to 0 as z tends to 0 because of the sandwich theorem. Proved. (I don't think my proof is valid).
    3)I'm not sure.
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  2. #2
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    For the first one argue with the series expansion at 0. Could $\displaystyle g(z)=\frac{f(z)}{z^n}$ be bounded around 0 if there was a coefficient $\displaystyle a_k\neq 0$ with $\displaystyle 0\leq k\leq n$ in the Taylor series for f?

    For the second one, if g is bounded near 0, could it have a pole or essential singularity there?

    For the third apply Liouville to g.
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  3. #3
    MHF Contributor arbolis's Avatar
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    I'm still clueless with 1) and 3).
    With 2), $\displaystyle -1 \leq g(z) \leq 1$. I just saw a theorem in my class notes which states that if g is analytic in $\displaystyle D(0,R)-z_0$, then $\displaystyle z_0$ is a pole of g if and only if $\displaystyle \lim _{z \to z_0} f(z)=\infty$. So clearly there is no pole at z=0 for g. I must still show that there isn't an essential singularity. Writing $\displaystyle f(z)=\sum _{m=0}^{+\infty} a_m z^m$ since it's analytic, it's obvious that dividing it by $\displaystyle |z|^n$ won't make appear an infinity of negative powers of z. Therefore since g(0) isn't defined, it can only be a removable singularity.
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  4. #4
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    Quote Originally Posted by arbolis View Post
    With 2), $\displaystyle -1 \leq g(z) \leq 1$.
    Be careful, you made the same mistake in your first post: We are dealing with complex numbers, which aren't ordered, so the inequality above is meaningless.

    For the first one note that $\displaystyle |g(z)| \leq 1$ and apply the same reasoning as in (2) and use the fact that $\displaystyle f(z)$ has a zero of order n at 0 iff $\displaystyle f(z)=z^nh(z)$ where h is analytic and $\displaystyle h(0)\neq 0$.

    For the third one, using the second one you get that $\displaystyle g(z)$ has an entire extension whose modulus is bounded by 1
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