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I am given that $$\vert f \vert\le A+B\vert z \vert^{3/2}$$ and I would like to show that $f$ is a linear polynomial. A generalization of Liouville's Theorem says that if $\vert f\vert\le C\vert z \vert^k$ for some positive integer $k$, then $f$ is a polynomial of degree at most $k$. Since there is a power of $3/2$, this fact doesn't immediately apply. So we can write $\vert fz^{1/2}\vert\le D\vert z^2\vert$ for $\vert z \vert>R$ for some sufficiently large $R$.

Hence $fz^{1/2}$ is a polynomial of degree at most $2$. Say, $fz^{1/2}=a+bz+cz^2$ ($c$ may be $0$). Then $f=az^{-1/2}+bz^{1/2}+cz^{3/2}$, which is not a linear polynomial for two reasons (it's not a polynomial in $z$ and it's not linear in $z$). Where am I going wrong?

The Substitute
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2 Answers2

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Hint: You don't quite have the hypothesis right for the "generalization of Liouville's Theorem". First show $f$ is a polynomial of degree at most $2$. Then show the coefficient of $z^2$ is $0$.

mrf
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Robert Israel
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You also could just reprove the lemma you are using in a more general version. You could use Cauchy’s formula. \begin{align} |f^{(2)} (z)| &= \Big| \frac{2!}{2πi} \int_{rγ} \frac{f(w)}{(w - z)^3} dw \Big| \\ &≤ C·r·\max \Big\{\frac{A}{|w - z|^3} + \frac{B·|w|^{3/2}}{|w - z|^3}; w ∈ rγ\Big\}\\ &≤ \frac{C'}{\sqrt r}, \end{align} where $γ\colon [0..1] → ℂ~,t ↦ \mathrm{e}^{it}$. As $r$ is arbitrary, $f^{(2)} = 0$, so …

Edit: Fixed a mistake. The last inequality is actually a little bit more work.

k.stm
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