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Showing that a family is not normal

Suppose $f : mathbb{D}setminus {0} rightarrow mathbb{C} $ is analytic and $z=0$ is an essential singularity of $f$. Show that the family ${f_n} $ defined by
$$ f_n(z) = f left( frac{z}{2^n} right), quad z in mathbb{D}setminus {0} $$
is not normal in $mathbb{D}setminus {0}$.

My attempt (inspired on this): Assume that ${f_n}$ is a normal family in $mathbb{D}setminus {0}$. So there exists from the sequence $(f_n)_n$ a convergent subsequence $(f_n)_{n_k}$ with limit $hat{f}$. Because $f$ is analytic, it follows that $hat{f}$ is analytic on $mathbb{D}setminus {0}$. We take an annulus $A$ of inner radius $1/3$ and outer radius $1/2$. Then we have that $hat{f}(A)$ is bounded and because $f_{n_k} rightarrow f$ as $k rightarrow infty$, we have that $f_{n_k} = f(frac{A}{2^{n_k}})$ is bounded for $k$ large enough. By the Cauchy formula for Laurent series $$ |a_{-j}| leq r^j max_{|z| = r}{|f|}.$$
We choose $r$ as the radius of the circle contained in $A/n_k$ and letting $krightarrow infty$. We then have to show that $|a_{-j}|=0$ for every $j leq -1$ such that we conclude that $f$ is an analytic function on $mathbb{D}$ which gives us a contradiction.

But in the last part I got stuck. Because if $k rightarrow infty$ then $r rightarrow 0$ and because $f$ has an essential singularity in $z=0$: $max_{|z| = r}{|f|} rightarrow infty $. So we have something of the form $0 cdot infty$ which is not defined.

Someone who can point out where I went wrong and help me out? Thanks!

Mathematics Asked by Kabouter9 on January 1, 2021

1 Answers

One Answer

You have shown that $max_{|z| = r_k}{|f|}$ is uniformly bounded for $$ r_k = frac{1}{2 cdot 2^{n_k}} $$ therefore choosing these radii in $$ |a_{-j}| leq r^j max_{|z| = r}{|f|}. $$ implies that $a_{-j} = 0$ for $j ge 1$.


Alternatively one can argue as follows: $hat f$ is bounded on the circle $|z| = 1/2$ and $f_{n_k} to hat f$ uniformly on that circle. It follows that $$ |f_{n_k}(z)| le M text{ for } |z| = frac 12 $$ with some $M> 0$ and sufficiently large $k$. Translated back to $f$ this means that $$ |f(z)| le M text{ for } |z| = frac{1}{2 cdot 2^{n_k}} , . $$ So $|f|$ is bounded by $M$ on those concentric circles. Using the maximum modulus principle it follows that the same estimate holds in the annuli between those circles: $$ |f(z)| le M text{ for } frac{1}{2 cdot 2^{n_{k+1}}} le |z| le frac{1}{2 cdot 2^{n_k}} , . $$ and that implies that $f$ is bounded in a neighborhood of $z=0$.

Using Riemann's theorem it follows that $f$ has a removable singularity at $z=0$.

Correct answer by Martin R on January 1, 2021

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