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Suppose $A \subset \mathbb{C}$ is a compact set and $ f_n: A \rightarrow \mathbb{C}$ is a sequence of continuous functions uniformly converging to $f$.

Suppose $f(z_0) = 0$ and $f(z) \ne 0$ when $z \ne z_0$. Observe that $f$ is continuous but not necessarily holomorphic in any open subset of $A$.

We define a sequence of functions $g_n: A \rightarrow \mathbb{C} $ given by $g_n(z) = |f_n(z)|^{1/n}$. With a limiting function being $g$.

What can we say - if anything - about the indeterminate $g(z_0)$.

My initial thoughts:

I thought $g(z) = 1 {z \in A, z \ne z_0}$ should be constant and therefore holomorphic and bounded in open punctured disk around $z_0$ and therefore we should be able to define $g(z_0) =1$ by Riemann’s Removable Singularity Theorem.

Then I wondered what would be the difference if we have $g_n(z) = f_n(z)^{1/\log(n)}$, $g_n(z) = f_n(z)^{1/\sqrt{n}}$, $g_n(z) = f_n(z)^{1/n^2}$, etc.

Is there a theorem that addresses this already?

Thanks in advance.

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  • $\begingroup$ "Observe that $f$ is continuous"... Obviously it was given that the $f_n$ are continuous, though you didn't include that here. And $f_n^{1/n}$ is undefined in $\Bbb R$ if $f_n(z) < 0$ and $n$ is even, so I assume it was also given that $f_n(z) \ge 0$. Again, not mentioned. Is there anything else you left out? $\endgroup$ Commented Feb 12, 2020 at 3:32
  • $\begingroup$ And you use the term "holomorphic" when talking about $f$ and $g$. But these are real-valued functions, and "holomorphic" is a term defined for functions from $\Bbb C \to \Bbb C$. Real-valued functions are never holomorphic. You are correct that $$g(z) = \begin{cases}0, &z = z_0\\1,&z \ne z_0\end{cases}$$ The result will be the same for $g_n = f_n^{a_n}$ for any sequence $a_n$ converging to $0$ (except for $g(z_0)$ being undefined if $a_n$ has an infinite number of negative values). $\endgroup$ Commented Feb 12, 2020 at 3:46
  • $\begingroup$ Correct that to $g(z_0)$ being indeterminant. I was thinking of all the $f_n$ being $0$ there, but that isn't the case. $\endgroup$ Commented Feb 12, 2020 at 4:03

1 Answer 1

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Suppose $\{a_n\}, \{b_n\}$ are sequences of real numbers with $a_n \to a$ and $b_n \to 0$ as $n \to 0$. Further suppose that all $a_n > 0$ and $a > 0$. Then

$$\lim_n a_n^{b_n} = 1$$

Because $a > 0$, there exists $N$ such that for $n \ge N, |a_n - a| < a/2$. Let $$0 < m < \min\left\{\frac a2, \frac 2{3a}, a_n, \frac 1{a_n}\mid n < N\right\}$$ and $$M > \max\left\{\frac {3a}2, \frac 2a, a_n, \frac 1{a_n}\mid n < N\right\}$$

Then we have that $m < a_n < M$ and also $m < 1/a_n < M$ for all $n$. Therefore $$m^{|b_n|} < a_n^{b_n} < M^{|b_n|}$$

But because $b_n \to 0$, so does $|b_n|$ and therefore by continuity of the exponential function, $$\lim_n m^{|b_n|} = \lim_n M^{|b_n|} = 1$$

By the squeeze theorem, $a_n^{b_n} \to 1$ also.

The condition $a_n > 0$ can be dropped, provided $b_n \ge 0$ or one doesn't mind a finite number of undefined values in the sequence $a_n^{b_n}$ (finite because $a > 0$ means that eventually $a_n > 0$). However, the condition $a > 0$ is required.

So your conditions about $f_n$ converging uniformly and compactness of the domain - or even the domain being in $\Bbb C$ do not matter. By simple pointwise convergence, if $g_n = f_n^{b_n}$, then $g(z) = 1$ for all $z$ with $f(z) > 0$.

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  • $\begingroup$ Thank you Paul. I’ve corrected the question based after reading your comments. I will need some time to read through the answer and see if I can think of any counter examples. I’ll accept the answer after i think it through. $\endgroup$
    – Shree
    Commented Feb 12, 2020 at 5:49
  • $\begingroup$ With the corrections to the question, do you think $g(z_0)$ can be assigned value 1? Or to be more accurate, can we define $G$ such that $G(z) = g(z)$ when $z\ne z_0$ and $G(z_0) = 1$? $\endgroup$
    – Shree
    Commented Feb 12, 2020 at 5:54
  • $\begingroup$ The main purpose of the exercise is to find conditions needed to get to a reasonable value of $g(z_0)$. Or rather figuring out what values it will take under various circumstances. $\endgroup$
    – Shree
    Commented Feb 12, 2020 at 6:31
  • $\begingroup$ Your changes introduces a problem: $z \mapsto z^{b_n}$ is not single-valued in $\Bbb C$, so you have to choose a branch to use before you can discuss $f_n^{b_n}$. However, unless you are cutting through $1$, It will still be the case that unless $f(z) = 0$ the limit will still be $1$. My answer for real-valued sequences still shows that the norm of $g$ is $1$. And the argument of $f_n^{b_n}$ will decrease to $0$ as $b_n \to 0$. So, yes, you can define $G$ to be $1$ everywhere and have it agree with $g$ except possibly at $z = z_0$ where the limit depends on the exact behavior of the $f_n$. $\endgroup$ Commented Feb 12, 2020 at 15:43
  • $\begingroup$ If we take the primary branch defined as positive real number to fix the multi-value issue and add a uniform continuity requirement for $g(z)$ (eg by verifying the derivative with respect to $z$ is bounded, actually it would probably have to be 0) will that prove beyond doubt that $g(z_0)$ can be assigned value 0? $\endgroup$
    – Shree
    Commented Feb 12, 2020 at 23:44

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