If $\,f^{7} $ is holomorphic, then $f$ is also holomorphic.

Question

I need some help with this problem:

Let $ \Omega $ be a complex domain, i.e., a connected and open non-empty subset of $ \mathbb{C} $. If $ f: \Omega \to \mathbb{C} $ is a continuous function and $ f^{7} $ is holomorphic on $ \Omega $, then $ f $ is also holomorphic on $ \Omega $.

Thanks in advance.

Answer 1

Let me provide an elementary proof:

First of all, either $g=f^7$ is identically zero or its roots are isolated. In the first case $f\equiv 0$, and there is nothing to prove. So assume that the roots of $g$ (if any) are isolated.

We need to show that, for every $z_0\in\Omega$, the limit $$ \lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0} \tag{1} $$ exists. But $$ \frac{f(z)-f(z_0)}{z-z_0}=\frac{f^7(z)-f^7(z_0)}{z-z_0}\cdot \frac{f(z)-f(z_0)}{f^7(z)-f^7(z_0)}=\frac{g(z)-g(z_0)}{z-z_0}\cdot\frac{1}{\sum_{k=0}^6 f^k(z)\,f^{6-k}(z_0)}. $$ Let $z_0\in\Omega$, such that $f(z_0)\ne 0$. This is equivalent to $g(z_0)\ne 0$, and as $f$ is continuous at $z_0$ (and non-vanishing in a neighborhood of it), then limit $(1)$ does exists as $$ \lim_{z\to z_0}\frac{g(z)-g(z_0)}{z-z_0}\cdot\frac{1}{\sum_{k=0}^6 f^k(z)\,f^{6-k}(z_0)} =\frac{g'(z_0)}{7f^6(z_0)}. $$ So if $Z=\{z:f(z)=0\}$, then $f\in\mathcal H(\Omega\smallsetminus Z)$, and as $f\in C(\Omega)$, and each point $w$ in $Z$ is isolated, then it is an isolated singularity. But as $f$ is continuous at $w$, then $w$ is a removable singularity of $f$.

Therefore, $f\in\mathcal H(\Omega)$.

Answer 2

We may assume $\Omega$ connected.

Let $g(z)\colon = [f(z)]^7$, a holomorphic function. Let $N$ the set of zeroes of $g$. If $N$ has an accumulation point in $\Omega$ then $g$ is identical $0$ and so is $f$. Otherwise, the set $N$ is discrete ( it may have an accumulation point in $\partial \Omega$).

The set $\Omega \backslash N$ is open and connected. The function $g$ maps $\Omega \backslash N$ to $\mathbb{C}\backslash\{0\}$. Let $p \colon \mathbb{C}\backslash\{0\} \to \mathbb{C}\backslash\{0\}$, $p(z) = z^7$.

$p$ is a holomorphic covering . We have $$p\circ f = g$$ that is, $f$ is a lift of $g$. Moreover, $f$ is continuous. It is a general fact that any continuous lift of a holomorphic map for a holomorphic covering is also holomorphic. We conclude that $f$ is holomorphic on $\Omega \backslash N$. The singularities $N$ are removable because $f$ is continuous on $\Omega$. So $f$ holomorphic on $\Omega$.

Obs: In fact it works with any nonconstant holomorphic function $\phi(z)$ instead of $z^7$. The problem can be treated locally in a similar way.

Answer 3

There is an easy proof if you assume the stronger statement that $f(z)$ is continuously differentiable, rather than just continuous. Indeed, $f(z)$ is holomorphic if and only if $df/d{\overline{z}} = 0$, and since $f(z)^7$ is holomorphic, $df/d{\overline{z}}\times f(z)^6 = 0$. Assume without loss of generality that $f(z)$ is not the zero function, so that it vanishes on a discrete subset $S$ of $\Omega$ (which is also the vanishing locus of the holomorphic function $f(z)^7$). On the complement of $S$ we can divide by $f(z)^6$, so we have $df/d{\overline{z}}=0$ on the complement of $S$. Since $S$ is discrete and $df/d{\overline{z}}$ is continous, then $df/d{\overline{z}}=0$ on all of $\Omega$.

Answer 4

There exist exactly $7$ holomorphic functions $g_0,...g_6$ such that $g_i^7=f^7$ (there are seven holomoprhic $7^{th}$ root functions which we can compose with $f^7$, or $7$ branches if you prefer). We need to show $f$ is equal to one of them. We know that everywhere it takes a value that is equal to one of the $7^{th}$ roots of $f^7$; Hence we can argue by continuity that its values lie on one of the branches only, and we are done.

Answer 5

It's true. If the zeros of f have a limit point, then so is $f^7$, but $f^7$ is holomorphic, so $f^7$ is zero every where. If all zeros of f is isolated, we can apply the Cauchy-Riemann equation and use the chain rule to get $ f^6 \cdot \frac{\partial f}{\partial \bar{z}} = 0$. Thus $f$ is holomorphic at $z$ whenever $ z$ is not a zero of $f$. Since $f$ is continuous and its zeros are isolated, we know from the removable singularity theorem that $f$ is also holomorphic at these zeros.

Answer 6

Let $g(z):=f^7(z)$. If $g(z_0)\ne0$ there is a neighborhood $U$ of $z_0$ and a holomorphic function $h:\>U\to{\mathbb C}$ with $g(z)=e^{h(z)}$ on $U$. We may assume that $U$ is so small that $|h(z)-h(z_0)|<1$ on $U$. Then there are exactly $7$ continuous functions $f_k:\>U\to {\mathbb C}$ with $f^7=g\restriction U$, namely the functions $$f_k(z)=e^{2k\pi i/7}\>e^{h(z)/7}\qquad(0\leq k<7)\ ,$$ and $f\restriction U$ is one of them. Since all $f_k$ are holomorphic so is $f\restriction U$. It follows that $f$ is holomorphic on $\Omega\setminus N$, where $N$ is the zero set of $g$. Since this set consists of isolated points (unless $g(z)\equiv0$), and $f$ is bounded in the neighborhood of each $z\in N$ it follows that $f$ is in fact holomorphic in all of $\Omega$.

Answer 7

This is a topological variant of Christian’s elegant solution.

---

Let $ g \stackrel{\text{df}}{=} f^{7} $. As $ g $ is holomorphic on $ \Omega $, either (i) it is $ 0 $ everywhere on $ \Omega $ or (ii) its set of roots is an isolated subset of $ \Omega $. If Case (i) occurs, then we are done. Hence, suppose that we are in Case (ii).

Let $ \mathcal{Z} $ denote the (closed) set of roots of $ g $. For any point $ p \in \Omega \setminus \mathcal{Z} \neq \varnothing $, there exist

• a (sufficiently small) open disk $ D $ centered at $ p $ and contained in $ \Omega \setminus \mathcal{Z} $, and
• a holomorphic function $ h: D \to \mathbb{C} $ such that $ g|_{D} = e^{h} $.

Now, for each $ z \in D $, we must have $$ f(z) \in \left\{ e^{2 \pi i k / 7} ~ e^{h(z) / 7} ~ \Big| ~ k \in \{ 0,\ldots,6 \} \right\}. $$ This is because the right-hand side is precisely the set of all complex $ 7 $-th roots of $ g(z) $. This means that we have a function $ k: D \to \{ 0,\ldots,6 \} $ such that $$ \forall z \in D: \quad f(z) = e^{2 \pi i \cdot k(z) / 7} ~ e^{h(z) / 7}. $$ We claim that $ k $ is locally constant on $ D $. Assume the contrary. Then there exists a $ q \in D $ such that $ k $ is not constant on any open neighborhood of $ q $. We can thus find a sequence $ (z_{n})_{n \in \mathbb{N}} $ in $ D \setminus \{ q \} $ converging to $ q $ such that $ k(z_{n}) \neq k(q) $ for all $ n \in \mathbb{N} $. By the Pigeonhole Principle, we can extract a subsequence $ (z_{n_{j}})_{j \in \mathbb{N}} $ such that all the $ k(z_{n_{j}}) $’s are equal to a single integer $ m \in \{ 0,\ldots,6 \} $. Then \begin{align} 0 & = \lim_{j \to \infty} \left[ f(q) - f(z_{n_{j}}) \right] \quad (\text{As $ f $ is continuous.}) \\ & = \lim_{j \to \infty} \left[ e^{2 \pi i \cdot k(q) / 7} ~ e^{h(q) / 7} - e^{2 \pi i \cdot k(z_{n_{j}}) / 7} ~ e^{h(z_{n_{j}}) / 7} \right] \\ & = e^{2 \pi i \cdot k(q) / 7} ~ e^{h(q) / 7} - e^{2 \pi i m / 7} ~ e^{h(q) / 7} \quad (\text{As $ h $ is continuous.}) \\ & = \left[ e^{2 \pi i \cdot k(q) / 7} - e^{2 \pi i m / 7} \right] e^{h(q) / 7} \\ & \neq 0. \quad (\text{As $ m \neq k(q) $.}) \end{align} We have a contradiction, so $ k $ must be locally constant on $ D $. As $ D $ is connected, $ k $ must be constant on $ D $. Consequently, $ f $ is a holomorphic function on $ D $. As $ p \in \Omega \setminus \mathcal{Z} $ is arbitrary, $ f $ is holomorphic on $ \Omega \setminus \mathcal{Z} $. (Note: Although holomorphicity has global consequences, it is a local property.) Finally, $ f $ is holomorphic on all of $ \Omega $ as it is continuous at each (isolated) root.