9

I wish to prove that if $f: \mathbb{C} \to \mathbb{C}$ is analytic and not of the form $az+b,$ then $z, f(z), f(f(z)), \dots$ are linearly independent functions over $\mathbb{C}.$

The cases $n=0, 1$ are trivial. To solve this problem, I'm starting on the first non-trivial scenario: assume $c_1 z + c_2 f(z) + c_3 f(f(z)) = 0$ for some $c_1, c_2, c_3.$ If I can solve this, I'll probably know how to solve the general case. But I'm getting nowhere.

The equations $z-f(z)-f(f(z))+f^3(z)=g(z)-g(g(z)) = 0$ for $f(z) = z+1, g(z) = |z|$ suggest we should impose analycity and non-linearity. Any hints or ideas?

Edit: I've obtained a major breakthrough, but one step is still missing. Define a function to be $n$-independent if $z, f(z), f(f(z)), \dots, f^{(n)}(z)$ are linearly independent over $\mathbb{C}$ on some non-empty open subset of $\mathbb{C}.$

Unproven Lemma: The set of $n$-dependent functions forms a vector space under addition.

$f$ being $n$-dependent easily implies $cf$ is $n$-dependent, so additivity is the only thing that stands in the way of proving this lemma. We need to figure out some way to handle terms like $(f+g) \circ (f+g) = f(f(z)+g(z))+g(f(z)+g(z))$ without messing up the rest of the terms.

First note that $h(z)=az+b$ is $2$-dependent since $z, h(z), h(h(z))$ are $3$ vectors in the $2$ dimensional vector space of polynomials with degree $\le 1.$

Suppose $f$ is analytic and non-linear. Let $n$ be minimal such that $f$ is $n$-dependent. Let $g(z) = f(z)-(f(1)-f(0))z-f(0).$ Since $f$ isn't linear, $n \ge 2,$ implying $g$ is $n$-dependepnt. Suppose $c_0 z + c_1 g(z) + \dots + c_n g^n(z) = 0.$ Setting $z=1,$ we get $c_0 = 0,$ so $c_1 z + \dots + c_n g^{n-1}(z) = 0$ on $\mathcal{O} = g(\mathbb{C}),$ which is open by the open mapping theorem since $g$ is analytic and non-constant.

Thus, $g$ is $n-1$ dependent. If $n \ge 3,$ this means $f$ is $n-1$ dependent, contradiction. If $n=2,$ then $g$ is $1$-dependent, so $g(z)=cz$ for some $c,$ which means $f$ is linear, contradiction.

Display name
  • 5,144
  • As the question is written, your claim is false. If you take $f(z)=|z|$, in fact, you obtain a counterexample, since $f(f(z))=f(z)$. As a side note, the fact that $f(\mathbb{C})$ is open does not follow from the continuity of $f$. e.g.: $f(z)=|z|;f(\mathbb{C})=\mathbb{R}$. Maybe you should add some hypotesis, like requiring that $f$ is holomorphic? –  Apr 22 '20 at 06:07
  • @Caffeine The problem I was trying to solve (I did not come up with it) was indeed false, and I tried to fix it by changing the hypotheses. For the record, the original, also false condition was $f$ is continuous and $f(z)/z$ is non-constant. But I'm not sure what to change them to. Making $f$ is holomorphic sounds like a good suggestion. – Display name Apr 22 '20 at 06:33
  • Your notation is confusing. Do you mean to say that $\operatorname{id}$, $f$, $f\circ f$, .. are linearly independent functions? – Hyperplane Apr 22 '20 at 09:54
  • @Hyperplane Yes. Which section was confusing? – Display name Apr 22 '20 at 09:59
  • Saying $z, f(z), f(f(z)), \ldots $ are linearly independent is a claim about complex numbers, whereas saying $f^{(0)}, f^{(1)}, f^{(2)}, \ldots$ are linearly independent is a claim about functions. – Hyperplane Apr 22 '20 at 10:02
  • 1
    Alright. Perhaps I should've said $\forall z$ then, or said "linearly independent functions" instead of just "linearly independent. – Display name Apr 22 '20 at 10:03
  • 1
    A special case would be: is there an analytic non-linear $f$ so that $f(f(z)) = z$? Now $f(z)=1/z$ is analytic except at one point. So we cannot relax the domain restriction: it must be all of $\mathbb C$. – GEdgar Apr 22 '20 at 13:58
  • By the way this question reminds me a bit of one of my own questions (cf. A possible converse to the Cayley-Hamilton theorem? and Can the conjugate $\phi^{-1} \circ A \circ \phi$ be linear for nonlinear $\phi$?), it seems to generally go into a similar direction: In light of the Cayley-Hamilton Theorem, what can we expect of a function which satisfies a polynomial functional equation? When is it necessarily linear? – Hyperplane Apr 22 '20 at 16:55
  • It should be noted that this is also false if one replaces $\mathbb C$ with $\mathbb C^n$, since $\operatorname{Aut}(\mathbb C)$ contains only linear functions, whereas $\operatorname{Aut}(\mathbb C^n)$ contains non-linear ones as well, such as shears: $(x,y) \mapsto (x+f(y), y)$. Then, due to Cayley-Hamilton, $f^{(k)}$ are not linearly independent if $f$ is linear, and we can compose $\phi^{-1}\circ f\circ \phi$ with some $\phi\in\operatorname{Aut}(\mathbb C^n)$ to get a non-linear solution. – Hyperplane Apr 26 '20 at 11:39

1 Answers1

5

Write $f^n(z)$ be the $n$-th iterate of $f$, also define $f^0(z) = z$.

Let $U\subset \mathbb{C}$ be an open connected set, $f:U\to U$ analytic. If $f$ is nonconstant and not injective, then $\{z,f,\cdots,f^n\}$ are linearly independent over $\mathbb{C}$.

Proof: Use induction on $n$. If $n=1$, and $a_0z+a_1f(z) = 0$ for all $z\in U$ with $a_i\in \mathbb{C}$. Since $f$ is not injective, there exists distinct $z_1,z_2\in U$, such that $f(z_1)=f(z_2)=c$, so $a_0z_i+a_1c=0 \implies a_0(z_1-z_2)=0$, so $a_0=0$, hence $a_1=0$ also.

Now if $\sum_{i=0}^n a_i f^i(z) = 0$ for all $z\in U$. Using $z_1,z_2$ obtained above, one can similarly show $a_0 = 0$. Let $V=f(U)$, then $$\sum_{i=0}^{n-1} a_{i+1} f^i(z) = 0 \qquad \forall z\in V$$ Since $V\subset U$ is open (open mapping theorem), the above equality holds for all $z\in U$ by analytic continuation. Induction hypothesis then shows all $a_i=0$. QED

Regarding your question, take $U=\mathbb{C}$. If we can show any injective entire function must be linear, then we're done. This is a consequence of Weierstrass-Casorati.

pisco
  • 18,983
  • I think the condition $f$ is nonzero should be replaced by $f$ is nonconstant. – Batominovski Apr 23 '20 at 11:24
  • I suggested because nonzero constant functions satisfy your hypothesis, but not the conclusion. The place where nonconstantness is important is when you used the Open Mapping Theorem. – Batominovski Apr 23 '20 at 11:32
  • @Batominovski Thank you very much for pointing this out. I missed the condition $f$ should be nonconstant when applying open mapping theorem. – pisco Apr 23 '20 at 11:36
  • I had just learned about the result on injective entire functions you cited a week ago. Now I'm disappointed that I didn't think of the proof, but I'm more disappointed that my lemma turned out to have been the trivial statement that the sum of 2 linear functions is linear. Also, how could I have missed the step of analytic continuation? Very nice job. – Display name Apr 23 '20 at 11:59