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I am almost positive I am incorrect, but I was hoping someone could explain why I am wrong because for the life of me I am at a loss. I am not asking for a counterexample or some contra-reasoning as to why my conclusion is incorrect, but rather I am hoping someone can point out the flaw in my actual argument.

Suppose $f\in C^{\infty}(\mathbb{C})$. Suppose $|f(z)|\to 0$ as $|z|\to \infty$. My claim is that $f\equiv 0$.

Reasoning: If $f$ is smooth on $\mathbb{C}$, then by definition $f$ is holomorphic, since $f$ is $\mathbb{C}$-differentiable. So then $f$ is entire. But if $|f(z)|\to 0$ as $|z|\to \infty$, then there exists some $N$ such that $|z|>N$ implies that $|f(z)|<1$. So then consider the closed disk of radius $N$. We know that $f$ is holomorphic on this disc. But such a disc is compact, so $f$ must achieve a maximum and minimum on this disc. So $f$ is bounded on this disc, and $f$ is also bounded outside of the disc by $1$. So $f$ is bounded and entire, so then by Liouville it is constant. But we know that it goes to $0$ as $|z|\to \infty$. So then $f\equiv 0$.

I am almost certain this is wrong somewhere, but to me my argument seems sounds. Any insight would be helpful.

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    What makes you think that you're wrong? – joriki Mar 19 '20 at 01:15
  • @joriki Well, it trivializes one of my homework problems to a degree which makes the problem statement utterly absurd. –  Mar 19 '20 at 01:17
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    @lagicol, ah... those fiendish homework-creators, always trying to confuse the heck out of you ;-) – vonbrand Mar 19 '20 at 01:25
  • @vonbrand Perhaps another, more intelligent reason why I suspect I am wrong is that my reasoning rests on the fact that $C^{\infty}(\mathbb{C})$ implies holomorphic. But if this were true, then the entire discussion in Chapter 5, Section 2 of Complex Analysis (Narasimhan) is completely strange; for example, Theorem 2 in this section describes a $u\in C^{\infty}(\mathbb{C})$ such that $\frac{\partial u}{\partial \overline{z}}=\phi$, where $\phi$ is a previously defined function. But this is absurd, since this forces $\phi\equiv 0$ in a very strange way... –  Mar 19 '20 at 01:31
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    I think I see. I am conflating two different notions of $C^{\infty}$ perhaps. Maybe my instructor and my text are using $C^{\infty}(\mathbb{C})$ slightly improperly to really mean $C^{\infty}(\mathbb{R^2})$. This makes everything make more sense perhaps. –  Mar 19 '20 at 01:33
  • I'm going with yes. That makes it constant. –  Mar 19 '20 at 01:56
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    usually $C^{\infty}$ means smooth in the sense of partial derivatives say (here one can still use complex terminology with partial derivatives $\partial_z, \partial_{\bar z}$); the problem itself though is true for any continuos function and your proof is just fine except that one only uses that a continuos function is bounded on a compact set, no need for maximum modulus etc results; note that one can have smooth functions with compact support in the plane so the function doesn't need to be constant – Conrad Mar 19 '20 at 02:04

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