1

The following mimics a proof of Qiaochu Yuan (Proof that $C\exp(x)$ is the only set of functions for which $f(x) = f'(x)$) of a similar theorem, there is just a difference that $f:\, \mathbb{C}\to\mathbb{C}$.

If $g(z)=f(z)e^{-z}$, then $$g'(z)=-f(z)e^{-z}+f'(z)e^{-z}=e^{-z}(f'(z)-f(z))=0,$$ so $g(z)$ is a constant. Since $g(0)=1$, that constant is $1$. This means that $f(z)=e^{z}$ is the unique solution (that is infinitely differentiable? – I am not sure if that condition is needed).

However, it seems that such proof uses the Mean value theorem in the background. But the Mean value theorem does not hold for complex numbers. Can any sense be made of the above proof?

Mistral GT
  • 35

3 Answers3

2

The mean value theorem is used to prove that any function $f:\mathbb{R}\to \mathbb{R}$ whose derivative is zero is constant. In fact, the one-dimensional mean-value theorem is enough to prove this fact for functions $f:\mathbb{C} \to \mathbb{C}$ as well.

Suppose that $f:\mathbb{C} \to \mathbb{C}$ is a differentiable function and $f'(z)=0$ for all $z$. Consider the path $\gamma(t)=\omega t$ for some nonzero complex number $\omega$. By the chain rule,

$$ \frac{d}{dt}\left[f(\gamma(t))\right]=f'(\gamma(t))\gamma'(t)=0 \cdot \omega = 0 \, . $$

So the function $f \circ \gamma$ is a function from $\mathbb{R}$ to $\mathbb{C}$, and the real and imaginary parts of this function (which are functions from $\mathbb{R}$ to $\mathbb{R}$) both have derivative zero. So the real and imaginary parts of this function are both constant.

We therefore have

\begin{align*} f(\omega)&=f\circ \gamma(1)\\ &=f\circ \gamma(0)&&\text{(because }f \circ \gamma\text{ is constant)}\\ &=f(0)&&\text{(because }\gamma(0)=0\text{)} \end{align*}

for any nonzero complex number $\omega$. So $f$ is constant.

Micah
  • 38,108
  • 15
  • 85
  • 133
  • "The mean value theorem is used to prove that any function $f:, \mathbb{R}\to\mathbb{R}$ whose derivative is zero is constant" – but isn't that equivalent to the basic derivative rule that $\frac{dc}{dx}=0$? – Mistral GT Jan 05 '20 at 18:49
  • I mean, a constant, by definition, never changes, so its derivative must be zero. – Mistral GT Jan 05 '20 at 19:24
  • 1
    @MistralGT: The fact that all constant functions have derivative zero is a basic rule of differentiation. However, you need the mean value theorem to know that there aren't any non-constant functions with derivative zero. – Micah Jan 05 '20 at 22:27
  • 1
    That is, the implication "$f$ is constant $\implies f'(x)\equiv 0$" is straightforward, but the implication "$f'(x)\equiv 0 \implies f$ is constant" requires the mean value theorem. The second implication is the one you need in your proof. – Micah Jan 05 '20 at 22:28
  • I thought it is obvious that $f'(x)=0\iff f(x)$ is constant. Like, the tangent must be, by definition, parallel to the $x$-axis everywhere, so $f$ must be a constant function; no other function obviously has these properties. A "potential" counterexample would be step functions or some other piecewise functions, but such functions are not infinitely differentiable. – Mistral GT Jan 06 '20 at 14:02
  • 1
    @MistralGT: I agree that what you're saying seems obvious, but that doesn't mean the proof is simple! When thinking about near-counterexamples, piecewise-defined functions are the right way to go. For example, the function $f(x)=\begin{cases} 0 & x < -1 \ 1 & x > 1\end{cases}$ has derivative zero on its entire domain ${x:|x|>1}$. This kind of example demonstrates that you need to know some fact about $\mathbb{R}$ as a whole to prove the theorem; just knowing what's going on at individual points isn't good enough. The mean value theorem is a way to get this kind of global knowledge. – Micah Jan 06 '20 at 16:23
1

If $f(z)$ is differentiable, then so is $g(z):=f(z)e^{-z}$ and $g'(z)=0$ makes $g(z)$ constant, by integration.

Note that $f''(z)=(f'(z))'=f'(z)=f(z)$ and by induction $f(z)$ is implicitly infinitely differentiable.

  • So is the condition on infinite differentiability required? Does it mean that the Mean value theorem is not needed? – Mistral GT Jan 05 '20 at 18:11
  • 1
    @MistralGT $f(z)$ is implicitly infinitely differentiable. –  Jan 05 '20 at 18:15
0

You are correct to note that in both the real and the complex case you need to know that a function whose derivative is identically $0$ is a constant.

In the real case that follows from the mean value theorem (or some equivalent assertion relying on the fact that the reals are complete).

In the complex case you can argue that a function that's (complex) differentiable is in fact analytic, so has a power series all of whose coefficients are $0$.

It's been a while since I looked at how that fact about analytic functions is proved. It probably depends somewhere on the completeness of the real numbers.

Ethan Bolker
  • 95,224
  • 7
  • 108
  • 199