A certain proof that $e^{z}$ is the only solution of $f'(z)=f(z), \, f(0)=1,\, z\in\mathbb{C}$

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.