If you didn't already know that $e^x$ is a fixed point of the derivative operator, could you still show that some fixed point would have to exist?

Question

Let's suppose you independently discovered the operator $\frac{d}{dx}$ and know only its basic properties (say, the fact it's a linear operator, how it works on polynomials, etc.) If you didn't know that $e^x$ was a fixed-point of this operator, would there be any way to (possibly nonconstructively) show that such a fixed point would have to exist? I'm curious because just given the definition of a derivative it doesn't at all seem obvious to me that there would be some function that's it's own derivative.

Note that I'm not asking for a proof that $\frac{d}{dx} e^x = e^x$, but rather a line of reasoning providing some intuition for why $\frac{d}{dx}$ has a fixed-point at all.

(And let's exclude the trivial fixed point of 0, since that follows purely from linearity rather than any special properties of derivatives).

Answer 1

See the wikipedia page on the theorem of Picard-Lindelöf. Your question follows by setting $f(t,y(t)) := y(t)$. The essential ingredients in the proof of this theorem are the fundamental theorem of calculus, the Banach fixed-point theorem and the fact that the space of continous functions on a compact interval equipped with the supremum norm is a Banach space. For global uniqueness of the solution you might also need Gronwall's inequality which features the exponential function, however if you are only interested in proving existence and local uniqueness you don't need Gronwall.

Answer 2

Suppose you have a small positive number $\epsilon$. Consider functions which map integer multiples of $\epsilon$ onto $\mathbb{R}$. These functions have an approximate derivative $$f'(i\epsilon)\approx \frac{f((i+1)\epsilon)-f(i\epsilon)}{\epsilon}$$ (for integer $i$)

I think that it is intuitively clear that for these functions and this approximate derivative, the approximate derivative has a fixed point. It can be constructed trivially as follows: define $f(0)=1$, $f((i+1)\epsilon)=f(i\epsilon)+\epsilon f'(i)$.

Of course, this approximate derivative approaches the true derivative in the limit $\epsilon\rightarrow 0$. If you expect that this line of reasoning still works "in the limit", then you expect that $\frac{d}{dx}$ has a fixed point.

Answer 3

You want a function $f$ such that $f'=f$. Let's hope that such function exists and has an inverse, $g$ (this is just for motivation: we will prove it in the long run).

We then have that, since $f \circ g =Id$, by the chain rule, $f'(g(x)) g'(x)=1$. Therefore,

$$g'(x)=\frac{1}{f'(g(x))}=\frac{1}{f(g(x))}=\frac{1}{x}.$$

The derivative of this inverse seems more manageable, since it does not depend on $g$ itself (The motivation ends here). Let's then define $g:\mathbb{R}_{>0} \to \mathbb{R}$:

$$g(x)=\int_1^x\frac{1}{t}dt.$$ By the FTC, $g$ satisfies what it needs. If we prove that $g$ is invertible, we are done. For that, first note that $g(xy)=g(x)+g(y)$. To see this, let $y$ be fixed. Then, $(g(xy))'=\frac{1}{xy}y=\frac{1}{x}$, and the derivative of the right side is also trivially $\frac{1}{x}$. Since both sides coincide in $1$ by a trivial checking, both must be equal. Since $y$ is arbitrary, the relation holds for any $x,y$ as we wanted.

Now, note that it follows that $g(2^n)=n g(2)$. But $g(2)>0$ by its very definition. Since $g$ is increasing, it follows that $g(x) \to \infty$ as $x \to \infty$. Since $g(1)=g(x)+g(\frac{1}{x})$, it follows that $g(x) \to -\infty$ as $x \to 0$, and we are finished proving that $g$ is a bijection due to the IVT.

Now, take the inverse $f$ of $g$. We will have by the chain rule that $$g'(f(x))f'(x)=1 \implies f'(x)=f(x).$$