I am a graduate student and I've been thinking about this fun but frustrating problem for some time. Let $d = \frac{d}{dx}$, and let $f \in C^{\infty}(\mathbb{R})$ be such that for every real $x$, $$g(x) := \lim_{n \to \infty} d^n f(x)$$ converges. A simple example for such an $f$ would be $ce^x + h(x)$ for any constant $c$ where $h(x)$ converges to $0$ everywhere under this iteration (in fact my hunch is that every such $f$ is of this form), eg. $h(x) = e^{x/2}$ or simply a polynomial, of course.

I've been trying to show that $g$ is, in fact, differentiable, and thus is a fixed point of $d$. Whether this is true would provide many interesting properties from a dynamical systems point of view if one can generalize to arbitrary smooth linear differential operators, although they might be too good to be true.

Perhaps this is a known result? If so I would greatly appreciate a reference. If not, and this has a trivial counterexample I've missed, please let me know. Otherwise, I've been dealing with some tricky double limit using tricks such as in this MSE answer, to no avail.

Any help is kindly appreciated.

$\textbf{EDIT}$: Here is a discussion of some nice consequences know that we now the answer is positive, which I hope can be generalized.

Let $A$ be the set of fixed points of $d$ (in this case, just multiples of $e^x$ as we know), let $B$ be the set of functions that converge everywhere to zero under the above iteration. Let $C$ be the set of functions that converges to a smooth function with the above iteration. Then we have the following:

$C$ = $A + B = \{ g + h : g\in A, h \in B \}$.

Proof: Let $f \in C$. Let $g$ be what $d^n f$ converges to. Let $h = f-g$. Clearly $d^n h$ converges to $0$ since $g$ is fixed. Then we get $f = g+h$.

Now take any $g\in A$ and $h \in B$, and set $f = g+h$. Since $d^n h$ converges to $0$ and $g$ is fixed, $d^n f$ converges to $g$, and we are done.

Next, here I'm assuming the result of this thread holds for a general (possibly elliptic) smooth linear differential operator $d : C^\infty (\mathbb{R}) \to C^\infty (\mathbb{R}) $. A first note is that fixed points of one differential operator correspond to solutions of another, i.e. of a homogeneous PDE. Explicitly, if $d_1 g = g$, then setting $d_2 = d_1 - Id$, we get $d_2 g = 0$. This much is simple.

So given $d$, finding $A$ from above amounts to finding the space of solutions of a PDE. I'm hoping that one can use techniques from dynamical systems to find the set $C$ and thus get $A$ after the iterations. But I'm approaching this naively and I do not know the difficulty or complexity of such an affair.

One thing to note is that once we find some $g \in A$, we can set $h(x) = g(\varepsilon x)$ for small $\varepsilon$ and $h \in B$. Conversely, given $h \in B$, I'm wondering what happens when set set $f(x) = h(x/\varepsilon)$, and vary $\varepsilon$. It might not coincide with a fixed point of $d$, but could very well coincide with a fixed point of the new operator $d^k$ for some $k$. For example, take $h(x) = cos(x/2)$. The iteration converges to 0 everywhere, and multiplying the interior variable by $2$ we do NOT get a fixed point of $d = \frac{d}{dx}$ but we do for $d^4$.

I'll leave it at this, let me know again if there is anything glaringly wrong I missed.

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