Is $1/F$ Schwartz if $F$ is "reverse Schwartz"?

Question

Let's call a positive function $F:\mathbb{R}\to\mathbb{R}$ "reverse Schwartz" if $F$ is smooth and

$$\forall n \forall k,\quad\lim_{x\to\infty}\frac{|x|^n}{|\partial_x^k F(x)|}=0\quad .$$

In other words, $F$ and all of its derivatives grow faster than polynomially for $|x|\to\infty$. Notice that, unlike $F$, $\partial_x^kF$ might have zeroes, but the condition only involves the limit for $x\to\infty$.

Question: if $F$ is reverse Schwartz, then does $1/F$ belong to the Schwartz class?

Remark: for the $n=0,\;k=1$ "usual" Schwartz estimate to hold for $1/F$, it's sufficient that $|F'|\ll F^2$. So, for example, is there a reverse Schwartz that doesn't satisfy the estimate $|F'|\ll F^2$?

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Motivation: I thought that functions such as $$g(x)=\frac{1}{\cosh(x)}$$ should "obviously" be in the Schwartz class. In order to prove it, one can notice that $g(x)=R(e^x)$, where $$R(t)=\frac{2t}{t^2+1}.$$ Considering any rational function $R=P/Q$, $$\partial_x^k (R(e^x))=[(t\partial_t)^kR(t)]|_{t=e^x},$$ and one is led to prove things like $$(t\partial_t)^k\left(\frac{P}{Q}\right)=\frac{H}{Q^{2^k}}$$ with $\deg H\leq p+(2^k-1)q$ where $p:=\deg P$ and $q:=\deg Q$. This is sufficient, but looks a bit contrived, and not very generalizable. Is this the way? Or, are there simpler and more general ways?

In two variables, the "obviously Schwartz-looking" function $$g(x,y)=\frac{1}{\cosh(x)+\cosh(y)}$$ is $R(e^x,e^y)$ where $$R(u,v)=\frac{2uv}{v(u^2+1)+u(v^2+1)}.$$ Computing derivatives leads to $$\partial_x^k\partial_y^\ell(R(e^x,e^y))=[(u\partial_u)^k(v\partial_v)^\ell R(u,v)]|_{u=e^x,\;v=e^y}$$ with similar control over the degrees. Again, is there a simpler way?

[I'm cross-posting this question from MathStackExchange, because it didn't get activity there.]

Answer 1

If one sets $F(x) := \sum_{n=1}^\infty a_n e^{a_n (x^2-n^2)}$, where $1 < a_1 < a_2 < \dots$ goes to infinity sufficiently quickly, then one can verify that $F$ is reverse Schwartz (all derivatives are non-negative by Taylor expansion, and go to infinity at a gaussian rate or faster), but (for sufficiently rapid $a_n$) one has $F(n) \sim a_n$ and $F'(n) \sim n a_n^2$, so $1/F$ is not going to be Schwartz.

ADDENDUM: The brevity of my answer may suggest that it was quite easy to resolve, but this problem actually took me a non-trivial amount of time to figure out. I thought it might be worth recording my thought processes that led to this answer while this process remains fresh in my mind, in case others find this helpful or interesting.

The OP already isolated the necessary condition $|F'| \ll F^2$ for this question. As a general rule, it is very rare to be able to control expressions involving higher derivatives in terms of expressions involving lower derivatives, unless some "elliptic regularity" phenomenon is in effect, such as that arising from the Cauchy–Riemann equations of complex analyticity. Since the reverse Schwartz condition certainly does not imply anything remotely resembling analyticity, this led me to initially suspect that the answer to this question should be negative, and to then try to disprove the claim $|F'| \ll F^2$.

My first attempt was to try to use a Baire category type argument to show that this type of failure was in fact generic. Typically, the key to this type of argument is to show that if one starts with a reverse Schwartz function $F$, a point $x_0$, and a constant $C$, that the pointwise inequality $|F'(x_0)| \leq C F(x_0)^2$ can be violated by an arbitrarily small "perturbation" of $F$ in a suitable metric. Usually, the way to proceed in such situations is to add a highly oscillatory perturbation such as $x \mapsto \varepsilon \sin(Nx)$ to $F$, with the small amplitude parameter $\varepsilon$ and large frequency parameter $N$ chosen so that one does not modify $F(x_0)$ much but instead modifies $F'(x_0)$ quite a bit. Here I ran into an unexpected difficulty: such modifications also modify higher derivatives $F''$, $F'''$, and so forth, and could potentially violate the rest of the reverse Schwartz conditions. I had not anticipated this, and now could see that these conditions do, in fact, allow for some way to control higher order terms by lower order terms, in contrast to my initial intuition.

After some further playing around, I noticed that it is difficult to introduce any oscillation at all, because the reverse Schwartz condition implies that all the derivatives of $F$ will stop changing sign for large enough $x$. In particular, $F$ becomes monotone for large $x$, either convex or concave for large $x$, and so forth. This does create some pointwise bounds: for instance, if we know that $F$ is positive, monotone increasing, and convex for $x \geq x_0$, then it is not difficult to use these hypotheses to obtain the bound $0 \leq F(x)-F(x-1) \leq F'(x) \leq F(x+1)-F(x)$ for $x \geq x_0+1$, mostly by using the convexity of $F$ on $[x-1,x+1]$. This indicated that the bound $|F'| \ll F^2$ that I was trying to disprove, could only fail when $F$ grew extremely rapidly. To put it another way, making $F'$ large at some location $x$ would cause $F$ to become even larger at some subsequent location $x+1$.

At this point I started experimenting with standard extremely rapidly growing functions like $\exp(\exp(x))$. I did notice that these functions had non-negative Taylor expansions and so it was very easy to verify the reverse Schwartz condition as $x \to +\infty$ (and one could easily modify the function to say $\exp(\exp(x^2))$ to get the reverse Schwartz condition at both ends). However they did not violate the condition $|F'| \ll F^2$ for large $x$. And indeed I knew from the (well known) finite time blowup of the Ricatti equation $F' = F^2$ that one should not be able to violate this condition everywhere; it should only be at some sparser set of points, such as the natural numbers $n=1,2,\dots$, where one could hope to construct a counterexample.

Given that oscillatory constructions seemed to be ruled out, and that I already knew that trying to control the derivative of $F$ at one point $n$ tended to cause undesirable problems at a later point $n+1$, this now suggested an iterative "procrastination" strategy in which one obtains the desired behavior at "time" $n$ by postponing all the resultant difficulties created by doing so to "time" $n+1$. Since none of the reverse Schwartz conditions involved any absolute upper bounds on $F$ and its derivatives, one could hope to simply incur higher and higher amounts of "debt" in this strategy that never actually gets "paid off" until after one has fully constructed the required counterexample.

So I started looking for functions that, for a given time $n$, had a large value of $F'(n)$ and a small value of $F(n)$, while being insignificant at previous times $1,\dots,n-1$, and not caring too much about what happened at later times $n+1,n+2,\dots$, other than that they satisfied the reverse Schwartz conditions, with the plan to add these all functions together to give the desired counterexample (I had noticed previously that the reverse Schwartz class seemed to be closed under addition if all the derivatives were positive). I initially played with various rapidly growing functions like $\exp(\exp(x))$, but eventually realized that the function $F_{a,b,n}(x) := a e^{b(x-n)}$ for $a,b,n>0$ would have all derivatives positive, and would have a derivative $F'_{a,b,n}(n) = ab$ at $n$ that was significantly larger than the value $F_{a,b,n}(n) = a$ at $n$ (or its square) if $b$ was chosen large compared to $a$. It was then a routine matter to combine such functions together to obtain a counterexample that obeyed the reverse Schwartz conditions for $x \to +\infty$; finally I replaced $x$ with $x^2$ and streamlined the argument to give the answer supplied above.

This discussion also suggests that the answer to the OP's question may become positive again if one also imposes some growth condition on $F$, e.g., an exponential growth upper bound, which would be good enough to recover the examples mentioned by OP. EDIT: it appears that such an exponential growth bound is insufficient, because one can simply "outlast" this bound by deferring the time $t_n$ at which one has the bound $F'(t_n) \geq n F(t_n)^2$ to be so late that an exponential bound provides no significant constraint on the behavior of $F$. Indeed, if $t_n$ is a sufficiently rapidly growing sequence, one can construct an $F$ so that $F$ is a polynomial of degree $n-1$ on $[t_{n-1}, t_n-1]$ with positive coefficients, and then by choosing $F^{(n)}$ to be a suitable large multiple of a bump function on $[t_n-1,t_n]$, arrange matters so that $F'(t_n) \geq n F(t_n)^2$, and that $F$ continues as a polynomial of degree $n$ from $t_n$ until $t_{n+1}-1$. If the $t_n$ are widely spaced enough (and if one also extends by symmetry to the negative axis), this creates a reverse Schwartz function which is of subexponential growth, but whose reciprocal still fails to be Schwartz.