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By "Schwartz class functions" I will be referring to the functions of the Schwartz space on $\mathbb{R}$, that is, smooth ($\mathcal{C}^\infty$) functions $f : \mathbb{R} \to \mathbb{R}$ satisfying this condition: $$\forall (k,l) \in \mathbb{N}^2, \exists M_{k,l} \in \mathbb{R_+},\forall x \in \mathbb{R}, \quad \left|x^k f^{(l)}(x)\right| \leq M_{k,l}$$

A function $f : I \subset \mathbb{R} \to \mathbb{C}$ will be said to be nowhere analytic if $f$ is not (real-)analytic in any $x \in I$, that is if, for any fixed $x$, there exists no open neighbourhood $D$ of $x$ on which $f$ is the pointwise limit of a power series centered in $x$. There are multiple known examples of smooth nowhere analytic functions, a concrete example figures on this Wikipedia page (this is a major difference with the theory of "complex-analytic" functions since it is well known that one-time differentiability/holomorphy and analyticity are equivalent for functions $g : \Omega \subset \mathbb{C} \to \mathbb{C}$, but I digress).

The question is then simply:

Does there exist a nowhere analytic Schwartz class function?

Note that the idea is not mine, but rather from this post that I answered to from user @algebroo, specifically their comment. I felt like giving this its own question was better due to OP's initial question being less precise, however feel free to call this a duplicate or close this if it's really necessary. I wish OP would have asked it themself to avoid this dilemma of "do I ask the question or not when it wasn't my idea?" but oh well. Hoping they won't mind.

Anyway, I tried looking for posts dealing with how derivatives of smooth nowhere analytic behave... which took two seconds since How badly-behaved are the derivatives of non-analytic smooth functions? figured in the "Related" section on algebroo's post, but what matters is that one of the answers mentions a theorem called Bernstein's Theorem, which says that a smooth function $f : I \to \mathbb{R}$ with $I$ an interval such that all its derivatives are positive is analytic on $I$. Someone provided a proof here: https://math.stackexchange.com/q/1193121/1104384.\ Yet, we have boundedness conditions on Schwartz class functions (a lot of them, even), thanks to their definition. I was therefore wondering if we could use this theorem to prove that, at least somewhere near infinity, there exist points at which our Schwartz class functions are analytic, therefore proving the non-existence of our functions of interest.

Of course, there are a few problems with trying to use that theorem (or at least I have a few problems with that):

  • Firstly, if we could apply Bernstein's theorem directly (through shifting, rescaling, etc...) then maybe it could lead to the conclusion that every Schwartz class function is analytic everywhere, which is false and was the premise of algebroo's initial question: just like user Alex Ortiz commented, compactly supported functions offer an easy counterexample, and my answer also should give counterexamples since it's just the "inverse" of that. But that does not mean there is no way to involve the theorem in some way, naturally!
  • Lastly, while at $l$ fixed we could probably "compare" the $M_{k,l}$, I fear that, to apply Bernstein's theorem, we would potentially need to bound the derivatives (or something tied to the derivatives) by a same $M \in \mathbb{R_+}$ and on the same interval, but that feels out-of-reach for now? But perhaps there is a way to do that or to circumvent that?

Of course, maybe the existence or non-existence of a nowhere analytic Schwartz class function can be proven in some other way. I wouldn't mind at all an abstract proof (like, I don't know, Baire's category theorem used cleverly on the correct space or something...), though a concrete example (in the case of existence) would obviously be very interesting.

(As a very last note, feel free to re-tag or edit if you judge it appropriate)

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    $\begingroup$ Any reason to assume that $e^{-x^2}\sum_{k\in \mathbb{N}} e^{-\sqrt{2^k}}\cos(2^k x)$ doesn't work? $\endgroup$
    – reuns
    Commented Feb 18, 2023 at 19:46
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    $\begingroup$ @reuns hmm indeed that would be an excellent candidate. I'll have to check that out. $\endgroup$
    – Bruno B
    Commented Feb 18, 2023 at 19:55
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    $\begingroup$ $f$ is analytic at $a$ iff $e^{-x^2} f$ is analytic at $a$ $\endgroup$
    – reuns
    Commented Feb 18, 2023 at 20:00
  • $\begingroup$ I had no doubts about that part, what remains to be seen is whether it is of the Schwartz class. $\endgroup$
    – Bruno B
    Commented Feb 18, 2023 at 20:22
  • $\begingroup$ @reuns Seems like that was the right call! Sometimes the solution is very simple, thanks a lot! If you want you can make it into an answer so that I can accept it, or I can write up something quickly. $\endgroup$
    – Bruno B
    Commented Feb 18, 2023 at 20:32

2 Answers 2

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Just like @reuns in the comments said, the function $f : x \mapsto e^{-x^2}F(x) := e^{-x^2}\sum_{k\in \mathbb{N}} e^{-\sqrt{2^k}}\cos(2^k x)$ is a nowhere analytic Schwartz class function (where $F$ is the Wikipedia example of a nowhere analytic function from the beginning of my question).

Indeed, if we let $S_n := \sum_{k\in \mathbb{N}} e^{-\sqrt{2^k}}(2^k)^n \in \mathbb{R}_+$ which bounds $|F^{(n)}|$ from above and $K_m$ be any real that bounds the function $x \mapsto \left|x^m e^{-x^2}\right|$ from above, then it is easy to see that there will exist $M_{m,n}$ a (finite) linear combination of the $K_{m'}$ and $S_{n'}$, $m' \leq m$ and $n \leq n'$, that will bound $x \mapsto \left|x^m f^{(n)}\right|$ from above by using the Leibniz product rule to derive $f = e^{-x^2} F$, and thus $f$ is a function of the Schwartz class.

But the class of functions analytic at a point $x$ is stable under multiplication, and $x \mapsto e^{-x^2}$ is analytic everywhere, therefore $F$ and $f$ are simultaneously nowhere analytic, and we are done.

Thanks a lot reuns.

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Take a look into the Rvachëv function.

It is a displaced version of the first lobe of the Fabius function which is an example of "an infinitely differentiable function that is nowhere analytic". The lobe got centered as zero becoming an even function, and it inherit the "nowhere analytical characteristic" at least for $|x|<1$.

The Rvachëv function is equal to zero outside $|x|<1$, so here it is matching a constant value so is arguable it is locally analytical there, even when globally a non-constant Power Series cannot match a constant value in a non-zero measure interval, since it violates the Identity theorem, so globally is not an analytic function.

Rvachëv function is of compact support and non-negative, and it is also an example of a smooth bump function, which are examples of Non-analytic smooth function. And as Wikipedia page for Schwartz space states "Any smooth function $f$ with compact support is in $\mathcal{S}(\mathbb{R}^n)$", then the Rvachëv function is in the Schwartz space.

In this way, at least you can say that the Rvachëv function is an example of nowhere analytical function in the Schwartz space $\mathcal{S}([-1,\ 1])$.

The Rvachëv function fulfills in $\mathbb{R}$ the Delayed differential equation: $$R'(x)=2R(2x+1)-2R(2x-1)$$

The Rvachëv function could be approximated in Desmos through the formula shown in this answer, or by a gross but simpler version through the function

$$f_1(x) = \begin{cases} 1,\quad x=0, \\ 0,\quad |x|\geq 0, \\ \dfrac{1}{1+\exp\left(\dfrac{1-2|x|}{x^2-|x|}\right)}, \text{otherwise}\end{cases}$$

approximations to the Rvachëv function

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    $\begingroup$ This is indeed a very interesting example of a smooth bump function, as it is also nowhere analytic on a whole interval (+1)! I wonder if you could make a function out of this one which truly answers my question by either extending periodically but lowering the amplitude fast enough, or by right-composing by some smooth bijection from $[-1,1]$ to $\mathbb{R}$ like $x \mapsto \tan\left(\frac{2x}{\pi}\right)$ or some other such bijection? $\endgroup$
    – Bruno B
    Commented Jul 16, 2023 at 6:05
  • $\begingroup$ @BrunoB I don't have enough background to made some. But maybe extending the Fabius function to non-negative values as mirroring which happens with the possitive ones would work: at least all derivatives will be bounded (I believe). $\endgroup$
    – Joako
    Commented Jul 16, 2023 at 17:49

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