5
$\begingroup$

Question: Given $f\in C^{\infty}$ which is not analytic on a bounded domain $\Omega \subseteq \mathbb{R}$. What can we say about the sequence $\lbrace f^{(m)} \rbrace _{m=1}^{\infty} $? Specifically - what can we say about the convergence and/or decay of its $L^2$ and $L^{\infty}$ norms? If nothing, why? Of special importance is to bound globally Taylor like terms of the form $f^{(m)}(\xi)/m! $ for some $\xi \in \Omega$.

Motivation, not Neccessary: My question is motivated by the residue terms of numerical integration formulas. We can usually have nice convergence rates for analytical functions due to the exponential decay in their taylor series coefficients $f^{(m)}/m! $.

I have read some of the math.stackexchange posts about it, and while this one does contain some possible leads, I couldn't find the answers to my questions there yet.

Thanks

Improve this question
$\endgroup$
5
  • 4
    $\begingroup$ Even for real analytic ones the $L^2$ or $L^\infty$ of the derivatives are not required to decay. Take for example the function whose Fourier transform is the characteristic function of the interval $[2,3]$. It is real analytic, but the corresponding sequences that you are looking at both blow up to infinity. $\endgroup$
    – Willie Wong
    Commented Dec 6, 2016 at 20:45
  • $\begingroup$ So the asymptotic exponential decay in real analytic functions is only pointwise? Also, do they blow up at a bounded rate? $\endgroup$
    – Amir Sagiv
    Commented Dec 6, 2016 at 20:58
  • 1
    $\begingroup$ maybe you forgot to divide by $m!$? $\endgroup$
    – Willie Wong
    Commented Dec 6, 2016 at 22:11
  • $\begingroup$ You're right. I'll change it. $\endgroup$
    – Amir Sagiv
    Commented Dec 7, 2016 at 7:52
  • 2
    $\begingroup$ I don't have an answer, but I have one interesting example to think about: the Fabius function, which is $C^\infty$ but nowhere analytic and satisfies a nice functional equation. $\endgroup$
    – Gro-Tsen
    Commented Dec 7, 2016 at 10:42

2 Answers 2

Reset to default
11
$\begingroup$

For the $L^\infty$ norms, nothing at all apart from Alexander Eremenko's answer, because of Borel's theorem: any sequence of real (resp. complex) numbers can be realized as the sequence of Taylor coefficients of some real (resp. complex) valued smooth function $f$ at any given point. By the Sobolev embedding theorem and the fact that $f$ may be chosen compactly supported in the interior of $\Omega$, one can bound the $L^\infty$ norm of $f$ from above by a positive constant times the sum of the $L^1$ norms of the $n$-th order derivatives of $f$, where $n$ is the dimension of the domain $\Omega$. Therefore, if the sequence of $L^\infty$ norms of the derivatives of $f$ blows up, so does the corresponding sequence of $L^1$ norms. Since you assumed $\Omega$ to be bounded, this implies the same property for the $L^p$ norms for all $1\leq p\leq\infty$ by Hölder's inequality.

(side remark: the case of Sobolev's embedding theorem which is relevant to the present context is an easy consequence of the fundamental theorem of Calculus applied once to each variable of $f$, which also shows that the constant above may be taken equal to 1)

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thanks. The interchange between the $L^{\infty}$ blowups to $L^1$ norms blowups is vie Holder inequality? $\endgroup$
    – Amir Sagiv
    Commented Dec 7, 2016 at 8:02
  • $\begingroup$ Also, in Alexander Eremenko's answer there is always a subsequanse that blows up. But it seems that from Borel's theorem we can build converging sequences of derivatives. How are the two compatible? $\endgroup$
    – Amir Sagiv
    Commented Dec 7, 2016 at 8:05
  • 2
    $\begingroup$ The argument I originally had in mind for the $L^1$ norms was not quite precise. I've just replaced it with a better one using the Sobolev embedding theorem. As for your second question, the particular kind of blowup provided in Alexander Eremenko's answer is meant to preclude a positive radius of convergence for the Taylor series of $f$ around any point of $\Omega$ by means of the root test, hence to prevent that f is analytic anywhere in $\Omega$. $\endgroup$
    – Pedro Lauridsen Ribeiro
    Commented Dec 7, 2016 at 13:03
  • 1
    $\begingroup$ On the other hand, Borel's theorem is not about that - the idea is that its proof consists in truncating the $j$-th term in the would-be Taylor series of $f$ at a point $x_0\in\mathrm{int}\Omega$ by cutoff functions supported in $\Omega$ which equal 1 in a neighborhood of $x_0$ and whose supports shrink very fast with $j$ (depending on how fast the $a_j$'s grow with $j$). Particularly, it's not guaranteed at all that the original Taylor series for f will converge (actually, for this particular $f$, it certainly won't converge to $f$ due to the non-analyticity of the cutoff functions). $\endgroup$
    – Pedro Lauridsen Ribeiro
    Commented Dec 7, 2016 at 13:04
8
$\begingroup$

We can say is that $\limsup |a_n|^{1/n}=\infty$, where $a_n=\max_{a\in K}|f^{(n)}(a)|/n!$, for every compact $K\subset\Omega$.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ Thanks. But If I take Borel's theorem, and pick $f^{(n)} = e^{-n}$, why is $\lim \sup |a_n|^{1/n}$ not bounded? Is it because Borel's Lemma applies only to a specific point $a_0 \in \Omega$? $\endgroup$
    – Amir Sagiv
    Commented Dec 7, 2016 at 20:48
  • 2
    $\begingroup$ Yes, exactly. Borel's Lemma is about one point. Once you restrict the growth of some norm on the interval, the function for which this norm does not grow fast enough will be analytic. $\endgroup$
    – Alexandre Eremenko
    Commented Dec 7, 2016 at 20:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .