2
$\begingroup$

I have already asked this question on MSE; now I repeat it on MO.

https://math.stackexchange.com/questions/4132346/on-which-subspace-w-subset-c-infty0-1-is-df-xfx-a-bounded-operator

First we introduce the concept "flat function":

Definition: A smooth function $f:\mathbb{R}\to \mathbb{R}$ is called a flat function at the origin if $f^{(k)}(0)=0$ for all $k=0,1,2,\ldots $.

Let $V=\{f\in C^{\infty}[0,1]\mid f \text{ is flat at the origin}\}$. We equip $V$ with $|\ \ |_{\infty}$. Is there an infinite dimensional subspace $W\subseteq V$ which is invariant under the differential operator $D$, $(Df)(x)=xf'(x)$, and $D$ is a bounded operator on $W$? The motivation comes from page 43, remark 2 of the following paper:

http://mcs.qut.ac.ir/article_243944.html

Improve this question
$\endgroup$
6
  • 2
    $\begingroup$ Do you require W to be complete (in the $L^\infty$ norm)? Of course, if $D$ is bounded on W, it will also be bounded on its completion, but the completion of W may not consist of "flat" functions. $\endgroup$
    – Michael Renardy
    Commented Jan 17, 2022 at 3:15
  • 1
    $\begingroup$ @MichaelRenardy No I mean a purely algebraic subspace D-invariant and D is bounded.But yourvcomment is interesting too could you plaease ellaborate it? $\endgroup$
    – Ali Taghavi
    Commented Jan 17, 2022 at 13:02
  • 2
    $\begingroup$ If W is assumed complete, such a space cannot exist. SInce W consists of flat functions, we can define an alternative norm in W by $\|f\|=\|f\|_\infty+\|f/x\|_\infty$, By the bounded inverse theorem, this norm must be equivalent to $\|f\|_\infty$. Therefore, if D is bounded, then the derivative operator is also bounded. But by Arzela-Ascoli, this implies the unit ball in W is compact, hence W must be finite dimensional. $\endgroup$
    – Michael Renardy
    Commented Jan 17, 2022 at 14:46
  • 1
    $\begingroup$ Bogdan Ziemian, in his Taylor formula for distributions, (English) Dissertationes Mathematicae (Rozprawy Matematyczne) 264, Warsaw: Polish Academy of Sciences (Polska Akademia Nauk - PAN), Institute of Mathematics (Instytut Matematyczny) 56 p. (1988), MR0931848, Zbl 0685.46025, gives a characterization of flat functions. $\endgroup$
    – Daniele Tampieri
    Commented Jan 18, 2022 at 14:37
  • 1
    $\begingroup$ In chapter 3, pp. 14-17 (particularly proposition 2, p. 16) he proves that the (modified) Mellin transform of a function of this kind is a particular entire function. Actually his result is more general since he is able to describe the Mellin transform of "only" measurable functions $f$ on $(0,1)$ for which $f(x)=O(x^\alpha)$ and there exists $r>0$ and $C>0$ such that $$\left|\int\limits_\varepsilon^1\frac{f(x)}{x^r}\mathrm{d}x\right| <C \text{ for all }0<\varepsilon <1,$$ where $\alpha$ is a positive constant, thus dealing with the case of "finitely" flat functions. $\endgroup$
    – Daniele Tampieri
    Commented Jan 18, 2022 at 14:55

1 Answer 1

Reset to default
3
$\begingroup$

This is not an answer, but it suggests that at least there are no simple examples. Suppose $f\in W$, with $\|f\|_\infty\le L$ and $D$ is bounded mapping $W$ into itself with norm bounded by $M$. Then $\|D^nf\|_\infty\le M^nL$. It follows that $f$ is an entire function of $\ln x$ of order (at most) 1. The order cannot be strictly less than 1, because in that case there is a lower bound (see e.g. Chapter VII,10 at Saks and Zygmund, Analytic Functions) which precludes $f$ from being flat. Moreover, $f$ cannot be an exponential function of $\ln x$, i.e. a power of $x$, since these are not flat either.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thank you and my (prevciously) +1. BTW "Entire function of ln(x)" This remind me of the concept of "Dulac series" in the investigation of the finteness part of the Hilbert 16th problem $\endgroup$
    – Ali Taghavi
    Commented Mar 10, 2023 at 17:09

You must log in to answer this question.

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