Timeline for On which subspace $W\subset C^{\infty}[0,1]$ is $(Df)(x)=xf'(x)$ a bounded operator provided all functions in $W$ are flat functions?
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| Jan 18, 2022 at 18:29 | answer | added | Michael Renardy | timeline score: 3 | |
| Jan 18, 2022 at 15:01 | comment | added | Daniele Tampieri | Finally, let me point out that he introduces his characterization exactly in order to study differential equations involving singular linear differential operator which, for the single variable case, have the following structure: $$ Q(x, D)u(x) = f(x)$$ where $D=x\frac{\mathrm{d}}{\mathrm{d}x}$ as in the case you are interested to. | |
| Jan 18, 2022 at 14:55 | comment | added | Daniele Tampieri | 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. | |
| Jan 18, 2022 at 14:37 | comment | added | Daniele Tampieri | 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. | |
| Jan 17, 2022 at 14:46 | comment | added | Michael Renardy | 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. | |
| Jan 17, 2022 at 13:02 | comment | added | Ali Taghavi | @MichaelRenardy No I mean a purely algebraic subspace D-invariant and D is bounded.But yourvcomment is interesting too could you plaease ellaborate it? | |
| Jan 17, 2022 at 3:15 | comment | added | Michael Renardy | 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. | |
| S Jan 16, 2022 at 15:38 | history | suggested | Dirk Werner | CC BY-SA 4.0 |
some language
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| Jan 16, 2022 at 15:22 | review | Suggested edits | |||
| S Jan 16, 2022 at 15:38 | |||||
| Jan 16, 2022 at 12:57 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
deleted 182 characters in body
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| Jan 16, 2022 at 12:26 | history | asked | Ali Taghavi | CC BY-SA 4.0 |