Timeline for Banach algebra of smooth functions
Current License: CC BY-SA 4.0
33 events
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| Nov 26, 2020 at 15:53 | answer | added | Ayman Moussa | timeline score: 1 | |
| Nov 14, 2020 at 17:38 | comment | added | Ayman Moussa | @juan Yes, but you can adapt this to the periodic case as well. Obviously you could produce (as I did above) a counter-example on $\mathscr{C}^0([0,\frac{1}{2}])$, the condition being now transformed into $\sum_p a_p <\infty$ so $a_p=\frac{1}{p^2}$ gives you in the same way a counter-example. Now, if you assume furthermore $a_p=0$ for odd integer $p$, you get an even function and the counter-example works the same way on $[-\frac{1}{2},\frac{1}{2}]$, with same values of all derivatives at end-points. By translation, you have therefore covered the case of $\mathbf{T}$. | |
| Nov 14, 2020 at 16:33 | comment | added | juan | @AymanMoussa Your function is not continuous at $\mathbf{T}$. | |
| Nov 14, 2020 at 16:15 | comment | added | Ayman Moussa | @juan For an explicit counter-example you can like this on $[0,1]$ : search $f$ of the form $f(x)=\sum_{p\geq 0} a_p x^p$ with $a_p\geq 0$ so that $\|f^{(n)}\|_{\infty} = f^{(n)}(1)$ for all $n\geq 0$. For such an $f$, we have $\sum_{n\geq 0} \frac{\|f^{n)}\|_\infty}{n!} = \sum_{n\geq 0} \frac{1}{n!}\sum_{p\geq n}a_p \frac{p!}{(p-n)!} = \sum_{p\geq 0} 2^p a_p$. Since $f'(x)=\sum_{p\geq 0} p a_{p+1} x^p$, you're searching for a sequence $(a_p)_p$ such that $\sum_{p} 2^p a_p<+\infty$ as well as $\sum_{p} 2^p p a_{p+1} = +\infty$. This is for instance satisfied for $a_p=\frac{1}{p^2 2^p}$. | |
| Nov 12, 2020 at 19:54 | comment | added | juan | @YemonChoi Thanks, I see. The examples $f$ will be weird functions. | |
| Nov 12, 2020 at 18:38 | comment | added | Yemon Choi | @juan it can't be closed under differentiation because of the Singer-Wermer theorem | |
| Nov 12, 2020 at 10:45 | history | edited | Ayman Moussa | CC BY-SA 4.0 |
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| Nov 12, 2020 at 10:45 | vote | accept | Ayman Moussa | ||
| Nov 11, 2020 at 19:20 | comment | added | juan | @AymanMoussa I think that it is not closed by differentiation. Although it is not easy to find $f\in A$ with $f\not\in A$. | |
| Nov 11, 2020 at 18:25 | comment | added | juan | @AymanMoussa Consider the case $d=1$, and let $A$ be the set of function of $\mathcal{C}^\infty$ such that $$\Vert f\Vert=\sum_{n=0}^\infty \frac{\Vert f^{(n)}\Vert_\infty}{n!}<+\infty.$$ It is not this an example of what you want? | |
| Nov 11, 2020 at 14:27 | history | edited | Ayman Moussa | CC BY-SA 4.0 |
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| Nov 11, 2020 at 13:57 | history | edited | Ayman Moussa | CC BY-SA 4.0 |
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| Nov 11, 2020 at 10:16 | history | edited | Ayman Moussa | CC BY-SA 4.0 |
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| Nov 11, 2020 at 0:23 | comment | added | Yemon Choi | @RomainGicquaud the subalgebra you describe is not Banach | |
| Nov 10, 2020 at 22:31 | comment | added | Qiaochu Yuan | @Romain: the natural "nondegeneracy" assumption here, I think, is to ask for a subalgebra that separates points. | |
| Nov 10, 2020 at 21:53 | comment | added | Yemon Choi | @QiaochuYuan your guess is correct BTW :) | |
| Nov 10, 2020 at 21:52 | comment | added | Yemon Choi | I am almost sure that the Singer-Wermer theorem gives a negative answer to your question, except for a small uncertainty (caused by my brain-tiredness rather than your wording) about whether your assumptions imply that $A$ is continuously embedded in $C^\infty$. I have written up what I can prove as an answer, but apologize in advance if I have missed some subtlety | |
| Nov 10, 2020 at 21:50 | answer | added | Yemon Choi | timeline score: 2 | |
| Nov 10, 2020 at 21:45 | comment | added | Olivier Bégassat | In the vein of @QiaochuYuan's proposal, what about the algebra of maps $\Bbb{U}\to\Bbb{C}$ that are restrictions of power series with radius of convergence = 2? I guess differentiation is discontinuous. | |
| Nov 10, 2020 at 21:40 | comment | added | Romain Gicquaud | With your assumptions, you still allow for the subalgebra of functions that do not depend, say on $x_1$. | |
| Nov 10, 2020 at 21:08 | history | edited | Ayman Moussa | CC BY-SA 4.0 |
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| Nov 10, 2020 at 20:50 | comment | added | Ayman Moussa | @DCM I don't know about Shilov, I'll look. This really is not my area of expertise. | |
| Nov 10, 2020 at 20:49 | comment | added | Ayman Moussa | @ChristianRemling you're right, I'll correct it, thanks. | |
| Nov 10, 2020 at 20:35 | comment | added | DCM | I should have said "Banach-ness, combined with being stable under differentiation" in my comment above ;) | |
| Nov 10, 2020 at 20:09 | comment | added | DCM | My suspicion is that 'being stable under differentiation' isn't compatible with how fast the Fourier coefficients have to decay for $A\subset C^\infty$ (thinking about the $d=1$ case). Stability under multiplication isn't inconsistent with $A\subset C^\infty$ (some Beurling algebras and some 'Dales-Davie' algebras consist entirely of $C^\infty$ functions). Shilov might be a good name to search if you're looking for results like this (although you might know that already). | |
| Nov 10, 2020 at 19:48 | comment | added | Qiaochu Yuan | When $d = 1$, what about the disk algebra of continuous functions on the closed unit disk which are holomorphic in the interior, equipped with the sup norm? I guess this might not be closed under differentiation though. | |
| Nov 10, 2020 at 18:48 | comment | added | Christian Remling | I take it you mean with the obvious algebraic operations (defined pointwise). In that case, $A$ would be automatically closed under multiplication, and this operation is continuous, so you don't need to ask for this separately. | |
| Nov 10, 2020 at 11:20 | history | edited | Denis Serre | CC BY-SA 4.0 |
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| Nov 10, 2020 at 9:30 | comment | added | Alexander Schmeding | I assume that the answer will be no i you assume that the algebra is rich enough (at least that might be a criterion to kick out something). I would think that something like the classical arguemnt why $C^\infty (\mathbb{R},\mathbb{R})$ is not a Banach space making differentiation continuous would work: n this case one finds that differentiation is a continuous linear operator with a countable unbounded family of eigenvalues (the exponentials). Perhaps this could be adapted... | |
| Nov 10, 2020 at 9:28 | history | edited | Ayman Moussa | CC BY-SA 4.0 |
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| Nov 10, 2020 at 9:28 | comment | added | Ayman Moussa | Hum, you're right. I will specify that I want it to be infinite dimensionnal. The goal was to be able to produce non trivial solutions ! Thanks for the correction ! | |
| Nov 10, 2020 at 9:22 | comment | added | user130903 | The answer is: yes. Take the algebra of constant functions. | |
| Nov 10, 2020 at 8:58 | history | asked | Ayman Moussa | CC BY-SA 4.0 |