Timeline for Integral representation of tempered distributions
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Apr 11, 2020 at 16:44 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Typo removal and minor improvement
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| Apr 9, 2020 at 14:32 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Added some remarks on what can be an integral representation of functionals
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| Apr 9, 2020 at 14:04 | comment | added | Daniele Tampieri | @IamWill surely yes, since $f\in\mathcal{S}(\Bbb R^d)$ implies it is $C^\infty$-smooth and rapidly decreasing, thus \eqref{1} has a precise, standard meaning as a $n$-linear functional. I am adding a few remarks on the topic of integral representations of functionals to my answer, by the way. | |
| Apr 9, 2020 at 13:57 | comment | added | JustWannaKnow | @DanieleTampieri thank you for the answer! You mentioned the answer to my previous question on the functional derivative. In this case (where $K\in \mathcal{S}'(\mathbb{R}^{d})$ is a derivative of some function $f: \mathcal{S}(\mathbb{R}^{d}) \to \mathbb{C}$, can we garantee the integral representation? This was my first post and it seems we can, but i not sure yet. | |
| Apr 9, 2020 at 12:47 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Minor addition and typo correction
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| Apr 9, 2020 at 6:46 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Adding an intuitive explanation of the kernel theorem and a reference
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| Apr 9, 2020 at 6:38 | comment | added | Daniele Tampieri | @leomonsaingeon: I agree with you, but $\delta$ it is a Radon measure, so it can be stated that $$ K(\varphi)=\int_{\Bbb R} \varphi \mathrm{d}\mu_{\delta_0}=\varphi(0)$$ at least by (though customary) abuse of notation: I wanted to give an example where the distribution, thought being slowly increasing, in not even a measure. | |
| Apr 9, 2020 at 6:31 | comment | added | leo monsaingeon | An even simpler counterexample: In dimension 1 (with $N=1$ and $n_1=1$) let $K$ be the Dirac mass distribution at the origin $K(\phi):=\phi(0)$. Then obviously $K$ is tempered, and it is a classical exercise to show that it cannot be written as an integral w.r.t. the Lebesgue measure $dx$ on the real line. No need for derivatives or tensor products... | |
| Apr 9, 2020 at 6:14 | history | answered | Daniele Tampieri | CC BY-SA 4.0 |