Timeline for Do you know important theorems that remain unknown?
Current License: CC BY-SA 4.0
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| Jun 7, 2019 at 21:41 | history | edited | Piotr Hajlasz | CC BY-SA 4.0 |
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| Jul 23, 2018 at 4:17 | history | edited | Piotr Hajlasz | CC BY-SA 4.0 |
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| Jun 5, 2018 at 19:45 | comment | added | Piotr Hajlasz | @MarkLewko Do you have a reference to the paper of Littlewood? I googled your quote and found it in notes of Tarry Tao, but there was no reference to a particular result to Littlewood. | |
| Jun 5, 2018 at 19:29 | comment | added | Mark Lewko | I believe this predates Hormander. I've seen this attributed to Littlewood as the "the higher exponents are always on the left" principle. | |
| Apr 23, 2018 at 19:11 | history | edited | Piotr Hajlasz | CC BY-SA 3.0 |
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| Apr 7, 2018 at 21:57 | history | edited | Suvrit | CC BY-SA 3.0 |
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| Apr 3, 2018 at 11:45 | history | edited | Martin Sleziak | CC BY-SA 3.0 |
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| Mar 27, 2018 at 18:05 | comment | added | paul garrett | @JohannesHahn, your terminological objection certainly makes sense, but/and, depending on notation, the visibly reasonable terminology can vary. E.g., from the viewpoint of your comment, $T\circ \tau =\tau\circ T$ makes $T$ look $\tau$ equivariant, indeed. But/and if we declare that the action of $\tau$ is by $\tau\circ T\circ \tau^{-1}$, then $T$ "becomes" $\tau$ invariant. In my own experience this comes up a lot, and I use "equivariant" and "invariant" indiscriminantly, in part because of the inevitable ambiguity. (Esp., e.g., on ${\mathrm Hom}(X,Y)$ spaces with bimodules...) | |
| Mar 27, 2018 at 16:02 | history | made wiki | Post Made Community Wiki by Todd Trimble | ||
| Mar 27, 2018 at 15:30 | history | edited | Piotr Hajlasz | CC BY-SA 3.0 |
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| Mar 27, 2018 at 15:11 | comment | added | Piotr Hajlasz | In harmonic analysis such operators are called translation-invariant, see reference [1] and many other books in harmonic analysis. | |
| Mar 27, 2018 at 15:07 | comment | added | Johannes Hahn | Are those operators really called translation invariant ? Shouldn't they be called equivariant ? "invariant" to me suggests an condition of the form $T\circ\tau_a = T$. | |
| Mar 27, 2018 at 14:58 | history | edited | Piotr Hajlasz | CC BY-SA 3.0 |
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| Mar 27, 2018 at 14:26 | history | answered | Piotr Hajlasz | CC BY-SA 3.0 |