Timeline for An integrable estimate of the Hölder constant of the map $x \mapsto \int_{\mathbb R^d} f(y) \partial_1 \partial_1 g_t (x-y) \, \mathrm d y$
Current License: CC BY-SA 4.0
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| Jan 19 at 6:06 | comment | added | Akira | @PietroMajer A well-known upper bound is $|\partial_1^2 g_t( y)| \lesssim \frac{g_{4t} (y)}{4t}$, so I guess it is not integrable. | |
| Jan 19 at 4:20 | vote | accept | Akira | ||
| Jan 18 at 23:39 | answer | added | Giorgio Metafune | timeline score: 4 | |
| Jan 18 at 20:28 | comment | added | Akira | @PietroMajer Thank you so much for your elaboration! I have updated my question to make it non-trivial. Please have a check on it. | |
| Jan 18 at 20:27 | history | edited | Akira | CC BY-SA 4.0 |
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| Jan 18 at 20:23 | history | undeleted | Akira | ||
| Jan 18 at 20:18 | history | deleted | Akira | via Vote | |
| Jan 18 at 20:10 | comment | added | Pietro Majer | I’d say the answer is yes, since F is a convolution of f with an $L^1 $ function G. Write $F(x)-F(x’) $ with the convolution integrands as $f(x-y)G(y)$ and $f(x’-y)G(y),$ and use the Hoelder inequality for $f.$ | |
| Jan 18 at 20:06 | history | edited | Akira | CC BY-SA 4.0 |
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| Jan 18 at 20:04 | history | edited | Akira | CC BY-SA 4.0 |
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| Jan 18 at 20:01 | history | edited | Akira | CC BY-SA 4.0 |
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| Jan 18 at 19:48 | history | asked | Akira | CC BY-SA 4.0 |