Timeline for How to show this Holder bound?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 5, 2012 at 16:59 | comment | added | user25266 | @Davide do you think I just create a new norm by adding on $[u_x]_\alpha$? It shouldn't break anything... | |
| Aug 4, 2012 at 17:30 | comment | added | user25266 | Hmm. Might be able to do something by adding and subtracting the same term in the numerator. | |
| Aug 4, 2012 at 17:13 | comment | added | user25266 | Sorry to keep asking, but do you know how I can bound $[u_x]_\alpha$? I don't think I can get anything from using the MVT or the mean value inequality (I get a $u_{xt}$ term which I can't do anything with). | |
| Aug 4, 2012 at 10:12 | comment | added | user25266 | I don't think that square should be there. From the source I'm using and from Krylov's book, the powers of |x-y| and |t-s| are different. I'm not really sure what the point of defining the space in such a way is though. | |
| Aug 3, 2012 at 18:16 | comment | added | Davide Giraudo | In your last expression (as in the first of the OP), a square is missing, I think ($|t-s|^{\color{red}2}$). This supremum can be bounded by a constant involving the diameter of the compact, namely, $\operatorname{diam}(S)^{1-\alpha}$. | |
| Aug 3, 2012 at 17:09 | comment | added | user25266 | Thanks. I believe you're right but I can't show how $[u]_\alpha$ and $[u_x]_{\alpha}$ are bounded above by something useful using the MVT. For $[u]_{\alpha}$, I get something like needing to show $\sup\left(\frac{(x-y)^2 + (t-s)^2}{(|x-y|^2 + |t-s|)^{\alpha}}\right)^{\frac{1}{2}} < \infty.$ Does that look about what you meant? | |
| Aug 3, 2012 at 17:06 | vote | accept | user25266 | ||
| Aug 3, 2012 at 10:29 | history | edited | Davide Giraudo | CC BY-SA 3.0 |
added 305 characters in body; added 55 characters in body
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| Aug 3, 2012 at 10:14 | comment | added | Davide Giraudo | I hope it's not a problem, because with mean value theorem we get an equivalent norm when we include them. | |
| Aug 3, 2012 at 10:02 | comment | added | user25266 | Thanks for the answer, but I don't see how you get the last inequality. The $C^{2,\alpha}$ norm of $u$ does not include the seminorms $[u_x]_{\alpha}$ and $[u]_{\alpha}$ that are present on the lhs unfortunately. | |
| Aug 3, 2012 at 9:52 | history | answered | Davide Giraudo | CC BY-SA 3.0 |