Timeline for Uniform boundedness of integral?
Current License: CC BY-SA 4.0
15 events
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Jul 27, 2020 at 19:48 | comment | added | Iosif Pinelis | @user64494 : The equality is correct, too, because of the $\sup$ on both sides of it. | |
Jul 27, 2020 at 19:46 | comment | added | user64494 | It should be $\le$ instead of $=$ in line 3. | |
Jul 27, 2020 at 19:45 | comment | added | Iosif Pinelis | @user64494 : Thank you. Have done this too. | |
Jul 27, 2020 at 19:43 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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Jul 27, 2020 at 19:43 | comment | added | user64494 | One more $y$ should be replaced in the above. | |
Jul 27, 2020 at 19:41 | comment | added | Iosif Pinelis | @user64494 : I have removed $y$ and rewritten everything only in terms of $b$ (and $x$). | |
Jul 27, 2020 at 19:40 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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Jul 27, 2020 at 19:28 | comment | added | Iosif Pinelis | @user64494 : $y$ is a dummy variable, and thus can be replaced by any symbol. Later, we substitute $|b|$ for $y$. I only did not want to use $b$ in place of $y$, because $b<0$ and hence $b\ne|b|$. | |
Jul 27, 2020 at 19:06 | comment | added | user64494 | Iosif Pinelis, can you do the right edit, in particular, defining $y$? I think, $y=b$ in your notation. | |
Jul 27, 2020 at 18:58 | comment | added | Iosif Pinelis | @user64494 : Oops! I have now inserted the missing factor $x^{-1/4}x^{-1/4}=x^{-1/2}$. Since this factor is bounded for $x\ge1$, it does not affect the conclusion. | |
Jul 27, 2020 at 18:54 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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Jul 27, 2020 at 18:40 | comment | added | user64494 | Sorry, I don't see any denominator in the above. Could you elaborate your comment? TIA. | |
Jul 27, 2020 at 18:13 | comment | added | user64494 | Do you take into account that $dt=\lambda^{1/4}\,ds$? I don't see $\lambda^{1/4}$ in the above. | |
Jul 27, 2020 at 17:01 | vote | accept | Sascha | ||
Jul 27, 2020 at 16:59 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |