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Timeline for Uniform boundedness of integral?

Current License: CC BY-SA 4.0

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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