Timeline for Does a weaker condition than vanishing derivative imply a function being constant?
Current License: CC BY-SA 3.0
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| Apr 30, 2020 at 0:51 | comment | added | LSpice | The link seems to be broken (and the text seems to suggest it's to worldcat.org/title/… , but I can't find a web page for that journal); but a paper by Aull provides a theorem by the same name: Aull - The first symmetric derivative. The result you cite is Theorem 1 of that paper; Theorem 2, saying that a continuous symmetric derivative of a continuous function implies the existence of the usual derivative, is also relevant here. | |
| Dec 10, 2012 at 16:45 | vote | accept | qianzhang | ||
| Dec 10, 2012 at 6:52 | history | edited | Jack Huizenga | CC BY-SA 3.0 |
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| Dec 10, 2012 at 5:17 | history | edited | Jack Huizenga | CC BY-SA 3.0 |
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| Dec 10, 2012 at 3:10 | comment | added | qianzhang | Thank you for your answer. Do you have any idea about the absolute continuity of $f$? An immediate corollary of the "quasi-mean value theorem" is that if $f^s$ is bounded, then $f$ is Lipschitz, but I still have no idea about the general situation. | |
| Dec 10, 2012 at 1:47 | history | answered | Jack Huizenga | CC BY-SA 3.0 |