Timeline for Chain rule for Newton-derivative
Current License: CC BY-SA 3.0
8 events
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Nov 16, 2017 at 9:02 | comment | added | Malte Winckler | Thank you both for your remarks and this extensive literature! I consider this question solved. | |
Nov 15, 2017 at 19:54 | comment | added | Christian Clason | That's a standard result that can be found in, e.g., the books by Ulbrich (Theorem 3.69), Ito and Kunisch (Lemma 8.4), or (shameless plug) my lecture notes (Theorem 9.2). Note that there is no chain rule for two Gâteaux differentiable functions. And a(!) Newton derivative is given by the pointwise multiplication with $v(x) = 1$ if $u(x)\in [-1,1]$ and $0$ else -- iff, as @Dirk writes, your function is considered as mapping to $L^p$ for $p<2$, see again the mentioned literature. (For both results, the proof is virtually the same as for Fréchet derivatives.) | |
Nov 15, 2017 at 17:42 | comment | added | Dirk | To anybody who has doubts: This is not the Frechet derivative and the $x+y$ in the argument of G is correct. More to the question: First, the Newton derivative is not unique. It could be that a Newton derivative in this case may be u where it is between -1 and +1 and zero elsewhere but I remember vaguely that there may be a problem with the norms here... (Buzzword "norm gap", probably in a paper by Kunisch or Ulbrich). | |
Nov 15, 2017 at 15:30 | comment | added | Malte Winckler | It was a typo. I edited the question. And regarding your comment with the Fréchet derivative, I think it's a little weaker and especially only a local property | |
Nov 15, 2017 at 15:27 | history | edited | Malte Winckler | CC BY-SA 3.0 |
fixed typos
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Nov 15, 2017 at 14:51 | comment | added | Federico Poloni | I assume that $v+y$ should be something else, because $v$ is undefined, but it looks like this is another name for the Fréchet dferivative. | |
Nov 15, 2017 at 14:42 | review | First posts | |||
Nov 15, 2017 at 14:52 | |||||
Nov 15, 2017 at 14:41 | history | asked | Malte Winckler | CC BY-SA 3.0 |