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Consider function from Hilbert space to real numbers. $F(x)=\| Ax\|$. My question how to find it's derivative $F'(x)$.

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TeX: It's better to use \| instead of \parallel because of spacing. Compare: $\|Ax\|$ and $\parallel Ax\parallel$. Also, you cannot write \parallelAx\parallel, since TeX then interprets \parrallelAx as a single name of a macro. You can try \parallel{}Ax\parallel or \parallel Ax\parallel. –  Martin Sleziak Jan 28 '12 at 7:48

1 Answer 1

I assume that $A$ is a bounded linear operator on a Hilbert space $X$, and we want to compute Frechet derivative. Consider two functions $$ G:X\to\mathbb{R}: x\mapsto\Vert x\Vert=\sqrt{\langle x,x\rangle} $$ $$ H:X\to X:x\mapsto A(x) $$ One may show that $G'(x)(h)=\frac{\langle x, h\rangle}{\Vert x \Vert}$ (for $x=0$, this derivative doesn't exist) and $H'(x)(h)=A(h)$ where $h\in X$. Then $$ F'(x)(h)=(G(H(x)))'(h)=(G'(H(x))\circ H'(x))(h)=G'(H(x))(H'(x)(h))= $$ $$ G'(H(x))(A(h))=\frac{\langle A(x), A(h)\rangle}{\Vert Ax\Vert} $$

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It's still wrong: You should have $\| Ax\|$ in the denominator. –  Christian Blatter Jan 28 '12 at 20:40
So why don't you haven't edited this post? –  no identity Jan 28 '12 at 20:41
Pointing out mathematical errors should always be welcomed. –  Jonas Meyer Jan 28 '12 at 20:50
This is a waste of time, because incorrect answer will hang until its author will come back. One should edit mistakes and (if he wants) mention about it in comments. And, of course, I appreciate Christian Blatter attention. –  no identity Jan 28 '12 at 20:53
@Norbert: I disagree. If someone wants to correct a mathematical error that could be very generous, but above what should be expected. In some cases it could be problematic, because the "corrector" could be wrong. But to me the main point here is that it is much better that it is pointed out in a comment than not corrected at all, and those pointing our errors are doing something positive for the site. (Regarding appropriate use of editing, there is a meta post where you could contribute: What is to be edited?) –  Jonas Meyer Jan 28 '12 at 21:00

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