I'm trying to solve this question below:
I need a hint to begin to solve this question.
Help with this question about rank of Hessians
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real-analysis
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0The answer to the first question in your title is certainly YES if $f$ and $\varphi$ are well-defined functions (which can be said without reading the context). – 2017-01-16
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0@GDumphart so did the author make a mistake? – 2017-01-16
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0True, $\varphi (a)=b$, but Hessian does not depend on the function's value _in that point alone_. – 2017-01-16
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0@IvanNeretin could you explain a little bit more? thank you. – 2017-01-16
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0See, they require $\varphi$ to be a diffeomorphism for a reason. Otherwise you'd be able to construct a pathological example (say, a function without inverse) which would violate the statement, despite still having $\varphi(a)=b$. – 2017-01-16