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In the proof of the stationary phase method (at least the one I have seen) Morse lemma shows up, which states: Let $g:\mathbb R^n\to \mathbb R$ be a function of class $C^\infty$ for which $0$ is a non-degenerate critical point ($\nabla g (0) =0$ and the Hessian at $0$ has trivial kernel). Then there exists a neighbourhood $U$ of $0$ and a $C^\infty$ diffeomorphism $$ \varphi = (x_1, \ldots , x_n): U \to V \subset \mathbb R^n\, , $$ with $\varphi (0)=0$ and such that the map $\tilde{g} = g\circ \varphi^{-1}$ (namely $\varphi$ in the "$x$-coordinates") takes the form $\tilde{g} (x) = g(0)- x_1^2 - \ldots - x_\lambda^2 + x_{\lambda+1}^2 + \ldots + x_n^2$.

Fix $\varepsilon>0$. I was wondering what information is necessary in order to guarantee that $U$ contains an $\varepsilon$-box around $\mathbf{0}$ ? Any comments or references are appreciated. Thank you.

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  • $\begingroup$ @reuns I am confused by your question... $\endgroup$
    – Johnny T.
    Commented Jan 6, 2018 at 13:18
  • $\begingroup$ Locally $g(x) = h \circ \phi(x)$ where $\phi$ is bi-smooth and $h(x)=h(0)+ a^\top x+ \frac12 x^\top B x$ is a quadratic polynomial. Then $\nabla_u h(x) = a^\top u+ x^\top B u$ and $\nabla_v g(x) = \nabla_{\Phi(x) v}h(x) =(a^\top + x^\top B) \Phi(x)v$ where $\Phi(x)_{ij} = \partial_j\phi(x)_i$. This should impose some conditions on $\nabla g$ (together with $\Phi(x)$ inversible and $\Phi(x)\cdot dx$ being an exact form) which when not satisfied means $g(x) = h \circ \phi(x)$ cannot be true globally. $\endgroup$
    – reuns
    Commented Jan 7, 2018 at 1:05

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