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Standard Morse lemma states that if the singularity of a smooth function $f$ is non-degenerate, one can choose coordinates such that function has a "quadratic" form $f = \sum \limits_i x_i^2 - \sum \limits_j x_j^2$. To prove this one uses an integral representation $f(x) - f(0) = \int \limits_0^1 \frac{d}{dt}f(tx) dt$ to get a quadratic form with in general non-constant coefficients $f(x) = \sum \limits_{ij} h_{ij} (x) x_i x_j$.

Now suppose that we have two functions $f$ and $g$ of two variables. We know that $g$ has a non-degenerate minimum at origin, while $f$ does not have any singularities. When can we choose coordinates such that $f = \alpha x, g = x^2 + y^2$, where $\alpha$ is constant?

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In general, this is not possible. One needs a neighbourhood $U$ of the origin such that on every level set $f^{-1}(c)\cap U$, the minimum of $g|_U$ is exactly $(c/\alpha)^2$. If this condition is satisfied, then regard $g$ as a fibrewise Morse function for the family $f\colon U\to V\subset\mathbb R$. Because the Morse lemma holds in families, one can now prove the claim.

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  • $\begingroup$ I am trying to visualize the condition and not succeeding, is there a simple explicit example where it does not hold? $\endgroup$
    – j.c.
    Mar 7, 2016 at 18:05
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    $\begingroup$ Take $f=x$ and $g=x^2+x^4+y^2$. The level sets of $f$ are vertical lines. In each level set over $x$, the minimum of $g$ is $x^2+x^4$, which is not proportional to $f(x)^2$. $\endgroup$ Mar 7, 2016 at 20:11

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