You are talking for one dimension; there are polynomials in two variables with degree 6 which are non-negative on $\mathbb R^2$ and not sum of squares of polynomials (same thing in three dimensions, degree 4).

@GiorgioMetafune Assuming that $u$ is even and real-valued, you get that the Fourier transform of $u\ast u$ is even and non-negative. Now an analytic non-negative function of one variable is a sum of squares, something which is not true in higher dimensions. Now, in one dimension you can write $u\ast u$ as a finite sum of $u_j\ast u_j$ where each $u_j$ is compactly supported; some details need to be completed, but there is clearly no hope to get something of that type in higher dimensions.

@ChristianRemling Thanks for your answer. It seems that in one dimension, it is possible to find an holomorphic square root, for instance if you know that $u\ast u$ is even real-valued with a non-negative Fourier transform. However, in more than one dimension, I guess that even with the more stringent assumption above, it is not possible to find an holomorphic square root.

Wouldn't that imply that $BMO$ is isomorphic to $L^{n/\alpha}$? Note that this is impossible since $L^p$ is reflexive for all $p\in (1,+\infty)$ whereas $BMO$ is not reflexive.

It seems correct, but the Morse lemma is truly the adequate answer (in my opinion), since you do have a normal form for your function near $0$; also the Morse lemma is not difficult to prove and its proof is shorter than you specific argument.