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Jan 19, 2018 at 10:57 comment added Dima Pasechnik Motzkin polynomial strikes again (the function given by David is the famous Motzkin's example of a non-SOS non-negative polynomial).
Dec 28, 2017 at 21:59 comment added David E Speyer Compare the constant terms on each side to get that the constant terms of all the $f_i$ are $0$. Then compare the quadratic terms to get that the linear parts of each $f_i$ are $0$. Then compare the degree $4$ terms to get that the degree $2$ parts of each $f_i$ are zero.
Dec 28, 2017 at 21:24 comment added Pietro Majer Why the $f_i$ vanish up the third degree?
Dec 28, 2017 at 18:12 comment added Pedro Lauridsen Ribeiro Moreover, one should probably add to your answer the (obvious but clarifying) fact that $x^4y^2+y^4z^2+z^4x^2−ax^2y^2z^2$, $a\in(0,3]$ only vanishes at the union of three straight lines $\{(t,0,0)\ |\ t\in\mathbb{R}\}\cup\{(0,t,0)\ |\ t\in\mathbb{R}\}\cup\{(0,0,t)\ |\ t\in\mathbb{R}\}$ due to the sharpness of the AM-GM inequality, and that one is considering the Taylor(-MacLaurin) expansion (with remainder) around $(0,0,0)$.
Dec 28, 2017 at 15:19 comment added Pedro Lauridsen Ribeiro In principle, the (counter)example you proposed should be compactly supported, but (as argued in my question) this can be easily achieved by multiplying everything by a squared bump function, so that's OK. That being said, recall that $f_j\rightarrow f$ in $\mathscr{D}(\mathbb{R}^n)$ if there is $K\subset\mathbb{R}^n$ compact such that $\mathrm{supp}f,\mathrm{supp}f_j\subset K$ for all $j$ and all derivatives of $f_j$ converge uniformly in $K$ to those of $f$, so one can start from a (counter)example in $\mathscr{C}^\infty$ with uniform convergence of all derivatives in compacta.
Dec 28, 2017 at 10:48 history answered David E Speyer CC BY-SA 3.0