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May 16, 2019 at 20:57 vote accept JohnA
May 14, 2019 at 21:02 answer added Dima Pasechnik timeline score: 4
May 14, 2019 at 20:51 comment added Dima Pasechnik well, if $g_j$ is SOS then $g_j(x)\leq 0$ iff $g_j(x)=0$, i.e. each individual square in the SOS decomposition of $g_j$ must be 0. So it's back to Lassere's case, with equality constraints only.
May 14, 2019 at 20:48 comment added JohnA Obvious typo fixed, thanks for pointing it out.
May 14, 2019 at 20:47 history edited JohnA CC BY-SA 4.0
edited body
May 14, 2019 at 20:46 comment added Dima Pasechnik if $g_j$ is an SOS then $g_j(x)\geq 0$ for any $x$, i.e. it can be just removed from the set of constraints.
May 14, 2019 at 20:33 history edited JohnA CC BY-SA 4.0
edited title
May 14, 2019 at 19:17 history asked JohnA CC BY-SA 4.0