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Mar 28, 2022 at 15:16 vote accept Puzzled
Mar 6, 2022 at 3:25 comment added Iosif Pinelis Do you have a further response to the answer below?
Mar 3, 2022 at 22:04 comment added Iosif Pinelis I am not sure if such a characterization is possible (except for a tautological one). Yet, you may want to post the characterization question separately, possibly elsewhere.
Mar 3, 2022 at 22:02 answer added Iosif Pinelis timeline score: 1
Mar 3, 2022 at 21:24 comment added Puzzled I got it. You are definitely right. Do you think it could be possible to characterize the set one has to remove from $\mathbb{R}^{18}$ for the positivity property I required to hold?
Mar 3, 2022 at 21:04 comment added Iosif Pinelis This does not address what I proposed. I said "close enough". not "proportional to". That is, what I proposed is $f_i=a_{i,0}+a_{i,1}x+a_{i,2}y+a_{i,3}xy+a_{i,4}x^2+a_{i,5}y^2$ for $i=0,1,2$, with $(a_{i,0},a_{i,1},a_{i,2},a_{i,3},a_{i,4},a_{i,5})$ close to $(1,0,0,0,1,1)$ for each $i=0,1,2$. The set of such triples $(f_0,f_1,f_2)$ contains a nonempty $18$-dimensional open ball.
Mar 3, 2022 at 19:52 comment added Puzzled I addressed your example in the new version of the question. I think the $G_i$ could be simply polynomials. For instance the $f_0,f_1,f_2$ in your example are all inside the zero locus of $9$ linear polynomials.
Mar 3, 2022 at 19:50 history edited Puzzled CC BY-SA 4.0
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Mar 3, 2022 at 14:15 comment added Iosif Pinelis Well, then clearly without loss of generality $r=1$. Also, are the $G_i$'s to be (piecewise-)smooth? If so, the answer is no: Take e.g. $a>1$ and $f_0,f_1,f_2$ each close enough to $x^2+y^2+1$.
Mar 3, 2022 at 13:59 comment added Puzzled "Random" would be the following: each $f_i$ depends on $6$ real parameters, so we can see the triple $(f_0,f_1,f_2)$ as a point in $\mathbb{R}^{18}$. There is a subset of the form $S = \{G_1 = \dots = G_r = 0\}\subset\mathbb{R}^{18}$, where the $G_i$ are functions of the coefficients of the $f_i$, such that for all $(f_0,f_1,f_2) \in\mathbb{R}^{18}\setminus S$ we have that $f(x_0,y_0)\geq 0$ for some $(x_0,y_0)\in\mathbb{R}^2$. I guess that from the probabilistic point of view this implies that the property I am requiring holds with probability $1$.
Mar 3, 2022 at 13:23 comment added Iosif Pinelis It is unclear what you mean by 'for a "random" choice': Do you mean with probability $1$ or with a nonzero probability? Also, it is unclear what you mean by "a suitable definition of random": what would be suitable for you and what would be not -- that should be clearly specified, using appropriate quantifiers. As your own example suggests, some definitions of random would be unsuitable to you. On MathOverflow, you should be specific enough so that it be quite clear what constitutes an answer to your question and what does not.
S Mar 3, 2022 at 9:59 review First questions
Mar 3, 2022 at 10:48
S Mar 3, 2022 at 9:59 history asked Puzzled CC BY-SA 4.0