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9
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Here is a counterexample: Take $n=2$ variables $X$ and $Y$. Let $L_1,\dots,L_5$ be linear polynomials such that
$$S:=\{ (x, y) \in {\mathbb R}^2 ~|~ L_i(xL_i(x,y) \ge 0\}$$ is a pentagon inscribed in the unit circle. Furthermore set $P:=1-X^2-Y^2$. Assume we could write $P$ as the sum of a globally nonnegative quadratic polynomial $Q$ and nonnegative linear combinations of the $L_i$ and $L_iL_j$. Now $P$ vanishes at the vertices of the pentagon and each $L_i$ is nonnegative at these vertices. Therefore $Q$ vanishes also at the vertices. But being a nonnegative quadratic polynomial, $S$ is a sum of squares of linear polynomials which all have also to vanish at the vertices and therefore are identically zero. This shows that $S$ is the zero polynomial. Now notice that each of the $L_i$ and $L_iL_j$ is strictly positive on at least one of the vertices of the pentagon (at which $P$ vanishes, of course). Since $P$ is a nonnegative linear combination of the $L_i$ and $L_iL_j$, this shows that $P=0$.
If the set $S$ defined by the $L_i$ has non-empty interior, then the convex cone of quadratic polynomials which can be written as a globally nonnegative quadratic polynomial $Q$ and nonnegative linear combinations of the $L_i$ and $L_iL_j$ is closed. In fact, this follows from a much more general result on truncated quadratic modules, see e.g. the book of Marshall cited below (Lemma 4.1.4). This implies that, in the above counterexample, even $P+\varepsilon$ for small $\varepsilon>0$ will fail though this polynomial is strictly positive on $S$.
However, there are a lot theorems going into the direction of what you want. You might want to have a look at the following books...
- Marshall: Positive polynomials and sums of squares
- Prestel: Positive polynomials
- Bochnak, Coste, Roy: Real algebraic geometry
- Basu, Pollack, Roy: Algorithms in real algebraic geometry
- Knebusch, Scheiderer: Einführung in die reelle Algebra
- Andradas, Bröcker, Ruiz: Constructible sets in real geometry
...and the following articles...
Also the so-called "S-procedure" could be of interest for you.
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8
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Here is a counterexample: Take $n=2$ variables $X$ and $Y$. Let $L_1,\dots,L_5$ be linear polynomials such that
$$S:={ $S:=\{ (x, y) \in {\mathbb R}^2 ~|~ L_i(x) \ge 0}$$ 0\}$$ is a pentagon inscribed in the unit circle. Furthermore set $P:=1-X^2-Y^2$. Assume we could write $P$ as the sum of a globally nonnegative quadratic polynomial $Q$ and nonnegative linear combinations of the $L_i$ and $L_iL_j$. Now $P$ vanishes at the vertices of the pentagon and each $L_i$ is nonnegative at these vertices. Therefore $Q$ vanishes also at the vertices. But being a nonnegative quadratic polynomial, $S$ is a sum of squares of linear polynomials which all have also to vanish at the vertices and therefore are identically zero. This shows that $S$ is the zero polynomial. Now notice that each of the $L_i$ and $L_iL_j$ is strictly positive on at least one of the vertices of the pentagon (at which $P$ vanishes, of course). Since $P$ is a nonnegative linear combination of the $L_i$ and $L_iL_j$, this shows that $P=0$.
If the set $S$ defined by the $L_i$ has non-empty interior, then the convex cone of quadratic polynomials which can be written as a globally nonnegative quadratic polynomial $Q$ and nonnegative linear combinations of the $L_i$ and $L_iL_j$ is closed. In fact, this follows from a much more general result on truncated quadratic modules, see e.g. the book of Marshall cited below (Lemma 4.1.4). This implies that, in the above counterexample, even $P+\varepsilon$ for small $\varepsilon>0$ will fail though this polynomial is strictly positive on $S$.
However, there are a lot theorems going into the direction of what you want. You might want to have a look at the following books...
- Marshall: Positive polynomials and sums of squares
- Prestel: Positive polynomials
- Bochnak, Coste, Roy: Real algebraic geometry
- Basu, Pollack, Roy: Algorithms in real algebraic geometry
- Knebusch, Scheiderer: Einführung in die reelle Algebra
- Andradas, Bröcker, Ruiz: Constructible sets in real geometry
...and the following articles...
Also the so-called "S-procedure" could be of interest for you.
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7
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Here is a counterexample: Take $n=2$ variables $X$ and $Y$. Let $L_1,\dots,L_5$ be linear polynomials such that
$S := { $S:={ (x, y) \in {\mathbb R}^2 | ~|~ L_i(x) \ge 0 }$ 0}$$ is a pentagon inscribed in the unit circle. Furthermore set $P:=1-X^2-Y^2$. Assume we could write $P$ as the sum of a globally nonnegative quadratic polynomial $Q$ and nonnegative linear combinations of the $L_i$ and $L_iL_j$. Now $P$ vanishes at the vertices of the pentagon and each $L_i$ is nonnegative at these vertices. Therefore $Q$ vanishes also at the vertices. But being a nonnegative quadratic polynomial, $S$ is a sum of squares of linear polynomials which all have also to vanish at the vertices and therefore are identically zero. This shows that $S$ is the zero polynomial. Now notice that each of the $L_i$ and $L_iL_j$ is strictly positive on at least one of the vertices of the pentagon (at which $P$ vanishes, of course). Since $P$ is a nonnegative linear combination of the $L_i$ and $L_iL_j$, this shows that $P=0$.
If the set $S$ defined by the $L_i$ has non-empty interior, then the convex cone of quadratic polynomials which can be written as a globally nonnegative quadratic polynomial $Q$ and nonnegative linear combinations of the $L_i$ and $L_iL_j$ is closed. In fact, this follows from a much more general result on truncated quadratic modules, see e.g. the book of Marshall cited below (Lemma 4.1.4). This implies that, in the above counterexample, even $P+\varepsilon$ for small $\varepsilon>0$ will fail though this polynomial is strictly positive on $S$.
However, there are a lot theorems going into the direction of what you want. You might want to have a look at the following books...
- Marshall: Positive polynomials and sums of squares
- Prestel: Positive polynomials
- Bochnak, Coste, Roy: Real algebraic geometry
- Basu, Pollack, Roy: Algorithms in real algebraic geometry
- Knebusch, Scheiderer: Einführung in die reelle Algebra
- Andradas, Bröcker, Ruiz: Constructible sets in real geometry
...and the following articles...
Also the so-called "S-procedure" could be of interest for you.
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6
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Here is a counterexample: Take $n=2$ variables $X$ and $Y$. Let $L_1,\dots,L_4$ L_1,\dots,L_5$ be the linear polynomials such that
$1\pm X$ and S := { (x, y) \in {\mathbb R}^2 | L_i(x) \ge 0 }$ 1\pm Y$ and is a pentagon inscribed in the unit circle. Furthermore set $P=2-X^2-Y^2$. P:=1-X^2-Y^2$. Assume we could write $P$ as the sum of a globally nonnegative quadratic polynomial $S$ Q$ and nonnegative linear combinations of the $L_i$ and $L_iL_j$. Now $P$ vanishes at the four points $(\pm 1,\pm 1)$ vertices of the pentagon and each $L_i$ is nonnegative at $(\pm 1,\pm 1)$these vertices. Therefore $S$ Q$ vanishes also at $(\pm 1,\pm 1)$the vertices. But being a nonnegative quadratic polynomial, $S$ is a sum of squares of linear polynomials which all have also to vanish at $(\pm 1,\pm 1)$ the vertices and therefore are identically zero. This shows that $S$ is the zero polynomial. Now notice that each of the $L_i$ and $L_iL_j$ is strictly positive on at least one of the points $(\pm 1,\pm 1)$ where vertices of the pentagon (at which $P$ vanishes, of course). Since $P$ is a nonnegative linear combination of the $L_i$ and $L_iL_j$, this shows that $P=0$.
If the set $S$ defined by the simultaneous inequalities $L_i\ge0$ L_i$ has non-empty interior, then the convex cone of quadratic polynomials which can be written as a globally nonnegative quadratic polynomial $S$ Q$ and nonnegative linear combinations of the $L_i$ and $L_iL_j$ is closed. In fact, this follows from a much more general result on truncated quadratic modules, see e.g. the book of Marshall cited below (Lemma 4.1.4). This implies that, in the above counterexample, even $P+\varepsilon$ for small $\varepsilon>0$ will fail though this polynomial is strictly positive on the set defined by the inequalities $L_i\ge0$.S$.
However, there are a lot theorems going into the direction of what you want. You might want to have a look at the following books...
- Marshall: Positive polynomials and sums of squares
- Prestel: Positive polynomials
- Bochnak, Coste, Roy: Real algebraic geometry
- Basu, Pollack, Roy: Algorithms in real algebraic geometry
- Knebusch, Scheiderer: Einführung in die reelle Algebra
- Andradas, Bröcker, Ruiz: Constructible sets in real geometry
...and the following articles...
Also the so-called "S-procedure" could be of interest for you.
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5
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Interesting question Here is a counterexample: Take $n=2$ variables $X$ and $Y$. Let $L_1,\dots,L_4$ be the polynomials $1\pm X$ and $1\pm Y$ and $P=2-X^2-Y^2$. Assume we could write $P$ as the sum of a globally nonnegative quadratic polynomial $S$ and nonnegative linear combinations of the $L_i$ and $L_iL_j$. Now $P$ vanishes at the four points $(\pm 1,\pm 1)$ and each $L_i$ is nonnegative at $(\pm 1,\pm 1)$. Therefore $S$ vanishes also at $(\pm 1,\pm 1)$. But being a nonnegative quadratic polynomial, $S$ is a sum of squares of linear polynomials which all have also to vanish at $(\pm 1,\pm 1)$ and therefore are identically zero. This shows that $S$ is the zero polynomial. Now notice that each of the $L_i$ and $L_iL_j$ is strictly positive on at least one of the points $(\pm 1,\pm 1)$ where $P$ vanishes. Since $P$ is a nonnegative linear combination of the $L_i$ and $L_iL_j$, this shows that $P=0$. If the set defined by the simultaneous inequalities $L_i\ge0$ has non-empty interior, then the convex cone of quadratic polynomials which can be written as a globally nonnegative quadratic polynomial $S$ and nonnegative linear combinations of the $L_i$ and $L_iL_j$ is closed. In fact, this follows from a much more general result on truncated quadratic modules, see e.g. the book of Marshall cited below (Lemma 4.1.4). This implies that, in the above counterexample, even $P+\varepsilon$ for small $\varepsilon>0$ will fail though this polynomial is strictly positive on the set defined by the inequalities $L_i\ge0$. However, there are a lot theorems going into the direction of what you want. You might want to have a look at the following books... http://www.math.uni-konstanz.de/~schweigh/publications/purestates.pdf
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4
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You said you have proved it in the one variable case but in my opinion it should be wrong already in that case: Take $L_1=1+X$, $L_2=1-X$ and $P=X^2$. Then each $L_i$ is linear, $P$ is quadratic and $P\ge0$ on the set where all $L_i$ are simultaneously nonnegative. On the other hand each product of some of the $L_i$ is positive at the origin, showing that the desired representation cannot exist.
If you allow however for $P$ being strictly positive, then you have much better chances Interesting question. You might want to have a look at the following books...
- Marshall: Positive polynomials and sums of squares
- Prestel: Positive polynomials
- Bochnak, Coste, Roy: Real algebraic geometry
- Basu, Pollack, Roy: Algorithms in real algebraic geometry
- Knebusch, Scheiderer: Einführung in die reelle Algebra
- Andradas, Bröcker, Ruiz: Constructible sets in real geometry
...and the following articles...
Also the so-called "S-procedure" could be of interest for you.
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3
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Interesting question You said you have proved it in the one variable case but in my opinion it should be wrong already in that case: Take $L_1=1+X$, $L_2=1-X$ and $P=X^2$. Then each $L_i$ is linear, $P$ is quadratic and $P\ge0$ on the set where all $L_i$ are simultaneously nonnegative. On the other hand each product of some of the $L_i$ is positive at the origin, showing that the desired representation cannot exist.
If you allow however for $P$ being strictly positive, then you have much better chances. You might want to have a look at the following books...
- Marshall: Positive polynomials and sums of squares
- Prestel: Positive polynomials
- Bochnak, Coste, Roy: Real algebraic geometry
- Basu, Pollack, Roy: Algorithms in real algebraic geometry
- Knebusch, Scheiderer: Einführung in die reelle Algebra
- Andradas, Bröcker, Ruiz: Constructible sets in real geometry
...and the following articles...
Also the so-called "S-procedure" could be of interest for you.
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2
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You said you have proved it in the one variable case but in my opinion it should be wrong already in that case: Take $L_1=1+X$, $L_2=1-X$ and $P=X^2$. Then each $L_i$ is linear, $P$ is quadratic and $P\ge0$ on ${L_i\ge0}$. On the other hand each product of some of the $L_i$ is positive at the origin, showing that the desired representation cannot exist.
If you allow however for $P$ being strictly positive, then you have much better chances Interesting question. You might want to have a look at the following books...
- Marshall: Positive polynomials and sums of squares
- Prestel: Positive polynomials
- Bochnak, Coste, Roy: Real algebraic geometry
- Basu, Pollack, Roy: Algorithms in real algebraic geometry
- Knebusch, Scheiderer: Einführung in die reelle Algebra
- Andradas, Bröcker, Ruiz: Constructible sets in real geometry
...and the following articles...
Also the so-called "S-procedure" could be of interest for you.
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1
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You said you have proved it in the one variable case but in my opinion it should be wrong already in that case: Take $L_1=1+X$, $L_2=1-X$ and $P=X^2$. Then each $L_i$ is linear, $P$ is quadratic and $P\ge0$ on ${L_i\ge0}$. On the other hand each product of some of the $L_i$ is positive at the origin, showing that the desired representation cannot exist.
If you allow however for $P$ being strictly positive, then you have much better chances. You might want to have a look at the following books...
- Marshall: Positive polynomials and sums of squares
- Prestel: Positive polynomials
- Bochnak, Coste, Roy: Real algebraic geometry
- Basu, Pollack, Roy: Algorithms in real algebraic geometry
- Knebusch, Scheiderer: Einführung in die reelle Algebra
- Andradas, Bröcker, Ruiz: Constructible sets in real geometry
...and the following articles...
Also the so-called "S-procedure" could be of interest for you.
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