Sufficient conditions for a real polynomial to be a sum of squares

0.15 MB·English

by  Jean B. Lasserre

SUFFICIENT CONDITIONS FOR A REAL POLYNOMIAL
TO BE A SUM OF SQUARES
7
0
0
2 JEAN B. LASSERRE

n
a Abstract. We provide explicit sufficient conditions for a polynomial
J f to be a sum of squares (s.o.s.), linear in the coefficients of f. All

1 conditions are simple and provide an explicit description of a convex
1 polyhedral subcone of the cone of s.o.s. polynomials of degree at most
2d. Wealsoprovideasimpleconditiontoensurethatf iss.o.s.,possibly
]
G after adding a constant.
A
.
h
t 1. Introduction
a
m
The cone Σ2 ⊂ R[X] of real polynomials that are sum of squares (s.o.s.)
[ and its subcone Σ2 of s.o.s. of degree at most 2d, play a fundamental role in
d

2 manyareas, andparticularly inoptimization; seeforinstanceLasserre[4,5],
v Parrilo[8]andSchweighofer[9]. Whenconsideredasaconvexconeofafinite
8
dimensionaleuclideanspace,Σ2 hasaliftedsemidefinite representation(such
5 d
3 sets are called SDr sets in [2]). Thatis, Σ2 is theprojection of a convex cone
d
2 ofaneuclideanspaceofhigherdimension,definedintermsofthecoefficients
1
of the polynomial and additional variables (the ”lifting”). However, so far
6
0 there is no simple description of Σ2 given directly in terms of the coefficients
d
/ of the polynomial. For more details on SDr sets, the interested reader is
h
t referred to e.g. Ben Tal and Nemirovski [2], Helton and Vinnikov [3], Lewis
a
et al. [7].
m
Of course, one could use Tarski’s quantifier elimination to provide a de-
:
v scription of Σ2, solely in terms of the coefficients, but such a description is
d
i
X likely hopeless to be simple; in particular, it could be sensitive to the degree
r d. Therefore, a more reasonable goal is to search for simple descriptions of
a subsets (or subcones) of Σ2 only. This is the purpose of this note in which
d
we provide simple sufficient conditions for a polynomial f ∈ R[X] of degree
at most 2d, to be s.o.s. All conditions are expressed directly in terms of the
coefficients (f ), with no additional variable (i.e. with no lifting) and define
α
a convex polyhedralsubcone of Σ2. Finally, we also provide a sufficient con-
d
dition on the coefficients of highest degree to ensurethat f is s.o.s., possibly
after adding a constant. All conditions stress the importance of the essen-
tial monomials (X2k) which also play an important role for approximating
i
nonnegative polynomials by s.o.s., as demonstrated in e.g. [5, 6].
1991 Mathematics Subject Classification. 12E05, 12Y05.
Keywordsandphrases. Realalgebraicgeometry;positivepolynomials;sumofsquares.
1
2 JEANB.LASSERRE
2. Conditions for being s.o.s.
For α ∈ Nn let |α| := n |α |. Let R[X] be the ring of real polynomials
in the variables X = (XP,i.=.1.,Xi ), and let R[X] the vector space of real
1 n 2d
polynomialsofdegreeatmost2d, withcanonicalbasisofmonomials (Xα) =
{Xα : α ∈ Nn; |α| ≤ 2d}. Given a sequence y = (y ) ⊂ R indexed in the
α
canonical basis (Xα), let L :R[X] → R be the linear mapping
y 2d
f(= f Xα) 7→ L (f) = f y , f ∈ R[X] ,
X α y X α α 2d
α α
and let M (y) be the moment matrix with rows and columns indexed in
d
(Xα), and defined by
(2.1) M (y)(α,β) := L (Xα+β) = y , α,β ∈ Nn : |α|,|β| ≤ d.
d y α+β
Let the notation M (y) (cid:23) 0 stand for M (y) is positive semidefinite. It is
d d
clear that
M (y) (cid:23) 0 ⇐⇒ L (f2) ≥ 0 ∀f ∈ R[X] .
d y d
The set Σ2 ⊂ R[X] of s.o.s. polynomials of degree at most 2d is a finite-
d 2d
dimensional convex cone, and
(2.2) f ∈ Σ2 ⇐⇒ L (f) ≥ 0 ∀y s.t. M (y) (cid:23) 0.
d y d
Remark 1. ToprovethatL (f)≥ 0forally suchthatM (y) (cid:23) 0itsuffices
y d
to prove that L (f) ≥ 0 for all y such that M (y) (cid:23) 0 and L (1) > 0 (and
y d y
equivalently, by homogeneity, for all y such that M (y) (cid:23) 0 and L (1) = 1).
d y
Indeed,supposethatL (f)≥ 0forally suchthatM (y) (cid:23) 0andL (1) >
y d y
0. Next, let y be such that M (y) (cid:23) 0 and L (1) = 0. Fix ǫ > 0 arbitrary
d y
andlety(ǫ) := y+(ǫ,0,...,0)sothatL (Xα)= y ifα 6= 0andL (1) =
y(ǫ) α y(ǫ)
ǫ > 0. Therefore M (y(ǫ)) (cid:23) 0 (because M (y) (cid:23) 0) and so 0 ≤ L (f) =
d d y(ǫ)
ǫf +L (f). As ǫ > 0 was arbitrary, letting ǫ ↓ 0 yields the desired result
0 y
L (f)≥ 0.
y
We first recall a preliminary result whose proof can be found in Lasserre
and Netzer [6].
Lemma 1 ([6]). With d ≥ 1, let y = (y ) ⊂ R be such that the mo-
α
ment matrix M (y) defined in (2.1) is positive semidefinite, and let τ :=
d d
max L (X2d). Then:
y i
i=1,...,n
(2.3) |L (Xα)| ≤ max[L (1), τ ], ∀α∈ Nn : |α| ≤ 2d.
y y d
We next complement Lemma 1.
Lemma 2. Let y = (y ) ⊂ R be normalized with y = L (1) = 1, and such
α 0 y
that M (y) (cid:23) 0. Let τ := max L (X2d). Then:
d d y i
i=1,...,n
(2.4) |L (Xα)|1/|α| ≤ τ1/2d, ∀α ∈ Nn : 1 ≤ |α| ≤ 2d.
y d
For a proof see §3.1.
SUFFICIENT CONDITIONS FOR A REAL POLYNOMIAL TO BE A SUM OF SQUARES3
2.1. Conditions for a polynomial to be s.o.s. With d ∈ N, let Γ ⊂ Nn
be the set defined by:
(2.5) Γ := {α ∈Nn : |α| ≤ 2d; α = 2β for some β ∈ Nn}.
We now provide our first condition.
Theorem 3. Let f ∈ R[X] and write f in the form
2d
n
(2.6) f = f + f X2d +h,
0 X i2d i
i=1
where h ∈ R[X] contains no essential monomial X2d. If
2d i
(2.7) f ≥ |f | − min[0,f ]
0 X α X α
α6∈Γ α∈Γ
|α| |α|
(2.8) min f ≥ |f | − min[0,f ]
i=1,...,n i2d X α 2d X α 2d
α6∈Γ α∈Γ
then f ∈ Σ2.
d
For a proof see §3.2. The sufficient conditions (2.7)-(2.8) define a polyhe-
dral convex cone in the euclidean space of coefficients (f ) of polynomials
α
f ∈ R[X] . This is because the functions,
2d
f 7→ min f , f 7→ min[0,f ], f 7→ −|f |,
i2d α α
i=1,...,n
are all piecewise linear and concave. The description (2.7)-(2.8) of this con-
vex polyhedral cone is explicit and given only in terms of the coefficients
(f ), i.e., with no lifting.
α
Notice that (2.7)-(2.8) together with f = 0 for some i, implies f = 0
i2d α
for all α 6∈ Γ, and f ≥ 0 for all α ∈ Γ, in which case f is obviously s.o.s.
α
Theorem 3 is interesting when f has a few non zero coefficients. When f
has a lot of non zero coefficients and contains the essential monomials X2k
i
for all k = 1,...,d, all with positive coefficients, one provides the following
alternative sufficient condition. With k ≤ d, let
(2.9) Γ1 := {α ∈ Nn : 2k−1 ≤ |α| ≤ 2k}
k
(2.10) Γ2 := {α ∈ Γ1 : α = 2β for some β ∈ Nn}.
k k
Corollary 4. Let f ∈ R[X] and write f in the form
2d
d n
(2.11) f = f +h+ f X2k,
0 XX i2k i
k=1i=1
4 JEANB.LASSERRE
where h ∈ R[X] contains no essential monomial X2k. If
2d i
f
0
(2.12) ≥ |f | − min[0,f ]
d X α X α
α∈Γ1\Γ2 α∈Γ2
k k k
|α| |α|
(2.13) min f ≥ |f | − min[0,f ]
i=1,...,n i2k X α 2k X α 2k
α∈Γ1\Γ2 α∈Γ2
k k k
for all k = 1,...,d, then f ∈ Σ2.
d
For a proof see §3.3. Notice that (2.12)-(2.13) together with f = 0 for
i2k
someiandsome k ∈ {1,...d}, implies f = 0for all α∈ Γ1\Γ2, and f ≥ 0
α k k α
for all α ∈ Γk.
2
Several variants of Corollary 4 can be derived; for instance, any other
way to distribute the constant term f as d f with f 6= f /d, is valid
0 Pk=1 0k 0k 0
and also provides another set of sufficient conditions. Consider also the case
when f can be written as
n
f = f +h+ f X2k,
0 X X i2k i
k∈K i=1
where K := {k ∈ {1,...,d} : min f > 0}, d∈ K, and h∈ R[X]
i=1,...,n i2k 2d
contains no essential monomial X2k, k ∈ K. Then one may easily derive a
i
set of sufficient conditions in the spirit of Corollary 4.
Finally, one provides a simple condition for a polynomial to be s.o.s.,
possibly after adding a constant.
Corollary 5. Let f ∈ R[X] and write f in the form
2d
n
(2.14) f = f +h+ f X2d,
0 X i2d i
i=1
where h ∈ R[X] contains no essential monomial X2d. If
i
(2.15) min f > |f |− min[0,f ]
i2d X α X α
i=1,...,n
α6∈Γ;|α|=2d α∈Γ;|α|=2d
with Γ as in (2.5), then f +M ∈Σ2 for some M ≥ 0.
d
Proof. Let −M := min[0, inf {L (f) : M (y) (cid:23) 0; L (1) = 1}]. We prove
y y d y
thatM < +∞. AssumethatM = +∞,andletyj beaminimizingsequence.
One must have τ := max L (X2d) → ∞, as j → ∞, otherwise if
jd i=1,...,n yj i
τ is boundedby, say ρ, by Lemma 1 one would have |L (Xα)| ≤ max[1,ρ]
jd yj
for all |α| ≤ 2d, and so L (f) would be bounded, in contradiction with
yj
L (f) → −∞. But then from Lemma 2, for sufficiently large j, one obtains
yj
SUFFICIENT CONDITIONS FOR A REAL POLYNOMIAL TO BE A SUM OF SQUARES5
the contradiction
L (f)
yj
0 > ≥ min f − |f | + min[0,f ]
τjd i=1,...,n i2d X α X α
α6∈Γ;|α|=2d α∈Γ;|α|=2d
(|α|−2d)/2d
− |f |τ ≥ 0,
X α jd
0≤|α|<2d
(|α|−2d)/2d
where the last inequality follows from (2.15) and τ → 0 as j → ∞.
jd
Hence, M < +∞ and so L (f +M) ≥ 0 for every y with M (y) (cid:23) 0, and
y d
L (1) = 1 . But then, in view of Remark 1, L (f +M) ≥ 0 for all y such
y y
that M (y) (cid:23) 0, which in turn implies that f +M is s.o.s. (cid:3)
d
In Theorem 3, Corollary 4 and 5, it is worth noticing the crucial role
played by the constant term and the essential monomials (Xα), as was al-
i
ready the case in [5, 6] for approximating nonnegative polynomials by s.o.s.
3. Proofs
The proof of Lemma 2 first requires the following auxiliary result.
Lemma 6. Let d ≥ 1, and y = (y ) ⊂ R be such that the moment matrix
α
M (y)defined in(2.1)ispositive semidefinite, and letτ := max L (X2d).
d d y i
i=1,...,n
Then: L (X2α) ≤ τ for all α ∈ Nn with |α| = d.
y d
Proof. The proof is by induction on the number n of variables. The case
n = 1 is trivial and the case n = 2 is proved in Lasserre and Netzer [6,
Lemma 4.2].
Let the claim betrue for k = 1,...,n−1 and consider the case n> 2. By
the induction hypothesis, the claim is true for all L X2α , where |α| = d
y
(cid:0) (cid:1)
and α = 0 for some i. Indeed, L restricts to a linear form on the ring
i y
of polynomials with n−1 indeterminates and satisfies all the assumptions
needed. So the induction hypothesis gives the boundedness of all those
values L X2α .
y
(cid:0) (cid:1)
Now take L X2α , where |α| = d and all α ≥ 1. With no loss of
y i
(cid:0) (cid:1)
generality, assume α ≤ α ≤... ≤ α . Consider the two elements
1 2 n
γ := (2α ,0,α +α −α ,α ,...,α )∈ Nn and
1 3 2 1 4 n
γ′ := (0,2α ,α +α −α ,α ,...,α ) ∈ Nn.
2 3 1 2 4 n
′ ′
We have |γ| = |γ | = d and γ = γ = 0, and from what precedes,
2 1
′
L (X2γ)≤ τ and L (X2γ ) ≤ τ .
y d y d
As M (y)(cid:23) 0, one also has
d
′ ′
L (X2α)2 = L (Xγ+γ )2 ≤ L (X2γ)·L (X2γ ) ≤ τ2,
y y y y d
which yields the desired result |L (X2α)| ≤ τ . (cid:3)
y d
6 JEANB.LASSERRE
3.1. Proof of Lemma 2. Theproofis by induction on d. Assumeit is true
for k = 1,...,d, and write M (y) in the following block form below with
d+1
appropriate matrices V,U ,V ,S :
i i i
M (y) U U V
d−2 1 2
UT S V V
1 2d−2 2d−1 2d
UT VT S V
2 2d−1 2d 2d+1
VT VT VT S
2d 2d+1 2d+2
When d= 1, the blocks M (y), and U ,U ,UT,UT,V disappear.
d−2 1 2 1 2
• The case |α| =2d+2 is covered by Lemma 6.
• Consider an arbitrary y with |α| = 2d. From the definition of the
α
moment matrix, one may choose a pair (i,j) such that the position (i,j) in
the matrix M (y) lies in the submatrix V , and the corresponding entry
d+1 2d
is y . From M (y) (cid:23) 0,
α d+1
M (y)(i,i)M (y)(j,j) ≥ y2,
d+1 d+1 α
As M (y)(i,i) is an element y of S with |β| = 2d−2, invoking the
d+1 β 2d−2
(2d−2)/2d
induction hypothesis yields M (y)(i,i) ≤ τ . On the other hand,
d+1 d
M (y)(j,j) is a diagonal element y of S with |β| = d + 1. From
d+1 2β 2d+2
Lemma 6, every diagonal element of S is dominated by τ , and so
2d+2 d+1
M (y)(j,j) ≤ τ . Combining the two yields
d+1 d+1
y2 ≤ τ(d−1)/dτ , ∀α: |α| = 2d.
α d d+1
Next, picking up the element α such that y = τ one obtains
α d
(3.1) τ2 ≤ τ1−1/dτ ⇒ τ1/d ≤ τ1/(d+1),
d d d+1 d d+1
and so,
y2 ≤ τ(d−1)/dτ ; |y |1/|α| ≤ τ1/(2d+2), ∀α : |α| =2d.
α d d+1 α d+1
• Next, consider an arbitrary y with |α| = 2d + 1. Again, one may
α
choose a pair (i,j) such that the position (i,j) in the matrix M (y)
d+1
lies in the submatrix V , and the corresponding entry is y . The en-
2d+1 α
try M (y)(i,i) corresponds to an element y of S with |β| = d, and
d+1 2β 2d
SUFFICIENT CONDITIONS FOR A REAL POLYNOMIAL TO BE A SUM OF SQUARES7
so, by Lemma 6, M (y)(i,i) ≤ τ ; similarly the entry M (y)(j,j) cor-
d+1 d d+1
responds to an element y of S with |β| = d+1, and so, by Lemma 6
2β 2d+2
again, M (y)(j,j) ≤ τ . From M (y) (cid:23) 0, we obtain
d+1 d+1 d+1
τ τ ≥ M (y)(i,i)M (y)(j,j) ≥ y2,
d+1 d d+1 d+1 α
which, using (3.1), yields |y |1/|α| = |y |1/(2d+1) ≤ τ1/(2d+2) for all α with
α α d+1
|α| = 2d+1.
• Finally, for an arbitrary y with 1 ≤ |α| < 2d, use the induction hy-
α
pothesis |y |1/|α| ≤ τ1/2d and (3.1) to obtain |y |1/|α| ≤ τ1/2(d+1). This
α d α d+1
argument is also valid for the case |α| = 2d, but this latter case was treated
separately to obtain (3.1).
It remains to prove that the induction hypothesis is true for d = 1. This
easily follows from the definition of themoment matrix M (y). Indeed, with
1
|α| = 1 one has y2 ≤ y ≤ τ (as L (1) = 1), so that |y | ≤ τ1/2 for all α
α 2α 1 y α 1
with |α| = 1. With |α| = 2, say with α +α = 2, one has
i j
τ2 ≥ L (X2)L (X2) ≥ L (X X )2 = y2,
1 y i y j y i j α
and so |y | ≤ τ for all α with |α| = 2. (cid:3)
α 1
3.2. Proof of Theorem 3. From (2.2), it suffices to show that L (f) ≥ 0
y
for any y such that M (y) (cid:23) 0, and by Remark 1, we may and will assume
d
that L (1) = 1.
y
So let y be such that M (y) (cid:23) 0 with L (1) = 1. Let τ be as in Lemma
d y d
1 and consider the two cases τ ≤ 1 and τ > 1.
d d
• The case τ ≤ 1. By Lemma 1, |L (Xα)| ≤ 1 for all α ∈ Nn with
d y
|α| ≤ 2d. Therefore,
L (f) ≥ f − |f | + min[0,f ] ≥ 0,
y 0 X α X α
α6∈Γ α∈Γ
where the last inequality follows from (2.7).
• The case τ > 1. Recall that L (1) = 1, and from Lemma 2, one has
d y
|L (Xα)|1/|α| ≤τ1/2d for all α∈ Nn with 1 ≤ |α| ≤ 2d. Therefore,
y d
L (f) ≥ f +( min f )τ
y 0 i2d d
i=1,...,n
|α|/2d |α|/2d
− |f |τ + min[0,f ]τ
X α d X α d
α6∈Γ α∈Γ
Consider the univariate polynomial t 7→ p(t), with
p(t)= f +( min f )t2d− |f |t|α|+ min[0,f ]t|α|,
0 i2d X α X α
i=1,...,n
α6∈Γ α∈Γ
and denote p(k) ∈ R[X], its k-th derivative.
8 JEANB.LASSERRE
By (2.8), min f ≥ 0 and so by (2.7), p(1) ≥ 0. By (2.8) again,
i2d
i=1,...,n
p′(1) ≥ 0. In addition, with 1≤ k ≤ 2d, (2.8) also implies
|α| (|α|−1) (|α|−(k−1))
min f ≥ |f | ···
i=1,...,n i2d X α 2d 2d−1 2d−(k−1)
α6∈Γ;|α|≥k
|α| (|α|−1) (|α|−(k−1))
− min[0,f ] ···
X α 2d 2d−1 2d−(k−1)
α∈Γ;|α|≥k
because |α|−j ≤ 2d−j, for all j = 1,...,k−1, and so
k−1 k−1
   
(2d−j) min f ≥ |f | (|α|−j)
Y i2d X α Y
i=1,...,n
j=0  α6∈Γ;|α|≥k j=0 
k−1
 
− min[0,f ] (|α|−j) ,
X α Y
α∈Γ;|α|≥k j=0 
which implies p(k)(1) ≥ 0. Therefore, p(k)(1) ≥ 0 for all k = 0,1,...,2d, and
so, as a special case of Budan-Fourier’s theorem, p has no root in (1,+∞);
see Basu et al [1, Theor. 2.36]. Hence, p ≥ 0 on (1,+∞) and as τ > 1,
d
L (f)≥ p(τ1/2d)≥ 0. (cid:3)
y d
3.3. Proof of Corollary 4. Let y be such that M (y) (cid:23) 0. Again in view
d
of Remark 1, we may and will assume that L (1) = 1.
y
d
Then L (f)≥ A , with
y X k
k=1
n
f
(3.2) A := 0 + f L (X2k)+ min[0,f ]L (Xα)
k d X i2k y i X α y
i=1 α∈Γ2
k
− |f ||L (Xα)|, k = 1,...,d.
X α y
α∈Γ1\Γ2
k k
Fix k arbitrary in {1,...,d} and consider the moment matrix M (y) (cid:23) 0,
k
which is a submatrix of M (y).
d
• Case τ ≤ 1. By Lemma 1 applied to M (y), |L (Xα)| ≤ 1 for all
k k y
α∈ Nn with |α| ≤ 2k. Therefore, with A as in (3.2),
k
f
0
A ≥ − |f | + min[0,f ] ≥ 0,
k d X α X α
α∈Γ1\Γ2 α∈Γ2
k k k
where the last inequality follows from (2.12).
SUFFICIENT CONDITIONS FOR A REAL POLYNOMIAL TO BE A SUM OF SQUARES9
• Case τ > 1. From Lemma 2 applied to M (y), |L (Xα)|1/|α| ≤ τ1/2k
k k y k
for all α with |α| ≤ 2k. Therefore, A ≥ p (τ1/2k), where p ∈R[t], and
k k k k
p (t) = f0 +t2k min f + min[0,f ]− |f |t|α|.
k d i=1,...,n i2k X α X α
 α∈Γ2  α∈Γ1\Γ2
k k k
(j)
As in the proof of Theorem 3, butnow using(2.12)-(2.13), one has p (1) ≥
k
0forallj = 0,1,...,2k. AsaparticularcaseofBudan-Fourier’stheorem,p
k
has no root in (1,+∞); see Basu et al [1, Theor. 2.36]. Therefore, p ≥ 0 on
k
1/2k
(1,+∞) which in turn implies A ≥ p (τ ) ≥ 0 because τ > 1. Finally,
k k k k
L (f)≥ d A ≥ 0, as A ≥ 0 in both cases τ ≤ 1 and τ > 1. (cid:3)
y Pk=1 k k k k
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LAAS-CNRS and Institute of Mathematics, LAAS, 7 avenue du Colonel
Roche, 31077 Toulouse cedex 4, France
E-mail address: [email protected]