Remarks on semi-algebraic functions

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1 Remarks on semi-algebraic functions Seiichiro Wakabayashi April the second version on August In this note we shall give some facts and remarks concerning semi-algebraic functions which we need in another paper. We think that the results given here are all well-known but we could not find any literature in which the main results here ( Theorems 5 10 and 11 below) are given and proved. Definition 1. Let S be a subset of R n. We say that S is semi-algebraic ( or a semi-algebraic set ( in R n )) if there is a finite family {A jk } 1 j m1 k r j of subsets of R n such that each A jk is defined by a real polynomial equation or inequality and m ( r j S = A jk ). Noting that we have the following m ( r j A jk ) = r 1 k 1 =1 r m ( m k m =1 A jk j ) Lemma 2. Let S 1 and S 2 be semi-algebraic sets in R n. Then S c 1 (= Rn \ S 1 ) S 1 S 2 and S 1 S 2 are semi-algebraic. Moreover if T is a semi-algebraic set in R m then S 1 T is semi-algebraic. The following theorem is called the Tarski-Seidenberg theorem ( see e.g. A.2 of [H]). Theorem 3 (Tarski-Seidenberg). Let S be a semi-algebraic set in R n+m. Then is semi-algebraic. S := {x R n ; (xy) S for some y R m }

2 Corollary. Let S and T be semi-algebraic sets in R n+m and R m respectively. Then the set Ŝ {x R n ; (xy) S for any y T } is semi-algebraic. Proof. Since Ŝ c (= R n \ Ŝ) = {x R n ; there is y T satisfying (xy) S c } = {x R n ; there is y R m satisfying (xy) S c (R n T )} Theorem 3 and Lemma 2 prove the corollary. Lemma 4. If S is a semi-algebraic set in R n then the closure S of S and the interior S of S are semi-algebraic. Remark. If S = {x R; x 2 (x 1) > 0} then S = (1 ) and S {x ; x 2 (x 1) 0} = S {0}. Proof. Put D := {(xεy) R 2n+1 ; ε > 0 y Sx R n and x y 2 < ε} E := {(xε) R n (0 ); there is y S satisfying x y 2 < ε}. Then D is semi-algebraic and E = {(xε) R n+1 ; there is y S satisfying (xεy) D}. From Theorem 3 E is semi-algebraic. Since S = {x R n ; (xε) E for any ε > 0} and S = (R n \ S) c S and S are semi-algebraic. Theorem 5. Let P(X) be a polynomial of X = (X 1 X n ) and put A {X R n ; P(X) 0}. Then the number of the connected components of A is finite and each component is semi-algebraic. Proof. We may assume that the coefficients of P(X) are real replacing P(X) by P Re (X) 2 + P Im (X) 2 if necessary where P(X) = P Re (X) + ip Im (X) and P Re (X) and P Im (X) are real polynomials. Let us prove the theorem by induction on n. If n = 1 then the theorem is trivial. Let L N(= {12 }) and suppose that the theorem is valid when n L. Let n = L + 1. We can write P(X) = P 1 (X) m1 P s (X) m s where the P j (X) are irreducible polynomials and mutually prime. Put Q(X) = P 1 (X) P s (X) 2

3 and denote by Q 0 (X) the principal part ( the terms of highest degree) of P(X). We may assume that Q 0 (0 01) 0 using linear transformation if necessary. Let D(X ) be the discriminant of the equation Q(X X n ) = 0 in X n where X = (X 1 X n 1 ). Then D(X ) 0 and by the assumption of induction there are N N and semi-algebraic sets A j in R n 1 ( 1 j N) such that the A j are mutually disjoint and coincide with the connected components of the set {X R n 1 ; D(X ) 0}. For each j N with 1 j N we can write Q(X) = Q 0 (0 01) l (X n λ k (X )) λ 1 (X ) < λ 2 (X ) < < λ r( j) (X ) Imλ k (X ) 0 ( r( j) + 1 k l) for X A j where l = deg Xn Q(X) and r( j) N since the equation Q(X X n ) = 0 in X n has only simple roots for X A j. Put A jk := {X A j R; there are λ 1 λ r( j) R and λ r( j)+1 λ l C such that λ 1 < λ 2 < < λ r( j) Imλ µ 0 ( µ = r( j) + 1 l) Q(X t) = Q 0 (0 01) l µ=1 (t λ µ ) as a polynomial of t and λ k 1 < X n < λ k if 2 k l X n < λ 1 if k = 1 and X n > λ r( j) if k = r( j) + 1} ( k = 12 r( j) + 1). Then the A jk are semi-algebraic and A (A j R) = r( j)+1 A jk. By Lemmas 2 and 4 B jk A jk A is semi-algebraic. Assume that there are disjoint open subsets C 1 and C 2 of B jk satisfying B jk = C 1 C 2 and C 2 A jk /0. Since A jk is connected C 1 A jk A where B denotes the boundary of B in R n for a subset B of R n. So we have C 1 = /0. This implies that B jk are connected. Since (A j R) A = r( j)+1 B jk we have A = N r( j)+1 B jk. Put Λ := {( jk) N N; 1 j N 1 k r( j) + 1}. 3

4 For ( jk)( j k ) Λ we say that ( jk) ( j k ) if there are ν N and ( j µ k µ ) Λ ( 1 µ ν) satisfying B jµ k µ B jµ+1 k µ+1 /0 ( 0 µ ν) where ( j 0 k 0 ) = ( jk) and ( j ν+1 k ν+1 ) = ( j k ). For ( jk) Λ we put A ( jk) := ( j k ) ( jk) B j k. Then A ( jk) is a connected component of A and semi-algebraic. Moreover we have A = ( jk) Λ A ( jk) which proves the theorem. Definition 6. Let f (X) be a real-valued function defined on R n. We say that f (X) is semi-algebraic ( or a semi-algebraic function) if the graph of f ( = {(Xy) R n+1 ; y = f (X)}) is a semi-algebraic set. Lemma 7. f (X) is semi-algebraic if and only if A {(Xy) R n+1 ; y f (X)} is a semi-algebraic set. Proof. Assume that f (X) is semi-algebraic. Then B {(Xyλ) R n+2 ; λ = f (X) and y λ} is a semi-algebraic set. Therefore Theorem 3 implies that A is semi-algebraic. Next assume that A is semi-algebraic. Then C {(Xyλ) R n+1 ; λ f (X) and y < λ} is semi-algebraic. Therefore Theorem 3 implies that D {(Xy) R n+1 ; y < f (X)} is semi-algebraic. Thus A \ D = {(Xy) R n+1 ; y = f (X)} is semi-algebraic. Definition 8. (i) Let f (X) be a complex-valued function defined on R n. We say that f (X) is semi-algebraic ( or a semi-algebraic function) if Re f (X) and Im f (X) are semi-algebraic. (ii) Let X 0 R n and let f (X) be a complex-valued function defined in a neighborhood of X 0. We say that f (X) is semi-algebraic at X 0 if there is r > 0 such that the sets {(Xy) R n+1 ; X X 0 < r and y = Re f (X)} and {(Xy) R n+1 ; X X 0 < r and y = Im f (X)} are semi-algebraic. (iii) Let U be an open subset of R n and let f (X) be a complex-valued function defined in U. We say that f (X) is semi-algebraic in U if f (X) is semi-algebraic at every X 0 U. Lemma 9. Let X 0 R n and let f (X) and g(x) be semi-algebraic ( resp. semi-algebraic at X 0 ). (i) α f (X) + βg(x) and f (X)g(X) are semi-algebraic ( resp. semi-algebraic at X 0 ) where αβ C. (ii) If g(x) 0 for X R n ( resp. g(x) 0 in a neighborhood of X 0 ) then f (X)/g(X) is semi-algebraic ( resp. semi-algebraic at X 0 ). (iii) If g(x) 0 for X R n ( resp. g(x) 0 in a neighborhood of X 0 ) then g(x) 1/l ( 0) is semi-algebraic ( resp. semi-algebraic at X 0 ) where l N. 4

5 Proof. Let us prove the first part of the assertion (i) in the case where f (X) and g(x) are semi-algebraic at X 0. The other assertions can be proved by the same argument. We may assume that f (X) and g(x) are real-valued. By assumption there is r > 0 such that A {(Xλ) R n+1 ; X X 0 < r and λ = f (X)} and B {(X µ) R n+1 ; X X 0 < r and µ = g(x)} are semi-algebraic sets. Since C := {(Xλ µy) R n+3 ; X X 0 < r λ = f (X) µ = g(x) and y = αλ + β µ} is semi-algebraic Theorem 3 implies that α f (X)+βg(X) is semi-algebraic at X 0. Theorem 10. Let X 0 R n and assume that f (X) is in C and semi-algebraic ( resp. semi-algebraic at X 0 ). Then there is a irreducible polynomial P(zX)( 0) of (zx) = (zx 1 X n ) satisfying P( f (X)X) 0 ( resp. P( f (X)X) = 0 in a neighborhood of X 0 ). Proof. Let us prove the theorem in the case where f (X) is semi-algebraic at X 0. We may assume that f (X) is real-valued. By assumption there is r > 0 such that f (X) C (B r (X 0 )) and the set S {(Xy) B r (X 0 ) R; y = f (X)} is semialgebraic where B r (X 0 ) = {X R n ; X X 0 < r}. First consider the case where n = 1. Let F(zX) be the product of all polynomials F jk (zx) except polynomials depending only on X that appear in the definition of the semi-algebraic set S in Definition 1 as A jk = {(zx) R n+1 ; F jk (zx) = 0 (resp. > 0)}. Then we have F( f (X)X) = 0 in B r (X 0 ) since S is a graph of f (X). Write F(zX) = F 1 (zx) m1 F s (zx) m s where the F j (zx) are irreducible polynomials and mutually prime. We put G(zX) = F 1 (zx) F s (zx) and denote by D(X) the discriminant of the equation G(zX) = 0 in z. Then D(X) 0. Let X 1 B r (X 0 ) and assume that D(X 1 ) 0. Since the roots of G(zX 1 ) = 0 in z are all simple f (X) is analytic at X 1 and there is j(x 1 ) N with 1 j(x 1 ) s such that F j(x 1 ) ( f (X)X) = 0 in a neighborhood of X 1. Next assume that D(X 1 ) = 0. Then there is δ > 0 such that D(X) 0 if 0 < X X 1 < δ. Moreover f (X) is equal to a convergent Puiseux series if 0 < ±(X X 1 ) < δ respectively modifying δ if necessary. Since f (X) is in C the Puiseux series are Taylor series and therefore f (X) is analytic at X 1. So f (X) is analytic in B r (X 0 ) and there is j N with 1 j s such that F j ( f (X)X) = 0 in B r (X 0 ). Next let us consider the case where n 2. Similarly there is a polynomial F(zX) ( 0) such that F( f (X)X) = 0 in B r (X 0 ). Write F(zX) = F 1 (zx) m1 F s (zx) m s 5

6 where the F j (zx) are irreducible polynomials and mutually prime. We put G(zX) = F 1 (zx) F s (zx) and denote by D(X) the discriminant of the equation G(zX) = 0 in z. We have D(X) 0. We may assume that D 0 (0 01) 0 where D 0 (X) denotes the principal part of D(X) using linear transformation if necessary. If D(X 0 ) 0 then f (X) is analytic at X 0 and we can choose j N with 1 j s so that F j ( f (X)X) = 0 in a neighborhood of X 0. Now assume that D(X 0 ) = 0. Choose X 1 R n 1 so that X 1 X 0 < r where X 0 = (X1 0 X n 0 ) and X 0 = (X1 0 X n 1 0 ). Since D(X1 X n ) 0 in X n applying the same argument for the case n = 1 we can see that f (X 1 X n ) is analytic in X n if (X 1 X n ) B r (X 0 ) and that there is j N with 1 j s satisfying F j ( f (X 1 X n )X 1 X n ) = 0 if (X 1 X n ) B r (X 0 ). On the other hand for each connected component A k of the set {X R n ; D(X) 0} there is j j(a k ) N with 1 j s satisfying F j ( f (X)X) = 0 in A k B r (X 0 ). Therefore there are δ > 0 and j N such that 1 j s and F j ( f (X)X) = 0 if X B r (X 0 ) and X X 0 < δ. Theorem 11. Let X 0 R n and assume that f (X) is a continuous function defined on R n ( resp. near X 0 ). Moreover we assume that there is a polynomial P(zX) ( 0) satisfying P( f (X)X) 0 ( resp. P( f (X)X) = 0 in a neighborhood of X 0 ). Then f (X) is semi-algebraic ( resp. semi-algebraic at X 0 ). Proof. Let us prove the theorem in the case where f (X) is defined in B r (X 0 ). We may assume that f (X) is real-valued and that P(zλ) is a real polynomial. Write P(zX) = P 1 (zx) m1 P s (zx) m s where the P j (zx) are irreducible and mutually prime. We put Q(zX) = P 1 (zx) P s (zx) and denote by D(X) the discriminant of the equation Q(zX) = 0 in z. Then we have D(X) 0. Put A := {X R n ; D(X) 0}. It follows from Theorem 5 that there are a finite number of semi-algebraic sets A 1 A N in R n such that the A j are the disjoint connected components of A and A = N A j. For each j N with 1 j N there are r( j) N with 1 r( j) m a polynomial c(x) and λ k (X) defined in A j ( 1 k m) such that c(x) 0 and Q(zX) = c(x) m (z λ k (X)) λ 1 (X) < λ 2 (X) < < λ r( j) (X) Imλ k (X) 0 ( r( j) + 1 k m) 6

7 for X A j where m = deg z Q(zX). Let j N satisfy 1 j N and A j B r (X 0 ) /0. Then there exists uniquely k( j) N satisfying 1 k( j) r( j) and λ k( j) (X) = f (X) in A j B r (X 0 ). Put E j := {(Xy) A j R; X B r (X 0 ) and there are a R and λ 1 λ m C such that Q(z X) = a m (z λ k ) λ 1 < < λ r( j) Imλ k 0 ( r( j) + 1 k m) and y = λ k( j) }. Then E j is semi-algebraic and Put E j = {(Xy) A j R; X B r (X 0 ) and y = f (X)}. Ẽ j := {(Xy) A j R; X B r (X 0 ) and y = f (X)}. Since Ẽ j = E j B r (X 0 ) R Ẽ j is semi-algebraic. So E j:a j B r (X 0 ) /0 Ẽ j is semi-algebraic. Note that N A j = R n and that A j B r (X 0 ) = /0 if A j B r (X 0 ) = /0. Then we have E = {(Xy) B r (X 0 ) R; y = f (X)}. References [H] L. Hörmander The Analysis of Linear Partial Differential Operators II Springer-Verlag Berlin-Heidelberg-New York-Tokyo

Modern Algebra 2: Midterm 2

Modern Algebra 2: Midterm 2 April 3, 2014 Name: Write your answers in the space provided. Continue on the back for more space. The last three pages are left blank for scratch work. You may detach them.