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Bernstein Polynomial
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Harmonic Analysis on Toric Varieties
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BERNSTEIN POLYNOMIALS AND BROWNIAN MOTION
BERNSTEIN POLYNOMIALS AND BROWNIAN MOTION,EMMANUEL KOWALSKI
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BERNSTEIN POLYNOMIALS AND BROWNIAN MOTION
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(2)
...Such polynomials are also important in Brownian motion [
21
]...
...This left open the case, see also [
21
], of smaller variance, that is polynomial and not exponential in d...
Ioannis Z. Emiris
,
et al.
Random polynomials and expected complexity of bisection methods for re...
...Some relations between Bernstein polynomials and Brownian motions are discussed in [
K
]...
...He defined it by using BergmanSzeg¨o kernels on a t
o
ric K¨ahler manifold whose moment polytope is a given Delzant polytope...
...He obtained an asymptotic expansion of the BergmanBernstein approximations by using the theory of Toeplitz operators and the harmonic analysis on toric
K
¨ahler manifolds developed in [STZ1], [STZ2], [SoZ1], and applied the BergmanBernstein approximations to obtain an asymptotic expansion of the DedekindRiemann sum...
...In [Z], Zelditch used a general
K
¨ahler metric on a toric K¨ahler manifold...
...In [Z], Zelditch used a general K¨ahler metric on a t
o
ric K¨ahler manifold...
...One may say that, when the polytope is Delzant, we use only the
K
¨ahler metric induced by the restriction of the FubiniStudy metric through a monomial embedding...
...(2) There exists a smooth function
K
: P o → Sym(m, R) such that for any f ∈ C(P), we...
...P f(z)
K
(x)(z − x)dBx(z), x ∈ P o , where Sym(m, R) denotes the space of all symmetric m ×m real matrices...
...We call the function
K
: P o → Sym(m, R) the defining matrix of the Bernstein measure B...
...(3) we have ∇m�(x) = m�(x)
K
(x)(� − x) for each x ∈ P o ...
...The defining matrix
K
(x), x ∈ P o , is given...
...by ([TZ])
K
(x) = � �...
...z�(z,x)dz = x, @x log�(z,x) =
K
(x)(z − x), (18)...
...where
K
(x) is a positive smooth function on (0,1) (the defining matrix of B) which we need to specify...
...We take a potential function '
of K on (0,1), that is ' ′′ = K. Inserting K = ' ′′...
...We take a potential function ' of K on (0,1), that is
9
; ′′ = K. Inserting K = ' ′′...
...We take a potential function ' of K on (0,1), that is ' ′′ = K. In
s
erting K = ' ′′...
...Note that the defining matrix (function)
K
must be positive on (0,1) (see Lemma 3.1), and hence x 7→ �(x) is a diffeomorphism from (0,1) onto its image...
...The defining function
K
(x) must be given by K(x) = 1/µ ′ (�(x)) = � ′ (x), x ∈ (0,1)...
...The defining function K(x) must be given by
K
(x) = 1/µ ′ (�(x)) = � ′ (x), x ∈ (0,1)...
...
K
(x) and its integral �(x) are determined uniquely by the equation (18)...
...Note that A(x) :=
K
(x) −1 is continuously extended to [0,1] with A(0) = A(1) = 0. But,...
...matrix
K
(x) ∈ Sym(m, R). The asymptotic expansion (21) can be differentiated any number of times and the resulting expansion holds locally uniformly on P o ...
...The matrix
K
(x) is not assumed to be nondegenerate...
...However, it follows that
K
(x) is nondegenerate with the inverse matrix A(x) given by...
...Note that, though the defining matrix
K
(x) is defined only for x in P o , the matrix A(x) in...
...The corresponding defining matrix is denoted by
K
: P o → Sym(m, R)...
...Lemma 3.1 For each x ∈ P o , the linear map A(x) : R m → R m defined by (23) is the inverse of the symmetric linear map of
K
(x) : R m → R m . Furthermore, A(x), and hence K(x) is positive definite for each x ∈ P o ...
...Lemma 3.1 For each x ∈ P o , the linear map A(x) : R m → R m defined by (23) is the inverse of the symmetric linear map of K(x) : R m → R m . Furthermore, A(x), and hence
K
(x) is positive definite for each x ∈ P o ...
...P h
K
(x)(z − x),u i z dBx(z) = Z P...
...P h
K
(x)(z − x),u i (z − x)dBx(z), which equals A(x)K(x)u by definition (23) of A(x)...
...which equals A(x)
K
(x)u by definition (23) of A(x)...
...P f(z)
K
(x)(z − x)dB N x ...
...f((z1 + � � � + zN)/N)
K
(x)(zj − x)dBx(z1) � � � dBx(zN)...
...&#
2
15; K(x)(z1 + � � � + zN − Nx)dBx(z1) � � � dBx(zN)...
...NP f(w/N)
K
(x)(w − Nx)d(B x ∗ � � � ∗ B x)(w)...
...Z P f(z)
K
(x)(z − x)dB N...
...P z ⊗
K
(x)(z − x)dB N x (z) = N Z P...
...P (z − x) ⊗
K
(x)(z − x)dB N (z), which means the desired formula...
...h z − x,u ih
K
(x)(z − x),� i (z − x)dBx(z), (35)...
...The following lemma shows an integrability of the defining matrix
K
, which will be used to prove Theorem 2.10...
...Lemma 3.5 We have [∇
K
(x)u]v = [∇K(x)v]u for any u,v ∈ R m and x ∈ P o ...
...Lemma 3.5 We have [∇K(x)u]v = [∇
K
(x)v]u for any u,v ∈ R m and x ∈ P o ...
...Proof. For x ∈ P o , we have
K
(x) = A(x) −1 , and hence, by (35), we obtain...
...Proof. For x ∈ P o , we have K(x) = A(x) −1 , and hence, by (35), we obtain [∇
K
(x)u]v = −K(x)[∇A(x)u]K(x)v...
...Proof. For x ∈ P o , we have K(x) = A(x) −1 , and hence, by (35), we obtain [∇K(x)u]v = −
K
(x)[∇A(x)u]K(x)v...
...Proof. For x ∈ P o , we have K(x) = A(x) −1 , and hence, by (35), we obtain [∇K(x)u]v = −K(x)[∇A(x)u]
K
(x)v...
...h
K
(x)(z − x),v ih K(x)(z − x),u i K(x)(z − x)dBx(z),...
...h K(x)(z − x),v ih
K
(x)(z − x),u i K(x)(z − x)dBx(z),...
...h K(x)(z − x),v ih K(x)(z − x),u i
K
(x)(z − x)dBx(z),...
...
K
: P o → Sym(m, R). For any multiindex � = (�1,...,�m) ∈ Z m0 , we write...
...Proposition 3.6 Let
K
: P o → Sym(m, R m ) be a smooth map...
...measure with the defining matrix
K
is, if it exists, unique...
...Proof. Let B : P → M(P) be a Bernstein measure with the defining matrix
K
: P o →...
...Sym(m, R). Then, the matrix
K
(x) is nondegenerate and the inverse matrix is given by A : P → Sym(m, R). We note that the inverse matrix A(x) defines the differential operators Lj := Xj + Dj, j = 1,...,m, where Xj is the multiplication operator: Xjf(x) = xjf(x)...
...polynomials on P is determined by the matrix
K
(x) because the differential operators Lj are defined in terms of A(x) = K(x) −1 . Since B : C(P) → C(P) is continuous, it is...
...polynomials on P is determined by the matrix K(x) because the differential operators Lj are defined in terms of A(x) =
K
(x) −1 . Since B : C(P) → C(P) is continuous, it is...
...polynomials on P is determined by the matrix K(x) because the differential operators Lj are defined in terms of A(x) = K(x) −1 . Since B : C(P) → C(P) is continuous, it is determined by
K
(x)...
...We keep the notation described before the statement of Theorem 2.10 in Section 2. First, we note that the functions mS,c,�, � ∈ S defined in (26) are continuous up to the boundary @P of P. Before giving a proof of this fact, we describe the values of mS,c,� at x ∈ @P. Let
K
be a (relatively open) face of P. Let...
...and let X ⊥
K
⊂ R m be the annihilator of XK. Then, define the map µK : X → K by...
...X �∈S∩
K
c(�)e h �,� i...
...P �∈S∩
K
c(�)e h �,� i �, � ∈ X. (47)...
...If �,� ′ ∈ X satisfy � − � ′ ∈ X ⊥
K
, then µK(�) = µK(� ′ ), and hence the above defines a map µK : XK ∼ = R m /X ⊥...
...
K
→ K, and it is a diffeomorphism from XK onto K ([F])...
...K →
K
, and it is a diffeomorphism from XK onto K ([F])...
...For any u ∈ R m , we define �(u) = miny∈P h y,u i . We fix u ∈ X ⊥
K
satisfying...
...K satisfying
K
= {y ∈ P ; h y,u i = �(u)}. Then, for any x = µK(�) ∈ K with � ∈ R m , we have limt→+∞ µS,c(� − tu) = x and...
...= ( 0 � 6∈ S ∩
K
, c(�)e h �,� i...
...P �∈S∩
K
c(�)e h �,� i � ∈ S ∩ K...
...�∈S∩K c(�)e h �,� i � ∈ S ∩
K
. (48)...
...Proof. Let
K
⊂ @P be a relatively open face of P. Let xn ∈ P o and x ∈ K...
...Proof. Let K ⊂ @P be a relatively open face of P. Let xn ∈ P o and x ∈
K
...
...
K
such that K = {y ∈ P ; h y,u i = �(u)} with �(u) = miny∈P h y,u i . For any A ⊂ S, we define �A(�) = P �∈A c(�)e h �,� i . We set...
...K such that
K
= {y ∈ P ; h y,u i = �(u)} with �(u) = miny∈P h y,u i . For any A ⊂ S, we define �A(�) = P �∈A c(�)e h �,� i . We set...
...Decompose �n ∈ R m according to the decomposition R m = XK ⊕ X ⊥ K as �n = �n +un, �n ∈ XK, un ∈ X ⊥
K
. Then,...
...Let B : P → M(P) be a finitely supported Bernstein measure, and let S = supp(B(x)) for some, and any x ∈ P o . Let
K
: P o → Sym(m, R) be the defining matrix of B. We write B(x) = P �∈S m�(x)��, x ∈ P. Then, by Lemma 3.1,...
...Let B : P → M(P) be a finitely supported Bernstein measure, and let S = supp(B(x)) for some, and any x ∈ P o . Let K : P o → Sym(m, R) be the defining matrix of B. We write B(x) = P �∈S m�(x)��, x ∈ P. Then, by Lemma 3.1, the matrix
K
(x) is nondegenerate...
...0
K
(b ∗ + s(x − b ∗ ))(x − b ∗ )ds, �b∗(x) = logV − Z 1...
...Lemma 3.5 shows that (d�b∗)x =
K
(x), ∇�b∗(x) = −�b∗(x), x ∈ P o . Since m�(x) > 0 on P o ,...
...(Note that the differential of �b∗ is
K
, which is positive definite on P o .) Denote its inverse by µb∗ : Im(�b∗) → P o . For any x ∈ P o and � ∈ Im(�b∗), we define f�(x) = h x,� i + �b∗(x)...
...Then, we have ∇f�(x) = � − �b∗(x) and ∇ 2 f�(x) = −
K
(x)...
...To describe the rate function (57) more concretely, we need the fact that the function �S,c in (50) is continuous on P. Let
K
be a (relatively open) face of the polytope P. Let �K : K → XK be the inverse of the map µK : XK → K defined in (47)...
...To describe the rate function (57) more concretely, we need the fact that the function �S,c in (50) is continuous on P. Let K be a (relatively open) face of the polytope P. Let �
K
: K → XK be the inverse of the map µK : XK → K defined in (47)...
...To describe the rate function (57) more concretely, we need the fact that the function �S,c in (50) is continuous on P. Let K be a (relatively open) face of the polytope P. Let �K :
K
→ XK be the inverse of the map µK : XK → K defined in (47)...
...When
K
= P o , we have µP o = µS,c and �P o = �S,c...
...As in the proof of Lemma 3.10, for each A ⊂ S, we set �A(�) = P �∈A c(�)e h �,� i and SK = S ∩ K. Define a function �
K
on K by...
...As in the proof of Lemma 3.10, for each A ⊂ S, we set �A(�) = P �∈A c(�)e h �,� i and SK = S ∩ K. Define a function �K on
K
by...
...�
K
(x) = log�SK (�K(x)) − h x,�K(x) i , x ∈ K. (58)...
...Lemma 3.14 The function �S,c is continuous on P, and its restriction to each face
K
is given by �K...
...Lemma 3.14 The function �S,c is continuous on P, and its restriction to each face K is given by �
K
...
...
K
such that K = {y ∈ P ; h y,u i = �(u)} with �(u) = minz∈P h z,u i . Then, first of all, we claim that, for x ∈ K, the following holds: lim t→+∞ �S,c(µS,c(�K(x) − tu)) = �K(x)...
...K such that
K
= {y ∈ P ; h y,u i = �(u)} with �(u) = minz∈P h z,u i . Then, first of all, we claim that, for x ∈ K, the following holds: lim t→+∞ �S,c(µS,c(�K(x) − tu)) = �K(x)...
...K such that K = {y ∈ P ; h y,u i = �(u)} with �(u) = minz∈P h z,u i . Then, first of all, we claim that, for x ∈
K
, the following holds: lim t→+∞ �S,c(µS,c(�K(x) − tu)) = �K(x)...
...K such that K = {y ∈ P ; h y,u i = �(u)} with �(u) = minz∈P h z,u i . Then, first of all, we claim that, for x ∈ K, the following holds: lim t→+∞ �S,c(µ
S
,c(�K(x) − tu)) = �K(x)...
...K such that K = {y ∈ P ; h y,u i = �(u)} with �(u) = minz∈P h z,u i . Then, first of all, we claim that, for x ∈ K, the following holds: lim t→+∞ �S,c(µS,c(�K(x) − tu)
)
= �K(x)...
...Let us prove (59). It is easy to show that, for any x ∈
K
,...
...h µ
S
,c(�K(x) − tu) − x i ,u ≤ Ce −tcK �S\S K (�K(x)) �SK (�K(x)) , (60)...
...Since x ∈
K
, we have h x,u i = �(u), and hence...
...Since x ∈ K, we have h x,u i = �(u), and hence �S,c(µ
S
,c(�K(x) − tu))...
...+ th µ
S
,c(�K(x) − tu) − x,u i + log(1 + RK(t,�K(x))/�SK (�K(x)))...
...Now, by (60), (61) and the fact that µ
S
,c(�K(x) − tu) tends to x ∈ K as t → +∞, the right hand side of (62) converges to �K(x), which shows (59)...
...Now, by (60), (61) and the fact that µS,c(�K(x) − tu) tends to
x ∈ K as t → +∞, the right hand side of (62) converges to �K(x), which shows (59)...
...Now, by (60), (61) and the fact that µS,c(�K(x) − tu) tends to x ∈ K as t → +∞, the right hand side of (62) converges
to �K(x), which shows (59)...
...Next, we take x ∈
K
, and {xn} ⊂ P o such that xn → x as n → ∞. Let � = �K(x) and �n = �n + un = �S,c(xn) with �n ∈ XK, un ∈ X ⊥...
...Next, we take x ∈ K, and {xn} ⊂ P o such that xn → x as n → ∞. Let � = �
K
(x) and �n = �n + un = �S,c(xn) with �n ∈ XK, un ∈ X ⊥...
...
K
. In the proof of Lemma 3.10, we have proved that �n → �. Let pK(n) = h z,un i with z ∈ K, which does not depend on the...
...choice of z ∈
K
. We note that �SK (�n + un) = e pK(n) �SK (�n)...
...Since �S,c(xn) = −h xn,�n + un i + log�S,c(�n), we conclude that �S,c(xn) → �
K
(x)...
...Remark 3.15 The function �
K
on the face K defined in (58) is continuous on K. This can be shown in the same way as in the above proof...
...Remark 3.15 The function �K on the face
K
defined in (58) is continuous on K. This can be shown in the same way as in the above proof...
...Remark 3.15 The function �K on the face K defined in (58) is continuous on
K
. This can be shown in the same way as in the above proof...
...The restriction of �
K
on a (relatively open) face L of K is given by �L...
...The restriction of �K on a (relatively open) face L of
K
is given by �L...
...Proposition 3.16 Let
K
be a relatively open face of P. Let x ∈ K. Then, the rate function I x (y) is given by the following:...
...Proposition 3.16 Let K be a relatively open face of P. Let x ∈
K
. Then, the rate function I x (y) is given by the following:...
...I x (y) = � +∞ y 6∈
K
, �K(x) − �K(y) + h x − y,�K(x) i y ∈ K. (63)...
...I x (y) = � +∞ y 6∈ K, �
K
(x) − �K(y) + h x − y,�K(x) i y ∈ K. (63)...
...I x (y) = � +∞ y 6∈ K, �K(x) − �
K
(y) + h x − y,�K(x) i y ∈ K. (63)...
...I x (y) = � +∞ y 6∈ K, �K(x) − �K(y) + h x − y,�
K
(x) i y ∈ K. (63)...
...I x (y) = � +∞ y 6∈ K, �K(x) − �K(y) + h x − y,�K(x) i y ∈
K
. (63)...
...Proof. Let
K
be a face, and we fix x ∈ K. By (48), (55), we have �x(�) = �...
...Proof. Let K be a face, and we fix x ∈
K
. By (48), (55), we have �x(�) = �...
...
K
. Then, for y ∈ K, we have h y,� + u i − log�x(� + u) = h y,� i − log�x(�) and hence IK(y) = JK(y) for each y ∈ K. Now, first of all, we assume y ∈ K. Then, for � ∈ XK, we have ∇� log�SK (�) = µK(�), which shows that � = �K(y) is a unique critical point of the function � 7→ h y,� i − log�SK (�)...
...K . Then, for y ∈
K
, we have h y,� + u i − log�x(� + u) = h y,� i − log�x(�) and hence IK(y) = JK(y) for each y ∈ K. Now, first of all, we assume y ∈ K. Then, for � ∈ XK, we have ∇� log�SK (�) = µK(�), which shows that � = �K(y) is a unique critical point of the function � 7→ h y,� i − log�SK (�)...
...IK(y) = h y,�K(y) i − log�SK (�K(y)) = −�K(y), (65) which shows (63) for y ∈
K
. We set F x...
...
K
(y) = �K(x)−�K(y)+h x−y,�K(x) i for y ∈ K, which is continuous in y ∈ K. Since I x (y) is lower semicontinuous in y, we have F x...
...K (y) = �
K
(x)−�K(y)+h x−y,�K(x) i for y ∈ K, which is continuous in y ∈ K. Since I x (y) is lower semicontinuous in y, we have F x...
...K (y) = �K(x)−�
K
(y)+h x−y,�K(x) i for y ∈ K, which is continuous in y ∈ K. Since I x (y) is lower semicontinuous in y, we have F x...
...K (y) = �K(x)−�K(y)+h x−y,�
K
(x) i for y ∈ K, which is continuous in y ∈ K. Since I x (y) is lower semicontinuous in y, we have F x...
...K (y) = �K(x)−�K(y)+h x−y,�K(x) i for y ∈
K
, which is continuous in y ∈ K. Since I x (y) is lower semicontinuous in y, we have F x...
...K (y) = �K(x)−�K(y)+h x−y,�K(x) i for y ∈ K, which is continuous in y ∈
K
. Since I x (y) is lower semicontinuous in y, we have F x...
...K (y) = �K(x)−�K(y)+h x−y,�K(x) i for y ∈ K, which is continuous in y ∈ K. Since I x (y) is lower semicontinuous in y, we have F x
K
(y) ≥ I x (y) for...
...any y ∈
K
. Now, assume that y is contained in a face L of K, and we show the inequality...
...any y ∈ K. Now, assume that y is contained in a face L of
K
, and we show the inequality...
...
K
(y) ≤ I x (y) for such a point y, which implies (63)...
...Since the supremum over � ∈ XL of the right hand side above is −�L(y) = −�
K
(y), we conclude IK(y) ≥ −�K(y) for y ∈ L. Finally, we show that I x (y) = +∞ when y 6∈ K. We...
...Since the supremum over � ∈ XL of the right hand side above is −�L(y) = −�K(y), we conclude IK(y) ≥ −�K(y) for y ∈ L. Finally, we show that I x (y) = +∞ when y 6∈
K
. We...
...h y,� i − log�SK (�) = −log�SK (0) + Z 1 0 h y − &#
1
81; K(t�),� i dt...
...Then, the function log�S,c is a
K
¨ahler potential for the FubiniStudy metric and −�S,c is its Legendre dual...
...ities. However, it is wellknown ([GKZ]) that, if P satisfies the Delzant condition, then MS,c is a nonsingular compact
K
¨ahler manifold...
...Consider the (restriction of the) FubiniStudy
K
¨ahler form !FS on MS,c...
...On the open orbit (C ∗ ) m ∼ �S,c((C ∗ ) m ), one can take a T m invariant
K
¨ahler potential '. Since ' is a function on (C ∗ ) m invariant under T m action, it defines a function on R m , which we denote by 'S,c...
...The symplectic potential associated to ' is the Legendre dual u' of the function 'S,c associ
a
ted to the K¨ahler potential ', which is defined by...
...To compare the BergmanBernstein approximation (66) with our Bernstein measures, we take the
K
¨ahler potential...
...∇˜ m S,N (x) = ˜ m S,N (x)
K
(x)( − Nx), (73)...
...where
K
(x) = AS,c(x) −1 is the defining matrix of the Bernstein measure BS,c : P → M(P)...
Tatsuya Tate
.
Bernstein measures on convex polytopes
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DAVAR KHOSHNEVISAN