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BERNSTEIN POLYNOMIALS AND BROWNIAN MOTION

BERNSTEIN POLYNOMIALS AND BROWNIAN MOTION,EMMANUEL KOWALSKI

BERNSTEIN POLYNOMIALS AND BROWNIAN MOTION   (Citations: 5)
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    • ...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. Emiriset 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 Bergman-Szeg¨o kernels on a toric K¨ahler manifold whose moment polytope is a given Delzant polytope...
    • ...He obtained an asymptotic expansion of the Bergman-Bernstein 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 Bergman-Bernstein approximations to obtain an asymptotic expansion of the Dedekind-Riemann 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 toric 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 Fubini-Study 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. Inserting 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 non-degenerate...
    • ...However, it follows that K(x) is non-degenerate 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)...
    • ...× 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 multi-index � = (�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 non-degenerate 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 non-degenerate...
    • ...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(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)...
    • ...Let us prove (59). It is easy to show that, for any x ∈ K,...
    • ...|h µS,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(µS,c(�K(x) − tu))...
    • ...+ th µS,c(�K(x) − tu) − x,u i + log(1 + RK(t,�K(x))/�SK (�K(x)))...
    • ...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)...
    • ...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 semi-continuous 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 semi-continuous 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 semi-continuous 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 semi-continuous 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 semi-continuous 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 semi-continuous 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 semi-continuous 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 − µ K(t�),� i dt...
    • ...Then, the function log�S,c is a K¨ahler potential for the Fubini-Study metric and −�S,c is its Legendre dual...
    • ...ities. However, it is well-known ([GKZ]) that, if P satisfies the Delzant condition, then MS,c is a non-singular compact K¨ahler manifold...
    • ...Consider the (restriction of the) Fubini-Study 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 associated to the K¨ahler potential ', which is defined by...
    • ...To compare the Bergman-Bernstein 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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