Images of some subspaces of L2(Rm) under Grushin and Hermite semigroup
Partha Sarathi Patra
(This is a joint work with Dr. D Venku Naidu)
Department of Mathematics Indian Institute of Technology Hyderabad, India
Images of some subspaces of L 2 ( R m ) under Grushin and Hermite - - PowerPoint PPT Presentation
Images of some subspaces of L 2 ( R m ) under Grushin and Hermite semigroup Partha Sarathi Patra (This is a joint work with Dr. D Venku Naidu) Department of Mathematics Indian Institute of Technology Hyderabad, India 6 th Fourier Analysis
(This is a joint work with Dr. D Venku Naidu)
Department of Mathematics Indian Institute of Technology Hyderabad, India
πe−|z|2dxdy, and the inner-product is defined by
πe−|z|2dxdy, and the inner-product is defined by
2 z2dx
1 2 ) −1 2 (−1)k dk
x2 2 ,
n
i=1 ∈ Nn 0. ◮ Hermite Operator: H = −∆x + x2
1 2 ) −1 2 (−1)k dk
x2 2 ,
n
i=1 ∈ Nn 0. ◮ Hermite Operator: H = −∆x + x2 ◮ φα’s are eigen vector of H corresponding to the eigen value
1 2 ) −1 2 (−1)k dk
x2 2 ,
n
i=1 ∈ Nn 0. ◮ Hermite Operator: H = −∆x + x2 ◮ φα’s are eigen vector of H corresponding to the eigen value
◮ Fourier transform Rn: Ff (ξ) = 1 √ 2π
◮ Grushin operator on Rn+1:G = −(∆x + |x|2 ∂2 ∂t2 ), ◮ Heat Equation Corresponding to this Grushin operator:
◮ The solution: u(x, t; s) = e−sGf (x, t).
◮ Grushin operator on Rn+1:G = −(∆x + |x|2 ∂2 ∂t2 ), ◮ Heat Equation Corresponding to this Grushin operator:
◮ The solution: u(x, t; s) = e−sGf (x, t).
◮ Grushin operator on Rn+1:G = −(∆x + |x|2 ∂2 ∂t2 ), ◮ Heat Equation Corresponding to this Grushin operator:
◮ The solution: u(x, t; s) = e−sGf (x, t).
◮ Grushin operator on Rn+1:G = −(∆x + |x|2 ∂2 ∂t2 ), ◮ Heat Equation Corresponding to this Grushin operator:
◮ The solution: u(x, t; s) = e−sGf (x, t). ◮ Fourier transform with respect to last variable Notation:
1 √ 2π
◮ Grushin operator on Rn+1:G = −(∆x + |x|2 ∂2 ∂t2 ), ◮ Heat Equation Corresponding to this Grushin operator:
◮ The solution: u(x, t; s) = e−sGf (x, t). ◮ Fourier transform with respect to last variable Notation:
1 √ 2π
◮ Parametrized hermite function: φλ α(x) = |λ|
n 4 φα(
◮ λ = 0 then {φλ α} forms an orthonormal basis for L2(Rn) and
α(x) = (2|α| + n)|λ|φλ α(x). ◮ Parametraized Hermite operator: Hλ = −∆x + |x|2λ2, λ = 0 ◮ Fourier transform on the last variable reduces the Grushin
◮ With the help of spectral resolution of Hλ we have
αL2(Rn)φλ α(x).
s (x, y)f (y)dy (using Mehlar’s formula).
s (x, y)
−n 2
sinh(2s|λ|)
2 e −|λ| 2
(x2+y2) coth(2s|λ|)e
|λ|xy sinh(2|λ|s) .
◮
s (x, y)f λ(y)dydλ ◮ e−sGf (x, t) can be extended to a holomorphic function in
◮ e−sG :
◮ we note that if u(x, t; s) is solution of (??) with initial
◮ i.e, for each x, u(x, t; s) = hs(f )(t) with f ∈
4t .
4(2t+1) + zw 2(2t+1)
1
2(2t+1) is a constant depending on t.
L2(R))e
y2 2(2t+1) .
L2(R))e
y2 2(2t+1) .
◮ This gives a growth condition for the elements of ht(
L2(R))e
y2 2(2t+1) .
◮ This gives a growth condition for the elements of ht(
◮ Let us consider the following Hilbert Space of holomorphic
holomorphic
2t+1
−y2 2t+1 dxdy < ∞
−z2 4(2t+1) , then the set A = {em : m = 0, 1, 2, · · · } is
+(R) = {f ∈
−(R) = {f ∈
+(R) ⊕
−(R) we can observe that,
+(R)) ⊕ ht(
−(R)) = B+ 2t+1(C) ⊕ B− 2t+1(C).
s (z) = 4n
2
s (Cn) where, Hλ s (Cn) is the space of holomorphic functions
s (z)dz < ∞.
s (z) =
0dη
s (z) is the weight of λ0-parametrized Hermite Bergman
◮ So it’s not possible to characterize the image of Grushin
◮ Now we take λ > 0 and consider the following formal
s (z, w) =
2 )2U
λ+ is
2
s
2 )ds.
s (z, ξ + iη) defined in the above equation (3.3) is
s
s (z, ξ + iη) = lim λ→0+
2 )2U
λ+ is
2
s
2 )ds (3.4)
z∈Q
s (z, ξ + iη)|dη < C.
s
s (z) =
s (z, iη)e−λ2dη.
+(Rn+1) = {f ∈
−(Rn+1) = {f ∈
+(Rn+1) ⊕
−(Rn+1).
s (Cn+1) containing all
s (z, w)|dzdw),
s (z, w)dzdw < ∞, and
2s+1(C).
s (Cn+1) by,
R→∞
s (z, w)dzdw,
s (Cn+1).
s (Cn+1).
s (Cn+1) := e−sG(
+(Rn+1)) is the Hilbert
s (Cn+1) with respect to the inner product , +
s (Cn+1) is isometrically isomorphic to
s (Cn+1).
s (z, w) when λ < 0,
s (Cn+1) can be
s (Cn+1) = e−sG(
−(Rn+1))
s (Cn+1) will be followed. And hence we
s (Cn+1) ⊕ B− s (Cn+1).
H (Rn) is defined as the
H (Rn) if and only if
H
◮ W µ,2 H (Rn) forms a Hilbert space under the inner product
◮ Define holomorphic Sobolev space
t
H (Rn)), t > 0.
t
H
◮ Clearly e−tH is an isometric isomorphism from W µ,2 H (Rn) to
t
◮ Let us define Fµ t , vector space of holomorphic functions F on
◮ We use following Caputo fractional derivative
t f (t) =
t is a pre-Hilbert space with respect to the sesquilinear form,
t (Ut(z))dz. Moreover there exist
t
t ≤ C2 FW µ,2 t
t
t (Cn).
G (Rn+1) as
−∞
α|2eλ2dλ < ∞.
◮ We denote
G+(Rn+1) as the Hilbert space of all those
G (Rn+1) such that,
α|2eλ2dλ < ∞,
s,+(Cn+1) := e−sG(
G,+(Rn+1)) is made into a
W µ,2
s,+ (Cn+1)
W µ,2
G,+(Rn+1)
αφλ α, gλeλ2dλ,
s
2
s
λ+ is
2
s
2
s
2 )2 dm
(λ+ is
2 )
θ
2 )|dsdθdzdw < ∞
s
s
R→∞
s
W µ,2
s,+ (Cn+1) ≤ FV+,µ s
(Cn+1) ≤ C2F W µ,2
s,+ (Cn+1),
G,+(Rn+1).
s,+(Cn+1) is the Hilbert completion of V+,µ s
s
s
s,−(Cn+1) is Hilbert completion of V−,µ s
s
s,+(Cn+1) ⊕
s,−(Cn+1).
s
s,+(Cn+1) ⊕
s,−(Cn+1).