Matrix representation of Toeplitz operators on Newton spaces

In this paper, we study several properties of an orthonormal basis { N n ( z ) } for the Newton space N 2 ( P ). In particular, we investigate the product of N m and N m and the orthogonal projection P of N n N m that maps from L 2 ( P ) onto N 2 ( P ). Moreover, we ﬁnd the matrix representation of Toeplitz operators with respect to such an orthonormal basis on the Newton space N 2 ( P )


Introduction
For any n ∈ N ∪ 0, let N n (z) denote the nth Newton polynomial, which is determined by the coefficients in the expansion: where |w| < 1 and z is any complex number in the complex plane C. Using the notations in [1,[12][13][14], N n (z) has the following expression where z n = z(z-1)(z-2)•••(z-(n-1)) n!
Let N 2 (P) be a Newton space as the closure of the set of polynomials in L 2 (C, μ) (see [3]).In [11], Markett, Rosenblum, and Rovnyak verified that N 2 (P) is a Hilbert space and the Newton polynomials {N n (z)} ∞ n=0 form an orthonormal basis for N 2 (P).Note that On the positive real line define a measure μ by dμ(t) = e -t dt.The measure μ is finite and has total mass (1) = 1.Consider the weighted Lebesgue space denoted by L 2 (μ), which comprises measurable complex-valued functions f defined with and let L ∞ (μ) be the set of all essentially bounded measurable functions in P. For f ∈ L 2 (μ), the weighted Mellin transform F on N 2 (P) of f is defined by For f , g ∈ L ∞ (μ), an inner product on N 2 (P) is defined by where F and G are weighted Mellin transforms of f and g, respectively, (see [11]).
Let P denote the orthogonal projection that maps L 2 (μ) onto N 2 (P) defined by where dA is an area measure on P. The reproducing kernel of N 2 (P) has the following form: For ϕ ∈ L ∞ (μ), the Toeplitz operator T ϕ on N 2 (P) is defined by From [9], it is known that the following properties of the Toeplitz operators T ϕ on N 2 (P) hold: (i) In [6,10], the authors studied the properties of composition operators on Newton space N 2 (P).Recently, Han [3] focused on the complex symmetric composition operators on Newton space N 2 (P) with respect to the specific conjugation.Furthermore, in [7,8], we considered the properties of Toeplitz operators on Newton space.From the above research point of view, we further investigate the properties of Newton space and Newton basis and study the matrix of Toeplitz operators on Newton space in this paper.
This paper is organized as follows.First, we study several properties of an orthonormal basis {N n (z)} for Newton space N 2 (P).In particular, we investigate the product of N m and N m and the orthogonal projection P of N n N m that maps from L 2 (P) onto N 2 (P).Next, we consider the matrix representation of Toeplitz operators with respect to such an orthonormal basis on Newton space N 2 (P).

Main results
In this section, we first study an orthonormal basis for the Newton space N 2 (P).We begin with the following lemma.

Lemma 2.1 ([10]
) Let the map : N 2 (P) → N 2 (P) defined by be the backwards unilateral shift on the orthonormal basis Lemma 2.2 For any m, n ≥ 0, the following equation holds     for n ≥ 1, it follows that Hence, we complete the proof.
we have Hence, it means that N 2 = I and so N -1 = N .
Let m, n ≥ 0 be nonnegative integers.Then, z m z n = z m+n holds on the Hardy space H 2 .However, N m N n is not N m+n on Newton space N 2 (P), in general.We next show that the product of N m and N n is the linear combination of {N j } (m+n)  j=max{m,n} .

Theorem 2.4 For any m
where . . .
Proof By the definition of N n (z), we have Thus, we can write N m (z)N n (z) as follows: 3), then we have 3) and we have as in Lemma 2.3.By repeating this method, we deduce that . By Lemma 2.3, we have the results.
From (2.2), we obtain the exact value of b j (m, n) as follows for the given m, n.We obtain the specific value of b(m, n) through a simple calculation. Therefore, From (2.2), for m = 2 and n = 2, we have 2), for m = 3 and n = 2, we have 2), for m = 4 and n = 2, we have In the Hardy space H 2 (T), z n z m is equal to z m-n , but in the weighted Bergmann space and for any n > 0, Therefore, Since Parseval's identity holds, we obtain that {N n (z)} ∪ {N m (z)} is complete (see [2]).Hence, for n ∈ N ∪ {0} and m ∈ N, {N n (z)} ∪ {N m (z)} forms an orthonormal basis for L 2 (P).
In [4], let P be an orthogonal projection of L 2 (D) onto the Bergamm space A 2 (D).Then, for nonnegative integers n, m, Next, we investigate the orthogonal projection P of N n N m .
Theorem 2.8 For any nonnegative integers m, n, where b m (n, mn + j) is the solution of the matrix equation as in Theorem 2.4 for 0 ≤ j ≤ n.
Proof For any nonnegative integer k, we obtain from Theorem 2.4 that for n ≥ k and Hence, if m ≥ n, then (2.6) becomes and if m < n, then P(N n N m ) = 0.
Remark 2.9 Note that b m (m, n) = b m (n, m) and b j (m, 0) = 1 for any nonnegative integer m, n, j.
(i) Since and by Theorem 2.8.
(ii) By Theorem 2.8 and Remark 2.4, we also obtain that and As an application of Theorem 2.8, we obtain the following corollary.

Corollary 2.10 For any nonnegative integers m, k with m
holds, where P denotes an orthogonal projection of L 2 (P) onto N 2 (P).
Proof By Theorem 2.8, we obtain that where b m (n, mk + j) denotes the solutions of the matrix equation as in Theorem 2.4 for 0 ≤ j ≤ n.
We finally find the matrices of Toeplitz operators T ϕ with harmonic symbols ϕ on the Newton spaces by using Theorems 2.4 and 2.8.In Theorem 2.11, we explain the characteristics of the entries of the Toeplitz matrix in Newton space.Applying this, by specifically using the coefficient of b j (m, n) in Corollary 2.14, it was found that the entries of the Toeplitz matrix in Newton space are expressed as a linear combination of the binomial coefficients of the given entries.

Theorem 2.11 For the harmonic symbol
and the adjoint of the matrix of T ϕ is given by , Proof For the harmonic symbol ϕ(z) = ∞ i=0 a i N i + ∞ i=1 a -i N i , the (m, n)th entry of the matrix of T ϕ with respect to orthonormal basis {N n } n≥0 of N 2 (P) is given by Then, there are two cases to consider.If m ≥ n, then (2.8) Thus, the first and third term of the right equation in (2.8) have no term of the form a i b m (i, n)N m .Hence, (2.7) becomes If m < n, then by a similar method, (2.7) gives

Thus, we have
where m and n are nonnegative integers.Hence, the matrix of T ϕ with respect to B = {N n } n≥0 is given by and the adjoint of the matrix of T ϕ is given by and, hence, we know that Proof The proof follows from Theorem 2.11.

Corollary 2.13 (i)
For the harmonic symbol ϕ(z) = a 1 N 1 + a 0 + a -1 N 1 , the matrix of T ϕ with respect to orthonormal basis B = {N 0 , N 1 } is given by the matrix of T ϕ with respect to orthonormal basis B = {N 0 , N 1 , N 2 } is given by a -3 N 3 be the harmonic symbol.Then, the matrix of T ϕ with respect to orthonormal basis B = {N 0 , N 1 , N 2 , N 3 } is given by Proof Since N 0 (z)N 0 (z) = b 0 (0, 0)N 0 (z) and N 1 (z)N 0 (z) = b 1 (1, 0)N 1 (z) by Theorem 2.4, we have b 0 (0, 0) = 1 and b 1 (1, 0) = 1.Moreover, since Corollary 2.14 Let ϕ(z) = n i=0 a i N i + n i=1 a -i N i be the harmonic symbol for even n.Then, the matrix of T ϕ with respect to orthonormal basis B = {N k } k=0,1,2,...,n is given by .
Proof The proof follows from Theorem 2.11 and Corollary 2.13.
Remark 2.15 Set the harmonic symbol.Then, the matrix of T ϕ with respect to orthonormal basis B = {N 0 , N 1 } is given by (ii) Let ϕ(z) = iN 2 -N 1 + 2 + iN 1 + 2N 2 be the harmonic symbol.Then, the matrix of T ϕ with respect to orthonormal basis B = {N 0 , N 1 , N 2 } is given by A conjugation on H is an antilinear operator C : H → H that satisfies C 2 = I and Cx, Cy = y, x for all x, y ∈ H.An operator T ∈ L(H) is complex symmetric if there exists a conjugation C on H such that T = CT * C.
Corollary 2.17 Assume that C and C μ,λ are conjugations on L 2 given by Cf (z) = f (z) and C μ,λ f (z) = μf (λz) for f ∈ N 2 (P) with |λ| = |μ| = 1, respectively.If for the harmonic symbol ϕ(z) = ∞ i=0 a i N i + ∞ i=1 a -i N i and the matrix of T ϕ with respect to orthonormal basis B = {N n } n≥0 , then the following statements are equivalent: Since the matrix of [T ϕ ] B is of the form as in Theorem 2.11, it follows that the matrix of C[T ϕ ] B C is the following: Then, [T ϕ ] B is complex symmetric with the conjugation C if and only if a i = a -i for i = 0, 1, 2, . . . .(ii) ⇔ (iii) Let ϕ(z) = ∞ i=0 a i N i + ∞ i=1 a -i N i be with respect to the basis B = {N n } ∞ n=0 .It is known from [5] that C μ,λ is unitarily equivalent to C 1,λ .Since the matrix of T ϕ is of the form as in Theorem 2.11, it follows that the matrix of C 1,λ T ϕ C 1,λ is the following: Then, [T ϕ ] B is complex symmetric with the conjugation C 1,λ if and only if a i = a -i for i = 0, 1, 2, . . . .

3 i=2 a i b 3
ϕ ] B is complex symmetric with the conjugation C.Example 2.19  Let C be a conjugation on L 2 given by Cf (z) = f (z) for f ∈ N 2 (P) and letB = {N 0 , N 1 , N 2 , N 3 }.(i) Let ϕ(z) = 2iN 3 + 2iN 2 + 3N 1 -7i + 3N 1 + 2iN 2 + 2iN 3be the harmonic symbol.If the matrix of T ϕ with respect to orthonormal basis B is given by[T ϕ ] B = ϕ ] B is complex symmetric with the conjugation C.(ii) If for the harmonic symbol ϕ(z) = 3 i=0 (N i + N i ), the matrix of T ϕ with respect to orthonormal basis B is given by [T ϕ ] B = ϕ ] B is complex symmetric with the conjugation C.