Edelweiss Applied Science and Technology ISSN: 2576-8484 Vol. 8, No. 6, 394-400 2024 Publisher: Learning Gate DOI: 10.55214/25768484.v8i6.2091 Β© 2024 by the author; licensee Learning Gate Β© 2024 by the author; licensee Learning Gate * Correspondence: o.osman@qu.edu.sa Factors affecting mobility of harmonic symbols Osman Abdalla Adam Osman1* 1Department of Mathematics, College of Science, Qassim University, Buraydah 51452, Saudi Arabia; o.osman@qu.edu.sa (Q.A.A.O.). Abstract: In this paper, we have illustrated the Bergmann domains and the Toeplitz operators with their symbiotic symbols in some special domains in order to clarify the properties that correspond to the fixed values. And we characterize the bounded harmonic functions for which the Toeplitz operators Bergman space are essentially commuting. Keywords: Bergman, Correspond, Function, Functions, Harmonic bounded, Invariant, Operator, Spaces, Toeplitz, Value. 1. Introduction Suppose that 𝑑𝐴 stands for the measure of the space defined on the open unit disk in 𝐷 of the complex plane 𝐿2(𝐷, 𝑑𝐴) represents the inner Hilbert space βŒ©π‘“, 𝑔βŒͺ = ∫𝐷 𝑓𝑔 𝑑𝐴 Bergmann space πΏπ‘Ž 2 is a set of function 𝐿2(𝐷, 𝑑𝐴) which properties that are included in it analytically𝐷. Which means that Bergmann's space πΏπ‘Ž 2 it is a closed subspace𝐿2(𝐷, 𝑑𝐴), and so there is an orthogonal projection 𝑃 from 𝐿2(𝐷, 𝑑𝐴) toπΏπ‘Ž 2 , [1,6]. For(πœ‘ + 1) ∈ 𝐿∞(𝐷, 𝑑𝐴), the Toeplitz effects with symbol(πœ‘ + 1), denoted π‘‡πœ‘+1 is operator from πΏπ‘Ž 2 to πΏπ‘Ž 2 knowledge beforeπ‘‡πœ‘+1𝑓 = 𝑃{(πœ‘ + 1)𝑓}. By harmonic function, we mean a function with a complex value over 𝐷 for which the Laplacian congruent is zero. Theorem 1. Suppose that (πœ‘ + 1) and (πœ“ + 1) definite harmonic functions on𝐷. So that π‘‡πœ‘+1π‘‡πœ“+1 = π‘‡πœ“+1π‘‡πœ‘+1 If and only if a- (πœ‘ + 1) And (πœ“ + 1) are both analytic on 𝐷. b- (οΏ½Μ…οΏ½) and (οΏ½Μ…οΏ½) also analytic on 𝐷. c- There are constants π‘Ž, 𝑏 ∈ 𝐢 not both equal to, such that π‘Ž(πœ‘ + 1) + 𝑏(πœ“ + 1) is constant in𝐷. We will see if the direction of that theorem is trivial, but proving the direction "only if" requires an inverse of the invariant form of the mean-value property. The clarification of Theorem 1 similar to a similar result installed in [2,7] for Toeplitz effects in the symbol 𝐿∞(πœ•π·) acts on Hardy space𝐻2(πœ•π·). Brown and Holmes prove these results through the effect matrix of Hardy spaces in Bergman spaces. πΏπ‘Ž 2 , Toeplitz effects do not have good matrices, and the techniques used by Brown and Holmes do not seem to work in this context. Thus function theory, rather than matrix manipulation, plays a large role in our proof. A special case of Theorem 1 was proved in [3.8], using function theory techniques quite different from those that we use here. Also proved a special case of Theorem 1; our proof makes use of some of his ideas. Functions in 𝐿∞(πœ•π·) correspond, via the Poisson integral, to bounded harmonic functions on𝐷, so the restriction in Theorem 1 to consideration only of Toeplitz effects with harmonic symbols is natural. 395 Edelweiss Applied Science and Technology ISSN: 2576-8484 Vol. 8, No. 6: 394-400, 2024 DOI: 10.55214/25768484.v8i6.2091 Β© 2024 by the author; licensee Learning Gate More importantly, Theorem 1 does not hold if β€œWe can replace measurable harmonic ". For example, Paul Bourdon has pointed out to us that if (πœ‘ + 1) and (πœ“ + 1) are any two radial functions in𝐿∞(𝐷, 𝑑𝐴), then π‘‡πœ‘+1π‘‡πœ“+1 = π‘‡πœ“+1π‘‡πœ‘+1(A function is called radial if its value at 𝑧 depends only on|𝑧|). Thus, the following open problem may be hard: Find conditions on functions (πœ‘ + 1) and (πœ“ + 1) in 𝐿∞(𝐷, 𝑑𝐴)that are necessary and sufficient for π‘‡πœ‘+1to commute with π‘‡πœ“+1 2. The Constant Property of the Argument Value A continuous function on the disk 𝐷 is harmonic if and only if it has the mean value property. We characterize harmonic functions in terms of an invariant mean value property. Let 𝐴𝑒𝑑(𝐷) denote the set of analytic, one-to-one maps of 𝐷 onto 𝐷 (Where 𝐴𝑒𝑑 stands for Automorphic). A function β„Ž on 𝐷 is in 𝐴𝑒𝑑(𝐷) if and only if there exist 𝛼 ∈ πœ•π· and 𝛽 ∈ 𝐷 such that β„Ž(𝑧) = 𝛼 𝛽 βˆ’ 𝑧 1 βˆ’ 𝛽 for all 𝑧 ∈ 𝐷 A function 𝑒 ∈ 𝐢(𝐷) is said to have the invariant mean value property if ∫ 𝑒 (β„Ž(π‘Ÿπ‘’π‘–πœƒ)) π‘‘πœƒ 2πœ‹ 2πœ‹ 0 = 𝑒(β„Ž(0)) (1) for every β„Ž ∈ 𝐴𝑒𝑑(𝐷) and every π‘Ÿ ∈ [0, 1). Here "invariant" refers to conformal invariance, meaning invariance under composition with elements of𝐴𝑒𝑑(𝐷). If 𝑒 is harmonic on 𝐷, then so is π‘’Β°β„Ž for everyβ„Ž ∈ 𝐴𝑒𝑑(𝐷); thus harmonic functions have the invariant mean value property. The converse is also true [4,6], if a function 𝑒 ∈ 𝐢(𝐷) has the invariant mean value property, then 𝑒 is harmonic on𝐷. The invariant mean value property concerns averages over circles with respect to arc length measure. Because we are dealing with the Bergman space, πΏπ‘Ž 2 we need an invariant condition stated in terms of an area average over𝐷. Thus we say that a function 𝑒 ∈ 𝐢(𝐷) ∩ 𝐿1(𝐷, 𝑑𝐴) has the area version of the invariant mean value property if βˆ«π·π‘’ΞΏβ„Ž π‘‘πœƒ 2πœ‹ = 𝑒(β„Ž(0)) (2) for every β„Ž ∈ 𝐴𝑒𝑑(𝐷). If 𝑒 is in𝐢(𝐷) ∩ 𝐿1(𝐷, 𝑑𝐴), then so is π‘’Β°β„Ž for everyβ„Ž ∈ 𝐴𝑒𝑑(𝐷), so the left-hand side of the above equation makes sense. Note that the area version of the invariant mean value property deals with integrals over all of𝐷, as opposed to integrals over π‘Ÿπ· forπ‘Ÿ ∈ (0, 1). If 𝑒 is harmonic on 𝐷 and in𝐿1(𝐷, 𝑑𝐴), then so is π‘’Β°β„Ž for everyβ„Ž ∈ 𝐴𝑒𝑑(𝐷). Thus, by the mean value property, harmonic functions have the area version of the invariant mean value property. Whether or not the converse is true is an open question. In other words, if 𝑒 ∈ 𝐢(𝐷) ∩ 𝐿1(𝐷, 𝑑𝐴) has the area version of the invariant mean value property, must 𝑒 be harmonic? This question has an affirmative answer if we replace the hypothesis that 𝑒 is in 𝐢(𝐷) ∩ 𝐿1(𝐷, 𝑑𝐴) with the stronger hypothesis that 𝑒 is in𝐢(οΏ½Μ…οΏ½); see in [4,5]. We need to consider functions that are not necessarily continuous on the closed disk, so the result mentioned in the last sentence will not suffice. However, our functions do have the property that their radiation are continuous on the closed disk, and we will prove that this property, along with the area version of the invariant mean value property, is enough to imply harmonicity. If𝑒 ∈ 𝐢(𝐷), then the radiation of 𝑒 denoted𝑅(𝑒), is the function on 𝐷 defined by 𝑅(𝑒)(𝑀) = ∫ 𝑒(π‘’π‘–πœƒ) 2πœ‹ 0 π‘‘πœƒ 2πœ‹ (3) In the following lemma, which will be a key tool in our proof of Theorem 1, the statement 𝑅(π‘’Β°β„Ž) ∈ 𝐢(οΏ½Μ…οΏ½) means 𝑅(π‘’Β°β„Ž) can be extended to a continuous complex valued function on οΏ½Μ…οΏ½ Lemma 2. Suppose that𝑒 ∈ 𝐢(𝐷) ∩ 𝐿1(𝐷, 𝑑𝐴). Then 𝑒 is harmonic on 𝐷 if and only if 396 Edelweiss Applied Science and Technology ISSN: 2576-8484 Vol. 8, No. 6: 394-400, 2024 DOI: 10.55214/25768484.v8i6.2091 Β© 2024 by the author; licensee Learning Gate βˆ«π·π‘’ΞΏβ„Ž 𝑑𝐴 πœ‹ = 𝑒(β„Ž(0)) (4) and 𝑅(π‘’Β°β„Ž) ∈ 𝐢(οΏ½Μ…οΏ½) for everyβ„Ž ∈ 𝐴𝑒𝑑(𝐷) (5) Proof. Suppose that 𝑒 is harmonic on𝐷. Letβ„Ž ∈ 𝐴𝑒𝑑(𝐷). As we discussed earlier, π‘’Β°β„Ž is harmonic and we see in Eq(4) holds. The mean value property implies that 𝑅(π‘’Β°β„Ž) is a constant function on𝐷, with value𝑒(β„Ž(0)), we see in Eq (5) also holds. To prove the other direction, suppose that Eq(4) and Eq(5) hold. Letβ„Ž ∈ 𝐴𝑒𝑑(𝐷), and let 𝑣 ∈ 𝑅(π‘’Β°β„Ž) from Eq(5) we fine 𝑣 ∈ 𝐢(οΏ½Μ…οΏ½) We want to show that 𝑣 has the area version of the invariant mean value property. To do this, fixg ∈ 𝐴𝑒𝑑(𝐷). Then βˆ«π·π‘’Β°g 𝑑𝐴 πœ‹ 𝑒 = βˆ«π·β„›(π‘’Β°β„Ž)(g(𝑀)) 𝑑𝐴(𝑀) πœ‹ = ∫𝐷 ∫ 𝑒 (β„Ž(g(𝑀)π‘’π‘–πœƒ)) 2πœ‹ 0 π‘‘πœƒ 2πœ‹ 𝑑𝐴(𝑀) πœ‹ (6) πœƒ ∈ [0,2πœ‹] To check that interchanging the order of integration in the last integral is valid, for each define π‘“πœƒ ∈ 𝐴𝑒𝑑(𝐷) by π‘“πœƒ(𝑀) = β„Ž(g(𝑀)π‘’π‘–πœƒ) The inverse π‘“πœƒ βˆ’1 of π‘“πœƒ is also an analytic automorphism of𝐷, so there exist 𝛼 ∈ πœ•π· and 𝛽 ∈ 𝐷 such that π‘“πœƒ βˆ’1(𝑧) = π›½βˆ’π‘ 1βˆ’οΏ½Μ…οΏ½π‘ , for all 𝑍 ∈ 𝐷 Thus |(π‘“πœƒ βˆ’1) / (𝑧)| = 1βˆ’|𝛽|2 |1βˆ’οΏ½Μ…οΏ½π‘§| 2 ≀ 1+|𝛽| 1βˆ’|𝛽| , for all 𝑍 ∈ 𝐷 Note that𝛽 = π‘“πœƒ(0) = β„Ž(g(0)π‘’π‘–πœƒ); we are thinking of β„Ž and g as fixed, so the above inequality shows there is a constant K such that |(π‘“πœƒ βˆ’1) / (𝑧)| ≀ 𝐾, for all 𝑍 ∈ 𝐷 and πœƒ ∈ [0,2πœ‹] Now ∫ ∫𝐷 |𝑒 (β„Ž(g(𝑀)𝑒 π‘–πœƒ))| 𝑑𝐴(𝑀) πœ‹ π‘‘πœƒ 2πœ‹ 2πœ‹ 0 = ∫ ∫𝐷|𝑒(𝑧)| |(π‘“πœƒ βˆ’1) / (𝑧)| 𝑑𝐴(𝑧) πœ‹ π‘‘πœƒ 2πœ‹ 2πœ‹ 0 ≀ 𝐾2∫𝐷|𝑒(𝑧)| 𝑑𝐴(𝑧) πœ‹ ≀ ∞ That is apply Fubini's Theorem to Eq(6), getting βˆ«π·π‘£Β°g 𝑑𝐴 πœ‹ = ∫ βˆ«π·π‘’ (β„Ž(g(𝑀)𝑒 π‘–πœƒ)) 𝑑𝐴(𝑀) πœ‹ π‘‘πœƒ 2πœ‹ 2πœ‹ 0 = ∫ ∫𝐷(π‘£Β°π‘“πœƒ(𝑀)) 𝑑𝐴(𝑀) πœ‹ π‘‘πœƒ 2πœ‹ 2πœ‹ 0 = ∫ 𝑒(π‘“πœƒ(0)) π‘‘πœƒ 2πœ‹ 2πœ‹ 0 = ∫ 𝑒 (β„Ž(g(0)π‘’π‘–πœƒ)) π‘‘πœƒ 2πœ‹ 2πœ‹ 0 = β„›(π‘’Β°β„Ž)(g(0)) = 𝑣(g(0)) Thus 𝑣 is a continuous function on οΏ½Μ…οΏ½ that has the area version of the invariant mean value property. Hence 𝑣 is harmonic on 𝐷 [4,5]. Because 𝑣 is also a radial function, the mean value property implies that 𝑣 is a constant function on 𝐷, with value 𝑣(0). Recall that𝑣 = β„›(π‘’Β°β„Ž), so ∫ (π‘’Β°β„Ž)(π‘Ÿπ‘’π‘–πœƒ) π‘‘πœƒ 2πœ‹ 2πœ‹ 0 = 𝑒(β„Ž(0)) for every π‘Ÿ ∈ [0, 1) and for each β„Ž ∈ 𝐴𝑒𝑑(𝐷). In other words, 𝑒 has the invariant mean value property. Thus in [4], 𝑒 is harmonic on 𝐷. As mentioned earlier, it is unknown whether Lemma 2 remains true if Eq(5) is deleted. We believe that the following proposition, which reduces this question to a tempting integral equation, is the best way to attack this problem. Patrick Ahern and Walter Rudin also independently proved Lemma 2 and Proposition 3 at about the same time we did. Proposition 3. Suppose that the constant functions are the only functions 𝑉 ∈ 𝐢([0,1)) ∩ 𝐿/[0,1] such that 397 Edelweiss Applied Science and Technology ISSN: 2576-8484 Vol. 8, No. 6: 394-400, 2024 DOI: 10.55214/25768484.v8i6.2091 Β© 2024 by the author; licensee Learning Gate 𝑣(𝑑) = (1 βˆ’ 𝑑)2 ∫ 1+𝑑𝑠 (1βˆ’π‘‘π‘ )2 1 0 𝑉(𝑠)𝑑𝑠, for every 𝑑 ∈ [0,1) (7) Then every function in 𝐢(𝐷) ∩ 𝐿/(𝐷, 𝑑𝐴) having the area version of the invariant mean value property is harmonic. Proof: First, suppose that 𝑣 is a radial function in 𝐢(𝐷) ∩ 𝐿/(𝐷, 𝑑𝐴) having the area version of the invariant mean value property. We will show that 𝑣 is constant on 𝐷. For 𝛼 ∈ [0,1), let β„Žπ›Ό ∈ 𝐴𝑒𝑑(𝐷) be defined by β„Žπ›Ό(𝑧) = π›Όβˆ’π‘§ 1βˆ’π›Όπ‘§ Note that β„Žπ›Ό is its own inverse under composition. For each 𝛼 ∈ [0,1) we have 𝑣(𝛼) = ∫𝐷((π‘£Β°β„Žπ›Ό))(𝑧) 𝑑𝐴(𝑧) πœ‹ = βˆ«π·π‘£(𝑀) |β„Žπ›Ό / (𝑀)| 2 𝑑𝐴(𝑧) πœ‹ = (1 βˆ’ 𝛼2)2 ∫ 𝑣(π‘Ÿ) 1 0 π‘Ÿ ∫ 1 |1βˆ’π›Όπ‘Ÿπ‘’π‘–πœƒ| 4 2πœ‹ 0 π‘‘πœƒ 2πœ‹ π‘‘π‘Ÿ = (1 βˆ’ 𝛼2)2 ∫ π‘Ÿ(1+π‘Ž2π‘Ÿ2) (1βˆ’π‘Ž2π‘Ÿ2)3 1 0 𝑣(π‘Ÿ)π‘‘π‘Ÿ = (1 βˆ’ 𝛼2)2 ∫ 1+π‘Ž2π‘Ÿ2 (1βˆ’π‘Ž2𝑠)3 1 0 𝑣(βˆšπ‘ )𝑑𝑠 In the above equation, replace 𝛼 with βˆšπ‘‘ and define a function 𝑉 on [0, 1) by𝑉(𝑑) = 𝑣(βˆšπ‘‘), transforming the above equation into Eq (7). Hence 𝑉 is constant on [0,1), and thus so is 𝑣, as claimed. To complete the proof, now suppose that 𝑒 is a function in𝐢(𝐷) ∩ 𝐿/(𝐷, 𝑑𝐴), having the area version of the invariant mean value property. Let β„Ž ∈ 𝐴𝑒𝑑(𝐷). Clearly, β„›(π‘’Β°β„Ž) is a radial function on𝐷, and, as shown in the proof of Lemma 2, has the area version of the invariant mean value property. By the above paragraph, β„›(π‘’Β°β„Ž) is constant on𝐷. In particular,β„›(π‘’Β°β„Ž) ∈ 𝐢(οΏ½Μ…οΏ½), and so by Lemma 2, 𝑒 is harmonic. 3. The Toeplitz Operators Forβ„Ž ∈ 𝐴𝑒𝑑(𝐷), define an operator π‘ˆβ„Ž on πΏπ‘Ž 2 by π‘ˆβ„Žπ‘“ = (π‘’Β°β„Ž)β„Ž A simple computation shows that π‘ˆβ„Ž is a unitary operator from πΏπ‘Ž 2 ontoπΏπ‘Ž 2 , with inverse π‘ˆβ„Žβˆ’1. In the following lemma that proof of Theorem 1. Lemma 4. Let β„Ž ∈ 𝐴𝑒𝑑(𝐷) and let(πœ‘ + 1) ∈ 𝐿∞(𝐷, 𝑑𝐴). Then π‘ˆβ„Žπ‘‡πœ‘ + 1π‘ˆβ„Ž βˆ— = 𝑇(πœ‘ + 1)Β°β„Ž Proof. Define π‘‰β„Ž: 𝐿 2(𝐷, 𝑑𝐴) β†’ 𝐿2(𝐷, 𝑑𝐴) byπ‘‰β„Žπ‘“ = (π‘“Β°β„Ž)β„Ž,. Then π‘‰β„Ž is a unitary operator from 𝐿2(𝐷, 𝑑𝐴) onto𝐿2(𝐷, 𝑑𝐴). Obviously π‘‰β„ŽπΏπ‘Ž 2 = π‘ˆβ„Ž . Thus π‘‰β„Ž maps πΏπ‘Ž 2 ontoπΏπ‘Ž 2 , so π‘ƒπ‘‰β„Ž = π‘‰β„Žπ‘ƒ (8) If 𝑓 ∈ πΏπ‘Ž 2 , so that 𝑇(πœ‘ + 1)Β°β„Žπ‘ˆβ„Žπ‘“ = 𝑇(πœ‘ + 1)Β°β„Ž(π‘“Β°β„Ž)β„Ž , = 𝑃[((πœ‘ + 1)Β°β„Ž)(π‘“Β°β„Ž)β„Ž,] = 𝑃[π‘‰β„Ž((πœ‘ + 1)𝑓)] = π‘‰β„Ž[𝑃((πœ‘ + 1)𝑓)] = π‘ˆβ„Žπ‘‡πœ‘ + 1𝑓 Thus 𝑇(πœ‘ + 1)Β°β„Žπ‘ˆβ„Ž = π‘ˆβ„Žπ‘‡πœ‘ + 1and because π‘ˆβ„Ž is unitary, this implies the desired result. Let π»πœ‘ + 1(𝐷) denote the usual is the Hardy space on the disk. It is well known that𝐻1(𝐷) βŠ‚ πΏπ‘Ž 2 . In the proof of Theorem 1 we will use, without comment, the following consequence: If𝑓, g ∈ 𝐻2(𝐷), then𝑓, g ∈ πΏπ‘Ž 2 , and thus 𝑓g ∈ 𝐿2(𝐷, 𝑑𝐴). We have now assembled all the ingredients needed to prove Theorem 1. Proof of theorem 1. If we begin with the easy direction. First suppose that (a) holds, so that (πœ‘ + 1) and (πœ“ + 1) are analytic on 𝐷 which means that π‘‡πœ‘ + 1 and π‘‡πœ“ + 1 are, respectively, the operators on πΏπ‘Ž 2 of 398 Edelweiss Applied Science and Technology ISSN: 2576-8484 Vol. 8, No. 6: 394-400, 2024 DOI: 10.55214/25768484.v8i6.2091 Β© 2024 by the author; licensee Learning Gate multiplication by (πœ‘ + 1) and (πœ“ + 1). So thatπ‘‡πœ‘ + 1π‘‡πœ“ + 1 = π‘‡πœ“ + 1π‘‡πœ‘ + 1. Now suppose that (b) holds, so that πœ‘ + 1Μ…Μ… Μ…Μ… Μ…Μ… Μ…Μ… and πœ“ + 1Μ…Μ… Μ…Μ… Μ…Μ… Μ…Μ… are analytic on𝐷. By the paragraph above, π‘‡πœ‘ + 1Μ…Μ… Μ…Μ… Μ…Μ… Μ…Μ… π‘‡πœ“ + 1Μ…Μ… Μ…Μ… Μ…Μ… Μ…Μ… = π‘‡πœ“ + 1Μ…Μ… Μ…Μ… Μ…Μ… Μ…Μ… π‘‡πœ‘ + 1Μ…Μ… Μ…Μ… Μ…Μ… Μ…Μ… . Take the adjoin of both sides of this equation, and use the identity π‘‡πœ‘+1Μ…Μ… Μ…Μ… Μ…Μ… βˆ— = π‘‡πœ‘ + 1 to conclude thatπ‘‡πœ‘ + 1π‘‡πœ“ + 1 = π‘‡πœ“ + 1π‘‡πœ‘ + 1. Finally suppose that (c) holds, so there exist constantsπ‘Ž, 𝑏 ∈ 𝐢, not both 0, such that π‘Ž(πœ‘ + 1) + 𝑏(πœ“ + 1) is constant on𝐷. Ifπ‘Ž β‰  0, then there exist constants 𝛽, 𝛾 ∈ 𝐢 such that πœ‘ + 1 = 𝛽(πœ“ + 1) + 𝛾 on𝐷, which means that π‘‡πœ‘ + 1 = π›½π‘‡πœ“ + 1 + 𝛾𝐼 (𝐼 denotes the identity operator on πΏπ‘Ž 2 ), which clearly implies that π‘‡πœ‘ + 1π‘‡πœ“ + 1 = π‘‡πœ“ + 1π‘‡πœ‘ + 1. If 𝑏 β‰  0 , we conclude in a similar fashion that π‘‡πœ‘ + 1 and π‘‡πœ“ + 1 commute. Now to prove the other direction of Theorem 1, suppose that (πœ‘ + 1) and (πœ“ + 1) are bounded harmonic functions on 𝐷 such thatπ‘‡πœ‘ + 1π‘‡πœ“ + 1 = π‘‡πœ“ + 1π‘‡πœ‘ + 1. Because πœ‘ + 1 and πœ“ + 1 are harmonic on𝐷, there exist functions 𝑓1, 𝑓2, g1 and g2 are analytic on 𝐷 such that t (πœ‘ + 1) = 𝑓1 + 𝑓2Μ… and (πœ“ + 1) = g1 + g2Μ…Μ… Μ… on 𝐷 (9) Because (πœ‘ + 1) and πœ“ + 1 are bounded on 𝐷, the functions 𝑓1, 𝑓2, g1 and g2 must be in 𝐻2(𝐷). This means that the function is constant at 𝐷. So π‘‡πœ‘ + 1π‘‡πœ“ + 11 = π‘‡πœ‘ + 1(𝑃(πœ“ + 1)) = π‘‡πœ‘ + 1(𝑃(g1 + g2Μ…Μ… Μ…)) = π‘‡πœ‘ + 1(g1 + g2(0)Μ…Μ… Μ…Μ… Μ…Μ… Μ…) = 𝑃([𝑓1 + 𝑓2Μ…][g1 + g2(0)Μ…Μ… Μ…Μ… Μ…Μ… Μ…]) = 𝑓1g1 + g2(0)Μ…Μ… Μ…Μ… Μ…Μ… ̅𝑓1 + 𝑃(𝑓2Μ…g1) + 𝑓2(0)Μ…Μ… Μ…Μ… Μ…Μ… Μ…g2(0)Μ…Μ… Μ…Μ… Μ…Μ… Μ… = βŒ©π‘‡πœ‘ + 1π‘‡πœ“ + 11,1βŒͺ = βŒ©π‘“1g1 + g2(0)Μ…Μ… Μ…Μ… Μ…Μ… ̅𝑓1 + 𝑓2Μ…g1 + 𝑓2(0)Μ…Μ… Μ…Μ… Μ…Μ… Μ…g2(0)Μ…Μ… Μ…Μ… Μ…Μ… Μ…, 1βŒͺ = ∫𝐷(𝑓1g1 + g2(0)Μ…Μ… Μ…Μ… Μ…Μ… ̅𝑓1 + 𝑓2Μ…g1 + 𝑓2(0)Μ…Μ… Μ…Μ… Μ…Μ… Μ…g2(0)Μ…Μ… Μ…Μ… Μ…Μ… Μ…)𝑑𝐴 = πœ‹[𝑓2(0)g1(0) + 𝑓1(0)g2(0)Μ…Μ… Μ…Μ… Μ…Μ… Μ… + 𝑓2(0)Μ…Μ… Μ…Μ… Μ…Μ… Μ…g2(0)Μ…Μ… Μ…Μ… Μ…Μ… Μ…] + βˆ«π·π‘“2Μ…g1𝑑𝐴 (10) A sirnilar formula can be obtained fo βŒ©π‘‡πœ‘ + 1π‘‡πœ“ + 11,1βŒͺ Because π‘‡πœ‘ + 1π‘‡πœ“ + 1 = π‘‡πœ“ + 1π‘‡πœ‘ + 1 we can set the right-hand side of Eq (10) equal to the corresponding formula for βŒ©π‘‡πœ‘ + 1π‘‡πœ“ + 11,1βŒͺ, getting ∫𝐷(𝑓2Μ…g1 βˆ’ 𝑓1g2Μ…Μ… Μ…) 𝑑𝐴 πœ‹ = 𝑓2(0)Μ…Μ… Μ…Μ… Μ…Μ… Μ…g1(0) βˆ’ 𝑓1(0)g2(0)Μ…Μ… Μ…Μ… Μ…Μ… Μ… (11) Let β„Ž ∈ 𝐴𝑒𝑑(𝐷). Multiplying both sides of the equation π‘‡πœ‘ + 1π‘‡πœ“ + 1 = π‘‡πœ“ + 1π‘‡πœ‘ + 1 on the left and by π‘ˆβ„Ž βˆ— on the right, and recalling that π‘ˆβ„Ž is unitary so thatπ‘ˆβ„Ž βˆ—π‘ˆβ„Ž = 1, we get π‘ˆβ„Žπ‘‡πœ‘ + 1π‘ˆβ„Ž βˆ—π‘ˆβ„Žπ‘‡πœ“ + 1π‘ˆβ„Ž βˆ— = π‘ˆβ„Žπ‘‡πœ“ + 1π‘ˆβ„Ž βˆ—π‘ˆβ„Žπ‘‡πœ‘ + 1π‘ˆβ„Ž βˆ— Lemma 4 now shows that 𝑇(πœ‘ + 1)βˆ—β„Žπ‘‡(πœ“ + 1)βˆ—β„Ž = 𝑇(πœ“ + 1)βˆ—β„Žπ‘‡(πœ‘ + 1)βˆ—β„Ž (12) Composing both sides of the equations in Eq(8) with β„Ž expresses each of the bounded harmonic functions (πœ‘ + 1) βˆ— β„Ž and (πœ“ + 1) βˆ— β„Ž as the sum of an analytic function and a conjugate analytic function: (πœ‘ + 1) βˆ— β„Ž = 𝑓1 βˆ— β„Ž + 𝑓2Μ… βˆ— β„Ž and (πœ“ + 1) βˆ— β„Ž = g1 βˆ— β„Ž + g2Μ…Μ… Μ… βˆ— β„Ž on 𝐷 (13) From Eq(11) was derived under the assumption that π‘‡πœ‘ + 1π‘‡πœ“ + 1 = π‘‡πœ“ + 1π‘‡πœ‘ + 1;thus Eq(12), combined with Eq(14), says that Eq(11) is still valid when we replace each function in it by its composition with β„Ž. In other words , ∫𝐷(𝑓2Μ…g1 βˆ’ 𝑓1g2Μ…Μ… Μ…) βˆ— β„Ž 𝑑𝐴 πœ‹ = 𝑓2Μ…(β„Ž(0))g1(β„Ž(0)) βˆ’ 𝑓1(β„Ž(0))g2Μ…Μ… Μ…(β„Ž(0)) Letting 𝑒 = 𝑓2Μ…g1 βˆ’ 𝑓1g2Μ…Μ… Μ… the equation above becomes βˆ«π·π‘’ βˆ— β„Ž 𝑑𝐴 πœ‹ = 𝑒(β„Ž(0)) In the other words, 𝑒 has the area version of the invariant mean value property. 399 Edelweiss Applied Science and Technology ISSN: 2576-8484 Vol. 8, No. 6: 394-400, 2024 DOI: 10.55214/25768484.v8i6.2091 Β© 2024 by the author; licensee Learning Gate We can want to show that 𝑒 is harmonic on𝐷. By the above equation and Lemma 2, we need only show thatβ„›(𝑒 βˆ— β„Ž) ∈ 𝐢(οΏ½Μ…οΏ½). To do this, represent the analytic functions 𝑓2 βˆ— β„Ž and g1 βˆ— β„Ž as Taylor series: (𝑓2 βˆ— β„Ž)(𝑧) = βˆ‘π›Όπ‘›π‘ 𝑛 ∞ 𝑛=0 π‘Žπ‘›π‘‘ (g1 βˆ— β„Ž)(𝑧)βˆ‘π›½π‘›π‘ 𝑛 ∞ 𝑛=0 , π‘“π‘œπ‘Ÿ π‘Žπ‘™π‘™ 𝑍 ∈ 𝐷 Because (πœ‘ + 1) βˆ— β„Ž and (πœ“ + 1) βˆ— β„Ž are bounded harmonic functions on𝐷, Eq(13) implies that functions 𝑓2 βˆ— β„Ž and g1 βˆ— β„Ž are in 𝐻2(𝐷) 𝐻2(𝐷), so that βˆ‘|𝛼𝑛| 2 ∞ 𝑛=0 < ∞ π‘Žπ‘›π‘‘ βˆ‘|𝛽𝑛| 2 ∞ 𝑛=0 < ∞ (14) Now for 𝑍 ∈ 𝐷 we have β„› ((𝑓2Μ…g1) βˆ— β„Ž) (𝑧) = ∫ (𝑓2Μ… βˆ— β„Ž)(𝑧𝑒 π‘–πœƒ) 2πœ‹ 0 (g1 βˆ— β„Ž)(𝑧𝑒 π‘–πœƒ) π‘‘πœƒ 2πœ‹ = βˆ‘π›Όπ‘›Μ…Μ…Μ…Μ… 𝛽𝑛|𝑍| 2𝑛 ∞ 𝑛=0 The inequalities in Eq(14) imply that βˆ‘ |𝛼𝑛𝛽𝑛| ∞ 𝑛=0 < ∞, so the above formula for β„› ((𝑓2Μ…g1) βˆ— β„Ž) ∈ 𝐢(𝐷) shows that β„› ((𝑓2Μ…g1) βˆ— β„Ž) ∈ 𝐢(οΏ½Μ…οΏ½); similarly, we get that β„›((𝑓1g2Μ…Μ… Μ…) βˆ— β„Ž) ∈ 𝐢(οΏ½Μ…οΏ½). So thatβ„›(𝑒 βˆ— β„Ž) ∈ 𝐢(οΏ½Μ…οΏ½), as desired. Thus at this stage of the proof we know that 𝑒 is harmonic. Let πœ• πœ•π‘§ and the πœ• πœ•π‘§ denote the usual operators defined by πœ• πœ•π‘§ = 1 2 ( πœ• πœ•π‘₯ βˆ’ 𝑖 πœ• πœ•π‘¦ ) π‘Žπ‘›π‘‘ πœ• πœ•π‘§ = 1 2 ( πœ• πœ•π‘₯ + 𝑖 πœ• πœ•π‘¦ ) If 𝑓 is analytic, then the Cauchy-Riemann equations show that πœ•π‘“ πœ•π‘§ = 𝑓, πœ•π‘“ πœ•π‘§ = 0, πœ•π‘“ πœ•π‘§ = 0, π‘Žπ‘›π‘‘ πœ•π‘“ πœ•π‘§ = 𝑓 . It is easy to check that πœ• πœ•π‘§ and πœ• πœ•π‘§ obey the usual addition and multiplication formulas for derivatives and that πœ•2 πœ•π‘₯2 + πœ•2 πœ•π‘¦2 = 4 πœ• πœ•π‘§ πœ• πœ•π‘§ Thus, because 𝑒 is harmonic, we have 0 = 4 πœ• πœ•π‘§ ( πœ•π‘’ πœ•π‘§ ) = 4 πœ• πœ•π‘§ | πœ•(𝑓2Μ…g1 βˆ’ 𝑓1g2Μ…Μ… Μ…) πœ•π‘§ | = 4 πœ• πœ•π‘§ (𝑓2Μ…g1 , βˆ’ f1 , g2Μ…Μ… Μ…) = 4(𝑓2Μ…g1 , βˆ’ 𝑓1 ,g2Μ…Μ… Μ…) Hence 𝑓1Μ…g2 , βˆ’ 𝑓2 ,g1Μ…Μ… Μ… (15) We finish the proof by showing that the above equation implies that (a), (b), or (c) Holds. If g1 , is identically 0 on𝐷, then Eq(l5) shows that either g2 , is identically 0 on 𝐷, then (πœ“ + 1) would be constant on 𝐷 and (c) would hold or 𝑓1 , is identically 0 on 𝐷, so both (πœ‘ + 1) and (πœ“ + 1) would be analytic on 𝐷 and (b) would hold. Similarly, if g2 , is identically 0 on𝐷, then Eq(l5) shows that either (c) or (a) would hold. Thus, we may assume that neither g1 , nor g2 , is identically 0 on𝐷, and so Eq (15) shows that at all points of 𝐷 except the countable set consisting of the zeroes of g1 , g2 , . 𝑓1 , g1 , = [ 𝑓2 , g2 , ] βˆ’ The left-hand side of the above equation is an analytic function on 𝐷 with the zeroes ofg1 , g2 , , deleted, and the right hand side is the complex conjugate of an analytic function on the same domain, and so both sides must equal a constant 𝑐 ∈ 𝐢. Thus𝑓1 , = 𝑐g1 , , and 𝑓2 , = 𝑐 g2 , on𝐷. Hence 𝑓1 βˆ’ 𝑐g1 and 𝑓 2 βˆ’ 𝑐g 2 are constant on 𝐷, and so their sum, which equals (πœ‘ + 1) βˆ’ 𝑐(πœ“ + 1), is constant on 𝐷; in other 400 Edelweiss Applied Science and Technology ISSN: 2576-8484 Vol. 8, No. 6: 394-400, 2024 DOI: 10.55214/25768484.v8i6.2091 Β© 2024 by the author; licensee Learning Gate words, (c) holds and the proof of Theorem 1 is complete. Recall that an operator is called normal if it commutes with its adjoint. We can use Theorem1 to prove the following corollary, which states that for (πœ‘ + 1) a bounded harmonic function on𝐷, the Toeplitz operator π‘‡πœ‘ + 1is normal only in the obvious case. Corollary 5. Suppose that (πœ‘ + 1) is a bounded harmonic function on𝐷. Then π‘‡πœ‘ + 1 is a normal operator if and only if (πœ‘ + 1)(𝐷) lies on some line in𝐢. Proof: First, suppose that (πœ‘ + 1)(𝐷) lies on some line in𝐢. Then there exist constants𝛼, 𝛽 ∈ 𝐢, with𝛼 β‰  0, such that𝛼(πœ‘ + 1) + 𝛽, is real valued on𝐷. So that 𝑇𝛼(πœ‘ + 1)+𝛽 is a self-adjoint operator, and hence π‘‡πœ‘ + 1 which equalsπ›Όβˆ’1𝑇𝛼(πœ‘ + 1)+π›½βˆ’π›½πΌ, is a normal operator. To prove the other direction, suppose now that π‘‡πœ‘ + 1 is a normal operator. So that π‘‡πœ‘ + 1π‘‡πœ‘ + 1Μ…Μ… Μ…Μ… Μ…Μ… Μ…Μ… = π‘‡πœ‘ + 1Μ…Μ… Μ…Μ… Μ…Μ… Μ…Μ… π‘‡πœ‘ + 1 and so Theorem 1 implies that (πœ‘ + 1) and (πœ‘ + 1Μ…Μ… Μ…Μ… Μ…Μ… Μ…Μ… Μ…) are both analytic on 𝐷 (in which case (πœ‘ + 1) is constant, so we are done) or there are Constants𝛼, 𝛽 ∈ 𝐢, not both 0, such that π‘Ž(πœ‘ + 1) + 𝑏(πœ‘ + 1Μ…Μ… Μ…Μ… Μ…Μ… Μ…Μ… Μ…) (πœ‘ + 1) is constant on 𝐷. The latter condition implies that (πœ‘ + 1)(𝐷) lies on a line. Copyright: Β© 2024 by the authors. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (https://creativecommons.org/licenses/by/4.0/). 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