Hardy spaces of Dirichlet series and function theory on polydiscs

Transcription

1 Hardy spaces of Dirichlet series and function theory on polydiscs Kristian Seip Norwegian University of Science and Technology (NTNU) Steinkjer, September 11 12, 2009

2 Summary Lecture One Theorem (H. Bohr) Suppose f is in H. Then sup f (σ + it) f ; σ>0

3 Summary Lecture One Theorem (H. Bohr) Suppose f is in H. Then sup f (σ + it) f ; σ>0 more generally, the Bohr lift F of f defines an analytic function in D c 0 and sup z D c 0 F(z) f.

4 Summary Lecture One Theorem (H. Bohr) Suppose f is in H. Then sup f (σ + it) f ; σ>0 more generally, the Bohr lift F of f defines an analytic function in D c 0 and sup z D c 0 F(z) f. Theorem (Cole & Gamelin 1986) A Dirichlet series f in H p defines an analytic function in C + 1/2 and f (s) p ζ(2σ) f p p.

5 Convergence of Dirichlet series In general, a Dirichlet series has several half-planes of convergence, as shown in the picture: Figure: Convergence regions for Dirichlet series

6 Abscissae of convergence for functions in H p So far we know that Dirichlet series in H has B 0

7 Abscissae of convergence for functions in H p So far we know that Dirichlet series in H has B 0 H p for p 2 has A 1/2

8 Abscissae of convergence for functions in H p So far we know that Dirichlet series in H has B 0 H p for p 2 has A 1/2 H p for 1 p < 2 has A 1/p (by Hausdorff Young).

9 Abscissae of convergence for functions in H p So far we know that Dirichlet series in H has B 0 H p for p 2 has A 1/2 H p for 1 p < 2 has A 1/p (by Hausdorff Young). A theorem of F. Bayart shows that in fact A 1/2 for all p 1:

10 Bayart s theorem Theorem (Bayart 2002) The map F(z) F(r 1 z 1, r 2 z 2,...) is bounded from H 1 (T ) to H 2 (T ) if #{r j : r j > 1/ 2} <.

11 Bayart s theorem Theorem (Bayart 2002) The map F(z) F(r 1 z 1, r 2 z 2,...) is bounded from H 1 (T ) to H 2 (T ) if #{r j : r j > 1/ 2} <. Thus, in particular, f (s) f (s + ε) is bounded from H 1 to H 2.

12 Bayart s theorem Theorem (Bayart 2002) The map F(z) F(r 1 z 1, r 2 z 2,...) is bounded from H 1 (T ) to H 2 (T ) if #{r j : r j > 1/ 2} <. Thus, in particular, f (s) f (s + ε) is bounded from H 1 to H 2. Moreover, we get the following interesting estimate on the coefficients of an f in H 1 : a n 2 2 Ω(n) f 2 1. n=1

13 Two integral means formulas Lemma If f is a Dirichlet polynomial, we have f p 1 T p = lim f (it) p dt. T T 0 The lemma gives us an alternate way to define H p.

14 Two integral means formulas Lemma If f is a Dirichlet polynomial, we have f p 1 T p = lim f (it) p dt. T T 0 The lemma gives us an alternate way to define H p. Theorem If f is in H, then f p p = lim σ 0 + ( lim T 1 T T 0 f (σ + it) p dt ).

15 Theorem on multipliers Theorem (Hedenmalm Lindqvist Seip 1997, Bayart 2002) The set of multipliers for H p (1 p < ) consists of those Dirichlet series that represent bounded analytic functions in C + ; the multiplier norm of a multiplier m is m H (C + ). Corollary: The space of multipliers can be identified with H and f = f H (C + ).

16 Bohr s problem on absolute convergence Our previous theorem also gives Bohr s inequality A B 1/2 (proved in 1913).

17 Bohr s problem on absolute convergence Our previous theorem also gives Bohr s inequality A B 1/2 (proved in 1913). Question (Bohr) How large can A B be?

18 Bohr s problem on absolute convergence Our previous theorem also gives Bohr s inequality A B 1/2 (proved in 1913). Question (Bohr) How large can A B be? Bohr was unable to exhibit even one example such that A U > 0.

19 Bohr s problem on absolute convergence Our previous theorem also gives Bohr s inequality A B 1/2 (proved in 1913). Question (Bohr) How large can A B be? Bohr was unable to exhibit even one example such that A U > 0. There was no progress on this problem until 1931 when Bohnenblust and Hille solved it completely by giving examples such that A U = 1/2.

20 The Bohnenblust Hille theorem on absolute convergence The Bohnenblust Hille theorem can be rephrased as: Theorem The infimum of those c > 0 such that an n c < + for every n=1 a nn s in H equals 1/2.

21 Refined version of the Bohnenblust Hille theorem Theorem The supremum of the set of real numbers c such that n=1 a n n 1 2 exp {c } log n log log n < for every n=1 a nn s in H equals 1/ 2. This is a refinement of a theorem of R. Balasubramanian, B. Calado, and H. Queffélec (2006); the theorem as stated above is contained in a preprint of F. Defant, L. Frerick, J. Ortega-Cerdà, M. Ounaïes, and K. Seip (2009), where the novelty is the precise bound for the constant c.

22 The Bohnenblust Hille inequality

23 The Bohnenblust Hille inequality Let P(z) = α =m a αz α be an m-homogeneous polynomial in n complex variables.

24 The Bohnenblust Hille inequality Let P(z) = α =m a αz α be an m-homogeneous polynomial in ( ) 1 n complex variables. Then trivially α =m a α 2 2 P, where P = sup z D n P(z).

25 The Bohnenblust Hille inequality Let P(z) = α =m a αz α be an m-homogeneous polynomial in ( ) 1 n complex variables. Then trivially α =m a α 2 2 P, where P = sup z D n P(z). Is it possible to have a similar inequality ( α =m a α p) 1 p C P for some p < 2 with C depending on m but not on n?

26 The Bohnenblust Hille inequality Let P(z) = α =m a αz α be an m-homogeneous polynomial in ( ) 1 n complex variables. Then trivially α =m a α 2 2 P, where P = sup z D n P(z). Is it possible to have a similar inequality ( α =m a α p) 1 p C P for some p < 2 with C depending on m but not on n? Bohnenblust Hille YES, and 2m/(m + 1) is the smallest possible p.

27 The Bohnenblust Hille inequality Let P(z) = α =m a αz α be an m-homogeneous polynomial in ( ) 1 n complex variables. Then trivially α =m a α 2 2 P, where P = sup z D n P(z). Is it possible to have a similar inequality ( α =m a α p) 1 p C P for some p < 2 with C depending on m but not on n? Bohnenblust Hille YES, and 2m/(m + 1) is the smallest possible p. It is of basic interest to know the asymptotic behavior of C when p = 2m/(m + 1) and m.

28 A multilinear inequality In 1930, Littlewood proved that for every bilinear form B : C n C n C we have ( i,j B(e i, e j ) 4/3 ) 3/4 2 sup z,w D n B(z, w).

29 A multilinear inequality In 1930, Littlewood proved that for every bilinear form B : C n C n C we have ( i,j B(e i, e j ) 4/3 ) 3/4 2 sup z,w D n B(z, w). This was extended to m-linear forms by Bohnenblust and Hille in 1931: ( i 1,...,i m B(e i1,..., e im ) 2m m+1 ) m+1 2m The exponent 2m/(m + 1) is best possible. 2 m 1 sup z i D n B(z 1,..., z m ).

30 The Bohnenblust Hille inequality A new result is that also the polynomial Bohnenblust Hille inequality is hypercontractive: Theorem (Defant, Frerick, Ortega-Cerdà, Ounaïes, Seip 2009) Let m and n be positive integers larger than 1. Then we have ( a α 2m ) m+1 m+1 2m e m( 2) m 1 sup a α z α α =m z D n α =m for every m-homogeneous polynomial α =m a αz α on C n.

31 The Bohnenblust Hille inequality A new result is that also the polynomial Bohnenblust Hille inequality is hypercontractive: Theorem (Defant, Frerick, Ortega-Cerdà, Ounaïes, Seip 2009) Let m and n be positive integers larger than 1. Then we have ( a α 2m ) m+1 m+1 2m e m( 2) m 1 sup a α z α α =m z D n α =m for every m-homogeneous polynomial α =m a αz α on C n. The novelty here is the hypercontractivity, i.e., the constant grows exponentially with m; known since the work of Bohnenblust Hille that the inequality holds with constant m m/2, modulo a factor of exponential growth.

32 Polarization There is an obvious relationship between the multilinear and polynomial inequalities, which goes via a one-to-one correspondence between symmetric multilinear forms and homogeneous polynomials.

33 Polarization There is an obvious relationship between the multilinear and polynomial inequalities, which goes via a one-to-one correspondence between symmetric multilinear forms and homogeneous polynomials. Definition We say that the m-linear form B is symmetric if B(e i 1,..., e im ) = B(e i σ(1),..., e i σ(m)) for every index set (i 1,..., i m ) and every permutation σ of the set {1,..., m}.

34 Polarization continued If we restrict a symmetric multilinear form to the diagonal P(z) = B(z,..., z), then P is a homogeneous polynomial.

35 Polarization continued If we restrict a symmetric multilinear form to the diagonal P(z) = B(z,..., z), then P is a homogeneous polynomial. Conversely: Given a homogeneous polynomial P : C n C of degree m, we may define a symmetric m-multilinear form B : C n C n C so that B(z,..., z) = P(z).

36 Polarization continued If we restrict a symmetric multilinear form to the diagonal P(z) = B(z,..., z), then P is a homogeneous polynomial. Conversely: Given a homogeneous polynomial P : C n C of degree m, we may define a symmetric m-multilinear form B : C n C n C so that B(z,..., z) = P(z). Namely, write P(z) = i 1 i m c(i 1,..., i m )z i1 z im, and let B be the symmetric m-multilinear form such that B(e i 1,, e im ) = c(i 1,..., i m )/ i when i 1 i m and i is the number of different indices that can be obtained from the index i = (i 1,..., i m ) by permutation.

37 Polarization continued If we restrict a symmetric multilinear form to the diagonal P(z) = B(z,..., z), then P is a homogeneous polynomial. Conversely: Given a homogeneous polynomial P : C n C of degree m, we may define a symmetric m-multilinear form B : C n C n C so that B(z,..., z) = P(z). Namely, write P(z) = i 1 i m c(i 1,..., i m )z i1 z im, and let B be the symmetric m-multilinear form such that B(e i 1,, e im ) = c(i 1,..., i m )/ i when i 1 i m and i is the number of different indices that can be obtained from the index i = (i 1,..., i m ) by permutation. (Note: i m!.)

38 Harris s lemma Lemma (Harris 1975) We have sup z i D n B(z 1,..., z m ) mm m! P.

39 Harris s lemma Lemma (Harris 1975) We have sup B(z 1,..., z m ) mm z i D n m! P. Since the number of coefficients of B obtained from one coefficient of P is bounded by m!, a direct application of Harris s lemma and the multilinear Bohnenblust Hille inequality gives ( ) 1 (m 1)/2m ( m! α =m a α 2m ) m+1 m+1 2m mm m! P ; we obtain then the afore-mentioned constant m m/2, modulo an exponential factor.

40 Harris s lemma Lemma (Harris 1975) We have sup B(z 1,..., z m ) mm z i D n m! P. Since the number of coefficients of B obtained from one coefficient of P is bounded by m!, a direct application of Harris s lemma and the multilinear Bohnenblust Hille inequality gives ( ) 1 (m 1)/2m ( m! α =m a α 2m ) m+1 m+1 2m mm m! P ; we obtain then the afore-mentioned constant m m/2, modulo an exponential factor. Therefore, to make the required improvement, one needs a refinement of the argument via multilinear forms.

41 Two lemmas Lemma (Blei 1979) For all sequences (c i ) i where i = (i 1,..., i m ) and i k = 1,..., n, we have ( n i 1,...,i m=1 ) c i 2m m+1 2m m+1 Lemma (Bayart 2002) 1 k m [ n ( i k =1 For any homogeneous polynomial P(z) = ( α =m i 1,...,i k 1,i k+1,...,i m c i 2 α =m ) 1 ] 1 2 m. a α z α on C n : a α 2) 1 2 ( 2) m a α z α L. 1 (T n ) α =m

42 The proof 1/3 We write again the polynomial P as P(z) = c(i 1,..., i m )z i1 z im. i 1 i m We have i 1 i m c(i 1,..., i m ) 2m/(m+1) ( c(i1,..., i m ) i 1,...,i m By Blei s lemma, the last sum is bounded by m [ n ( 2 c(i 1,..., i ) m) 1/2 ] 1/m m [ n ( m i k=1 i k =1 i k k=1 i k =1 i 1/2 i k ) 2m/(m+1) i k c(i 1,..., i m) 2 i 2 ) 1/2] 1/m.

43 The proof 2/3 We now freeze the variable i k and group the terms to make a polynomial again: ( i k c(i 1,..., i m ) 2 ) 1/2 ( ) 1/2 = i 2 i k B(e i1,..., e im ) 2 = Pk 2. i k i k where P k (z) is the polynomial P k (z) = B(z,..., z, e i k, z,..., z). Now we use Bayart s estimate and get ( i k c(i 1,..., i m ) 2 i 2 i k ) 1/2 m 1 2 T n B(z,..., z, e i k, z,..., z).

44 The proof 3/3 We replace e i k by λe i k with λ = 1. We take τ k (z) = λ k (z)e i k in such a way that n ( i k =1 i k c(i 1,..., i m ) i 2 i k ) 1/2 m 1 2 e m 2 m 1 P, T n B(z,..., τ k (z),..., z) where in the last step we used Harris s lemma. Finally, c(i 1,..., i m ) 2m/(m+1) e m m 1 2 m P. i 1 i m

45 The proof 3/3 We replace e i k by λe i k with λ = 1. We take τ k (z) = λ k (z)e i k in such a way that n ( i k =1 i k c(i 1,..., i m ) i 2 i k ) 1/2 m 1 2 e m 2 m 1 P, T n B(z,..., τ k (z),..., z) where in the last step we used Harris s lemma. Finally, c(i 1,..., i m ) 2m/(m+1) e m m 1 2 m P. i 1 i m (The factor e m can be reduced to e by use of a refined version of Harris s lemma.)

46 Consequences of the hypercontractive BH inequality The hypercontractive version of the polynomial Bohnenblust Hille inequality may seem marginal improvement, but it has several interesting consequences: It leads to precise asymptotic results regarding certain Sidon sets, Bohr radii, as well as our result on absolute convergence of Dirichlet series.

47 Sidon sets Definition If G is an Abelian compact group and Γ its dual group, a subset of the characters S Γ is called a Sidon set if a γ C a γ γ γ S γ S The smallest constant C(S) is called the Sidon constant of S.

48 Sidon sets Definition If G is an Abelian compact group and Γ its dual group, a subset of the characters S Γ is called a Sidon set if a γ C a γ γ γ S γ S The smallest constant C(S) is called the Sidon constant of S. We estimate the Sidon constant for homogeneous polynomials: Definition S(m, n) is the smallest constant C such that the inequality α =m a α C P holds for every m-homogeneous polynomial in n complex variables P = α =m a αz α.

49 The Sidon constant for homogeneous polynomials Since the ) number of different monomials of degree m is, Hölder s inequality gives: ( n+m 1 m Corollary Let m and n be positive integers larger than 1. Then S(m, n) e m( ( ) m 1 n + m 1 2) m 1 2m. m

50 The Sidon constant for homogeneous polynomials Since the ) number of different monomials of degree m is, Hölder s inequality gives: ( n+m 1 m Corollary Let m and n be positive integers larger than 1. Then S(m, n) e m( ( ) m 1 n + m 1 2) m 1 2m. m (We also have the trivial estimate (n ) + m 1 S(m, n), m so the corollary is of interest only when log n m.)

51 The n-dimensional Bohr radius Definition The n-dimensional Bohr radius K n is the largest r > 0 such that all polynomials α c αz α satisfy sup c α z α sup c α z α. z rd n z D n α α

52 The n-dimensional Bohr radius Definition The n-dimensional Bohr radius K n is the largest r > 0 such that all polynomials α c αz α satisfy sup c α z α sup c α z α. z rd n z D n α α When n = 1, this is the classical Bohr radius studied by H. Bohr in 1913; M. Riesz, I. Schur and F. Wiener proved that K 1 = 1/3.

53 The n-dimensional Bohr radius Definition The n-dimensional Bohr radius K n is the largest r > 0 such that all polynomials α c αz α satisfy sup c α z α sup c α z α. z rd n z D n α α When n = 1, this is the classical Bohr radius studied by H. Bohr in 1913; M. Riesz, I. Schur and F. Wiener proved that K 1 = 1/3. When n > 1, the precise value of K n is unknown. Problem Determine the asymptotic behavior of K n when n.

54 Footnote on F. Wiener The initial F is not a misprint: F. Wiener is the mathematician Friedrich Wilhelm Wiener, born in 1884, and probably a casualty of World War One. See Boas and Khavinson s biography / v1.pdf.

55 Asymptotic behavior of K n The problem was studied by Boas and Khavinson in They showed that 1 1 log n 3 n K n 2 n.

56 Asymptotic behavior of K n The problem was studied by Boas and Khavinson in They showed that 1 1 log n 3 n K n 2 n. In 2006, Defant and Frerick showed that: log n c n log log n K n.

57 Asymptotic behavior of K n The problem was studied by Boas and Khavinson in They showed that 1 1 log n 3 n K n 2 n. In 2006, Defant and Frerick showed that: log n c n log log n K n. Theorem (DFOOS 2009) The n-dimensional Bohr radius satisfies c log n n log n K n 2 n.

58 Proof of theorem on the Bohr radius We use a well known lemma of F. Wiener. Lemma Let P be a polynomial in n variables and P = m 0 P m its expansion in homogeneous polynomials. If P 1, then P m 1 P 0 2 for every m > 0.

59 Proof of theorem on the Bohr radius We use a well known lemma of F. Wiener. Lemma Let P be a polynomial in n variables and P = m 0 P m its expansion in homogeneous polynomials. If P 1, then P m 1 P 0 2 for every m > 0. We assume that sup D n a α z α 1. Observe that for all z in rd n, aα z α a 0 + r m a α. m>1 α =m

60 End of proof If we take into account the estimates (log n) m ( ) n + m 1 m! and e m( 1 + n ) m, n m m then we use the Sidon estimate and F. Wiener lemma: r m a α r m e m(2 e) m( n ) m/2(1 a0 2 ). log n m>1 m>1 α =m log n Choosing r ε n with ε small enough, we obtain aα z α a 0 + (1 a 0 2 )/2 1 whenever a 0 1.

61 Remark: What is the Bohr subset of D n? A more difficult problem would be to find the Bohr subset of D n, i.e., the set of points z in D n for which c α z α sup c α z α α z D n α holds for all polynomials α c αz α.

62 Estimates on coefficients of Dirichlet polynomials For a Dirichlet polynomial Q(s) = N a n n s, n=1 we set Q = sup t R Q(it) and Q 1 = N n=1 a n. Then S(N) is the smallest constant C such that the inequality Q 1 C Q holds for every Q.

63 Estimates on coefficients of Dirichlet polynomials For a Dirichlet polynomial Q(s) = N a n n s, n=1 we set Q = sup t R Q(it) and Q 1 = N n=1 a n. Then S(N) is the smallest constant C such that the inequality Q 1 C Q holds for every Q. Theorem (Konyagin Queffélec 2001, de la Bretèche 2008, DFOOS 2009) We have S(N) = { ( 1 N exp 2 + o(1) ) } log N loglog N when N.

64 Historical account of the estimate for S(N) The inequality S(N) { ( 1 N exp 2 + o(1) ) } log N loglog N was established by R. de la Bretèche, who also showed that S(N) { ( 1 N exp o(1)) } log N loglog N follows from an ingenious method developed by Konyagin and Queffélec. The same argument, using the hypercontractive BH inequality at a certain point, gives the sharp result.

65 More on the function theory of H So far, our discussion has centered around the Bohnenblust Hille inequality. In the remaining part of the talk, we will look at properties of boundary limit functions of elements in H and the related spaces H p. Now we do not use the Bohnenblust Hille inequality, but the Bohr correspondence is still of basic importance.

66 Carlson s theorem Suppose f is in H. By a direct computation, we get what is known as Carlson s theorem 1 : 1 T lim f (σ + it) 2 dt = a n 2 n 2σ. T T 0 n=1 1 Named after the Swedish mathematician Fritz Carlson ( ).

67 Carlson s theorem Suppose f is in H. By a direct computation, we get what is known as Carlson s theorem 1 : 1 T lim f (σ + it) 2 dt = a n 2 n 2σ. T T 0 n=1 Thus a n 2 f 2 <. n=1 1 Named after the Swedish mathematician Fritz Carlson ( ).

68 Carlson s theorem Suppose f is in H. By a direct computation, we get what is known as Carlson s theorem 1 : 1 T lim f (σ + it) 2 dt = a n 2 n 2σ. T T 0 Thus n=1 a n 2 f 2 <. n=1 Note that Cauchy-Schwarz gives A 1/2, and thus we have a simple proof of Bohr s inequality A U 1/2. 1 Named after the Swedish mathematician Fritz Carlson ( ).

69 Ergodicity We may think of the line t σ 0 + it as a quasi-periodic motion, or, more precisely, for each prime p, we associate a periodic motion on the circle of radius p σ 0: t p it p σ 0. Thus we think of vertical translations as a motion on a system of circles, or, if you like, on an infinite-dimensional torus.

70 Ergodicity We may think of the line t σ 0 + it as a quasi-periodic motion, or, more precisely, for each prime p, we associate a periodic motion on the circle of radius p σ 0: t p it p σ 0. Thus we think of vertical translations as a motion on a system of circles, or, if you like, on an infinite-dimensional torus. We use the product measure of the normalized arc length measures for the circles. Then the flow is ergodic (by the afore-mentioned approximation theorem of Kronecker), and we may use the Birkhoff Khinchin ergodic theorem.

71 Ergodicity We may think of the line t σ 0 + it as a quasi-periodic motion, or, more precisely, for each prime p, we associate a periodic motion on the circle of radius p σ 0: t p it p σ 0. Thus we think of vertical translations as a motion on a system of circles, or, if you like, on an infinite-dimensional torus. We use the product measure of the normalized arc length measures for the circles. Then the flow is ergodic (by the afore-mentioned approximation theorem of Kronecker), and we may use the Birkhoff Khinchin ergodic theorem. In particular, this gives us a fancier way of proving Carlson s theorem: 1 T lim f (σ + it) 2 dt = T T 0 a n 2 n 2σ. n=1

72 Carlson s theorem on the boundary? An interesting question raised by Hedenmalm (2003) is whether Carlson s theorem extends to the imaginary axis.

73 Carlson s theorem on the boundary? An interesting question raised by Hedenmalm (2003) is whether Carlson s theorem extends to the imaginary axis. To place it in context, I will mention a few related matters before we answer that question.

74 A basic inequality The following estimate is a mean value inequality that appears in analytic number theory: y+t y f (1/2 + it) 2 dt CT f 2 2.

75 A basic inequality The following estimate is a mean value inequality that appears in analytic number theory: y+t y f (1/2 + it) 2 dt CT f 2 2. So f is locally in H 2 (C + 1/2 ), or, if you like, f (s)/s belongs to H 2 (C + 1/2 ).

76 Extension of Carlson s theorem?

77 Extension of Carlson s theorem? For f in H 2, we get 1 lim T T T 0 f (1/2 + it) 2 dt = a n 2 n 1 n=1 by first proving it for Dirichlet polynomials (trivial) and then extending it to general f using the basic inequality.

78 Extension of Carlson s theorem? For f in H 2, we get 1 lim T T T 0 f (1/2 + it) 2 dt = a n 2 n 1 n=1 by first proving it for Dirichlet polynomials (trivial) and then extending it to general f using the basic inequality. However, for H, Carlson s theorem does not extend: Theorem (Saksman Seip 2009) There exists an f in H such that 1 lim T T T 0 f (it) 2 dt does not exist. For every ε > 0 there is a singular inner function f in H such that f 2 < ε.

79 Construction of counter-examples Again we use the Bohr correspondence and move to a polydisc. We use a beautiful construction of Rudin ( Function Theory in Polydiscs, 1969): If ψ is positive, bounded, and lower semi-continuous on T m, then ψ is a. e. the radial boundary limit of the modulus of a function in H (D m ).

80 Construction of counter-examples Again we use the Bohr correspondence and move to a polydisc. We use a beautiful construction of Rudin ( Function Theory in Polydiscs, 1969): If ψ is positive, bounded, and lower semi-continuous on T m, then ψ is a. e. the radial boundary limit of the modulus of a function in H (D m ). Choose m = 2, so that we only deal with two variables, say 2 s and 3 s. Consider ir as a subset of T 2, and cover it by an open set E of measure < ε/2. Take as ψ a function being 1 on E and ε outside E. The challenge is to prove that Rudin s construction leads to a function with radial limit 1 almost everywhere on ir (a subset of T 2 of measure 0).

81 The embedding problem for H p Recall the estimate of Cole and Gamelin (1986): f in H p satisfies f (σ + it) C(σ 1/2) 1/p for σ > 1/2, just as functions in H p (C + 1/2 ).

82 The embedding problem for H p Recall the estimate of Cole and Gamelin (1986): f in H p satisfies f (σ + it) C(σ 1/2) 1/p for σ > 1/2, just as functions in H p (C + 1/2 ). Natural to ask: Question Fix an exponent p > 0, not an even integer. Does there exist a constant C p < such that 1 for every Dirichlet polynomial f? 0 ( ) 1 p f 2 + it dt C p f p H p

83 The embedding problem for H p Recall the estimate of Cole and Gamelin (1986): f in H p satisfies f (σ + it) C(σ 1/2) 1/p for σ > 1/2, just as functions in H p (C + 1/2 ). Natural to ask: Question Fix an exponent p > 0, not an even integer. Does there exist a constant C p < such that 1 for every Dirichlet polynomial f? 0 ( ) 1 p f 2 + it dt C p f p H p Trivial for p = 2k, k a positive integer, because we may apply the case p = 2 to f k. The problem is probably very difficult.

84 Embedding problem for p = 1 and weak factorization For p = 1, the embedding would be implied by an extension to T of the Ferguson Lacey weak factorization theorem. (The problem of finding such an extension of the weak factorization theorem was raised by H. Helson (2005).)

85 The embedding problem for H p weak version Question Assume that 2 < q < p < 4. Is it true that ( 1 ( ) 1 q 1/q dt) f 2 + it C q f H p 0 for every Dirichlet polynomial f? Is this true at least for one such pair of exponents?

86 Fatou theorems Since ir has measure 0 when viewed as a subset of T, care has to be taken if we want to speak about the restriction to ir of a function in L p (T ). Set for τ = (τ 1, τ 2,...) T and θ 0 The Kronecker flow on D : b θ (τ) := (p θ 1 τ 1, p θ 2 τ 1,...). T t ((z 1, z 2,...)) := (p it 1 z 1, p it 2 z 2,...). We equip T (z) := {T t (z) : t R} with the natural linear measure.

87 Fatou theorems Since ir has measure 0 when viewed as a subset of T, care has to be taken if we want to speak about the restriction to ir of a function in L p (T ). Set for τ = (τ 1, τ 2,...) T and θ 0 The Kronecker flow on D : b θ (τ) := (p θ 1 τ 1, p θ 2 τ 1,...). T t ((z 1, z 2,...)) := (p it 1 z 1, p it 2 z 2,...). We equip T (z) := {T t (z) : t R} with the natural linear measure. Theorem (Saksman Seip 2009) For every F in H (D ) we may pick a representative F for the boundary function of F on T such that F(τ) = lim θ 0 + F(b θ (τ)) for a.e. τ T. In fact, for every τ T, we have F(τ ) = lim θ 0 + F(b θ (τ )) for a.e. τ T (τ).

88 Fatou theorem for H p Set T 1/2 := b 1/2(T ). We need to make sense of the restriction F F T 1/2 as a map from H p (D ) to L p (T 1/2 ). For F a polynomial, we must have F T 1/2 (τ) = F(b 1/2 (τ)). Write this as a Poisson integral and use that polynomials are dense in H p (D ). Theorem (Saksman Seip 2009) For every F in H p (D ) (p 2) we may pick a representative F 1/2 for the restriction F T 1/2 on T such that F 1/2 (τ) = lim θ 1/2 + F(b θ (τ)) for a.e. τ T. In fact, for every τ T, we have F 1/2 (τ ) = lim θ 1/2 + F(b θ (τ )) for a.e. τ T (τ).

89 Strong ergodic theorem only when p = 2, 4, 6,...? Arguing as for p = 2, we now get: If F is in H p (D ) (p 2) and the embedding holds, then for every τ in T 1/2 1 T lim F(T t τ) p dt = F T T 1/2 p L p (T ). 0 Conversely, by the closed graph theorem (fix T = 1), the embedding would follow from such a strong ergodic theorem.

90 Strong ergodic theorem only when p = 2, 4, 6,...? Arguing as for p = 2, we now get: If F is in H p (D ) (p 2) and the embedding holds, then for every τ in T 1/2 1 T lim F(T t τ) p dt = F T T 1/2 p L p (T ). 0 Conversely, by the closed graph theorem (fix T = 1), the embedding would follow from such a strong ergodic theorem. So our strong variant of the Birkhoff Khinchin ergodic theorem for functions in H p (D ) is known to hold only when p = 2, 4, 6,...!

91 A problem concerning Riesz projection on T

92 A problem concerning Riesz projection on T Riesz projection P + (orthogonal projection from L 2 (T ) to H 2 (T )) is obviously unbounded on L p for any p 2, and therefore the embedding problem can not be solved in the standard way by interpolation.

93 A problem concerning Riesz projection on T Riesz projection P + (orthogonal projection from L 2 (T ) to H 2 (T )) is obviously unbounded on L p for any p 2, and therefore the embedding problem can not be solved in the standard way by interpolation. Question Is there a p > 2 such that P + is bounded from L (T ) to L p (T )?

94 A problem concerning Riesz projection on T Riesz projection P + (orthogonal projection from L 2 (T ) to H 2 (T )) is obviously unbounded on L p for any p 2, and therefore the embedding problem can not be solved in the standard way by interpolation. Question Is there a p > 2 such that P + is bounded from L (T ) to L p (T )? I only know, by construction of an example, that any such p must be a little less than 4...