SEVERAL COMPLEX VARIABLES THE L 2, -THEORY OF HORMANDER. Course for talented undergraduates Beijing University, Spring Semester 2014.

Transcription

1 SEVERAL COMPLEX VARIABLES THE L 2, -THEORY OF HORMANDER. JOHN ERIK FORNÆSS Course for talented undergraduates Beijing University, Spring Semester 204 Contents. Introduction 2 2. Hormander, Section Hormander,.2, Hormander Hormander.6, Hormander 2., 2.2, Hormander 2.5, Hormander 2.6, 4.-Banach spaces Hormander 4.-Banach spaces Hormander 4.-L p spaces 3. Hormander 4.-L p, L 2 spaces Hormander Hormander 4.2 in L 2 loc, Levi problem Hormander, Acta paper Hormander Acta cont., Chen s proof of Ohsawa-Takegoshi Chen s proof of Ohsawa-Takegoshi, cont Appendix, on polydiscs Appendix 2, Solution of T f = v Appendix 3, Ohsawa-Takegoshi in L p Appendix 4, The strong openness conjecture in L p. 82 Date: June, 204.

2 2 JOHN ERIK FORNÆSS. Introduction These lectures will give an introduction to several complex variables. We will generally follow the classical book by Hormander, An Introduction to complex analysis in several variables. The notes will add some more details to the text of Hormander, especially after the introductory material. We generally follow the numbering of results as in Hormanders book, but results in Hormander might be broken up into smaller steps, for example, Lemma 4..3 in Hormanders book is broken up into 4..3 a to 4..3 j in these notes. We follow Chapter.,.2 and.6 and then Chapter 2., 2.2, 2.5 and 2.6. Afterwards we move to Chapter 4., 4.2. We deviate a little by considering L p spaces for general p for a while, before restricting to L 2 spaces. We prove the existence of solutions to on pseudoconvex domains in C n in L 2 loc and give the solution of the Levi problem. At the end we go through the recent proof by Bo-Yong Chen of the Ohsawa-Takegoshi Theorem. This requires first that we discuss a version of Hormanders solution of the equation in L 2 as in Theorem..4 in the 965 Acta paper of Hormander. Some additional material that is not needed for the presentation are in the Appendices. For example, the solution of in a polydics of section 2.3 is in an appendix, since it is not needed in the proof of the L 2 theorem of Hormander. There are also some remarks on L p spaces there, such as Ohsawa-Takegoshi in L p and the strong openness conjecture in L p. The author thanks the Beijing International Center for Mathematical Research, BICMR, for its hospitality during the Spring Semester of 204. The course can also be downloaded from Preliminaries for the course is some knowledge of one complex variable and some functional analysis. 2. Hormander, Section.-.2 Chapter I: Analytic functions of one complex variable.

3 SCV 3. Preliminaries: Here holomorphic (=analytic) functions are introduced. These are C functions on domains in C which satisfy the Cauchy-Riemann equation u z = 0 The set of all such functions is denoted by A(). Here u z u z = ( ) u 2 x i u y = ( ) u 2 x + i u y Generally, du = u x dx + u y u u dy = dz + z z dz. For analytic functions, du = u z dz, i.e. du and dz are paralell. For analytic functions we write u z = u. Examples of analytic functions are all polynomials P (z) = i a iz i and the exponential function e z = e x (cos y + i sin y). We have that products, compositions and inverses of analytic functions are analytic..2 Cauchy s integral formula and its applications. Let ω be a bounded open set in C with boundary consisting of finitely many C Jordan curves. For ease of reference, we list the results using the same numbering as in Hormander. The proofs can be read easily in Hormanders book. Cauchy integral formula for general functions: Theorem.2. Let u C (ω). Then for ζ, ( u(ζ) = ) u u(z) 2πi z ζ dz + z dz dz z ζ ω Theorem.2.2 If µ is a measure with compact support in C, the integral dµ(z) u(ζ) = z ζ defines an analytic C function outside the closed support of µ. In any open set ω where dµ = ϕ(z)dz dz 2πi for some ϕ C k (ω), k, we have that u C k (ω) and u z = ϕ. ω

4 4 JOHN ERIK FORNÆSS Corollary.2.3 Every u A() is in C. Also u A() if u A(). Theorem.2.4 For every compact set K and every open neighborhood ω of K there are constants C j, j = 0,,... such that (.2.4) sup u (j) (z) C j u L (ω) z K for all u A(), where u (j) = j u. Exercises Recall that a function f(z) = u(x, y) + iv(x, y) is analytic if f C and f z = 0. ) Show that the equation f z = 0 is equivalent to the classical Cauchy Riemann equations u x = v y u y = v x 2) Show that the function defined by f(z) = e (z 4) for z 0 and f(0) = 0 satisfies the Cauchy Riemann equations at every point. Is f C? 3) Show that ( ) f z ( ) f z = f z and = f z

5 SCV 5 3. Hormander,.2,.6 Corollary.2.5 If u n A() and u n u when n, uniformly on compact subsets of, it follows that u A(). Proof. Pick a point ζ and choose a disc (w, r) with ζ (w, r). For each n, we have by theorem.2.. that u n (ζ) = u n (z) 2πi z ζ dz. z w =r Using the uniform convergence we then get u is continuous and u(ζ) = u(z) 2πi z ζ dz. z w =r From this formula we see that u is C and analytic on (w, r). Corollary.2.6 If u n A() and the sequence u n is uniformly bounded on every compact subset of, there is a subsequence u nj converging uniformly on every compact subset of to a limit u A(). Corollary.2.7 The sum of a power series 0 a nz n is analytic in the interior of the circle of convergence. Theorem.2.8 If u is analytic in = {z; z < r}, we have u(z) = u (n) (0)z n /n! with uniform convergence on every compact subset of. Uniqueness of analytic continuation: 0 Corollary.2.9 If u A() and there is some point z where u (k) (0) = 0 for all k 0, it follows that u = 0 in if is connected. Corollary.2.0 If u is analytic in the disc = {z; z < r} and if u is not identically 0, one can write u in one and only onoe way in the form u(z) = z n v(z) where n 0 is an integer and v A(), v(0) 0 (which means that /v is also analytic in a neighborhood of 0. Theorem.2. If u is analytic in {z; z z 0 < r} = and if u(z) u(z 0 ) when z, then u is constant in. Maximum Principle: Corollary.2.2 Let be bounded and let u C() be analytic in. Then the maximum of u in is attained on the boundary.

6 6 JOHN ERIK FORNÆSS.6 Subharmonic Functions. Definition: A C 2 function is said to be harmonic if h = 4 2 h z z = 0. This is equivalent to the equation 2 h + 2 h = 0. If h = Re(f), f analytic, then h x 2 x 2 is harmonic: 2 h z z = 2 ( f+f 2 ) z z = 2 f 2 z z + 2 f 2 z z = ( ) f + 2 z z 2 z = ( ) f 2 z z = 0 ( ) f z Conversely, suppose that u is C 2 and harmonic on a disc D. Then for each Jordan curve γ D bounding a domain U, we get by Stokes theorem that γ ( u ydx + u x dy) = U u xx + u yy = 0. Hence the function v(q) = q z 0 u y dx + u x dy defines a function v on D. This function satisfies v x = u y and v y = u x so u + iv is analytic. So on a disc, we see that h is harmonic if and only if h is the real part of an analytic function. Definition.6. A function u defined in an open set C with values in [, > is called subharmonic if a) u is upper semicontinuous, that is {z ; u(z) < c} is open for every real number c. b) For every compact K and every continuous function h on K which is harmonic in the interior of K and such that u h on K, we have u h on K. Note: The function u is called subharmonic in this text. Note: A function is u upper semicontinuous if and only if there exists a sequence of continuous functions u j such that u j u.

7 SCV 7 4. Hormander.6 Note: Harmonic functions are subharmonic: Let u be harmonic on and let K be a compact subset of. Let h be a continuous function on K which is harmonic on the interior of K. Also suppose that u h on K. Suppose there exists a point p the interior K such that u(p) > h(p). Then if ϵ > 0 is small enough we have that the function v = u h + ϵ z 2 satisfies v(p) > sup K v. We can choose p to be a point where v takes on a maximum value. Note however that v xx + v yy = 4ϵ > 0. This contradicts that p is a max point. Theorem.6.2 If u is subharmonic and 0 < c R, it follows that cu is subharmonic. If u α, α A, is a family of subharmonic functions, then u = sup α u α is subharmonic if u < and u is upper semicontinuous, which is always the case if A is finite. If u, u 2,... is a decreasing sequence of subharmonic functions, then u = lim j u j is subharmonic. Theorem.6.3 Let u be defined with values in [, > and assume that u is upper semicontinuous. Then each of the following conditions are necessary and sufficient for u to be subharmonic: (i) If D is a closed disc in and f is an analytic polynomial such that u Re(f) on D, then it follows that u Re(f) on D. (ii) If δ = {z ; d(z, c ) > δ}, we have 2π (.6.) u(z)2π dµ(r) u(z + re iθ )dθdµ(r), z δ 0 for every positive finite measure dµ on the interval [0, δ]. (iii) For every δ > 0 and every z δ there exists some positive finite measure dµ with support in [0, δ] such that dµ has some mass outside the origin and (.6.) is valid. As pointed out by one the students in class, statement (ii) is not the right one. The correct statement should be (ii) Let z and assume that {z + re iθ ; θ [0, 2π], 0 r s}. Then (.6.) holds where µ is supported on [0, s]. Corollary.6.4 If u, u 2 are subharmonic, then u + u 2 is subharmonic Proof. We use (.6.). Corollary.6.5 A function u defined in an open set C is subharmonic if every point in has a neighborhood on which u is subharmonic. We use property (iii) of Theorem.6.3. Corollary.6.6 If f A(), then log f is subharmonic. Proof. Use property (i) in Theorem.6.3. So suppose that log f Re(g) on the boundary of a disc, where g is a holomorphic polynomial. Then

8 8 JOHN ERIK FORNÆSS f e Re(g) = e g on the boundary of the disc. Hence fe g on the boundary of the disc. Hence also on the inside. Therefore log f Re(g) on the whole disc. Theorem.6.7 Let ϕ be a convex increasing function on R and set ϕ( ) = lim x ϕ(x). Then ϕ(u) is subharmonic if u is subharmonic. Definition A function ϕ(x) is convex if for every a < b, and every t < 0, >, ϕ(ta + ( t)b) tf(a) + ( t)f(b), i.e. the graph lies under the chord. Observation: An immediate consequence is that out side (a, b) the graph lies above the straight line continuing the chord. Lemma.6.7a Suppose that ϕ(x) is convex and suppose that x 0 R. Then there exists a constant k R so that ϕ(x) ϕ(x 0 ) + k(x x 0 ) for all x. Also ϕ is continuous. Proof of the Lemma. Take any sequence a n < x 0 < b n where both converge to x 0. Let k be any limit for the slopes of the chords from a n to b n. To prove continuity, suppose that x n x 0. Fix a < x 0 < b. Considering the chord from a to x n shows that lim inf ϕ(x n ) ϕ(x 0 ). Considering the chord from x 0 to b shows that the lim sup ϕ(x n ) ϕ(x 0 ). A similar argument applies for x n x 0. Proof of theorem.6.7: Let x 0 R and let k be as in Lemma.6.7a. Let x = u(z + re iθ ). Then Hence 2π 2π 0 We want to show that 2π If 2π Let x 0 = 2π ϕ(u(z + re iθ )) ϕ(x 0 ) + k(u(z + re iθ ) x 0 ). ϕ(u(z + re iθ ))dθ ϕ(x 0 ) + k( 2π u(z + re iθ )dθ x 0 ). 2π 0 2π 2π 0 ϕ(u(z + re iθ ))dθ ϕ(u(z)). 0 u(z + re iθ )dθ =, this is clear. So assume the integral is finite. 2π 0 u(z + re iθ )dθ. Then 2π 2π 2π 0 ϕ(u(z + re iθ ))dθ ϕ( 2π u(z + re iθ )dθ). 2π 0 Since 2π 0 u(z +re iθ )dθ u(z) and since ϕ is increasing, ϕ( re iθ )dθ) ϕ(u(z)). Hence 2π ϕ(u(z + re iθ ))dθ ϕ(u(z)). 2π 0 2π 2π 0 u(z +

9 SCV 9 It follows from Theorem.6.3 (iii) that ϕ(u) is subharmonic. Corollary.6.7b If u is subharmonic, then e u is subharmonic. If f is analytic, f is subharmonic. The first part follows from Theorem.6.7 since e x is a convex increasing function. Since log f is subharmonic by Corollary.6.6, it follows that f = e log f also is subharmonic. Corollary.6.8 Let u, u 2 be nonnegative and assume that log u j is subharmonic in. Then log(u + u 2 ) is subharmonic. Theorem.6.9 Let u be subharmonic in the open set and not identically in any connected component of. Then u is integrable on all compact subsets of (we write u L loc ()), which implies that u > almost everywhere. Proof. Suppose that u(z) >. Pick a closed disc D centered at z contained in. If we let µ = rdr we get from (.6.) that u(z)a D uda. Note that u is bounded above on D. Hence it follows that u is in L on D. It follows that u > a.e. on D. Hence we can repeat the argument for points z near the boundary of D. It follows that the set U of points z where u is integrable in some neighborhood, is open and closed. By hypotheses U is nonempty. Hence U =. Exercises ) Show that the function sup 0<ϵ< ϵ log z fails to be subharmonic. 2) Suppose that u is a C 2 subharmonic function on C. Let f : C be an analytic function defined on an open set in C. Show that u f is subharmonic on. 3) Let u a subharmonic function in { z < 2}. Suppose that u(z) = 0 for all z, < z < 2. Show that u Hormander.6, 2. Theorem.6.0 If u is subharmonic in and not identically in any component of, then we have that (.6.3) u vdλ 0 for all v C0 2 () with v 0. Here λ denotes Lebesgue measure.

10 0 JOHN ERIK FORNÆSS Proof. Let 0 < r < d(supp(v), c ). Then by.6. we have for every z supp(v) that 2πu(z) 2π 0 u(z + re iθ )dθ. Since v 0, we get for every z supp(v) that 2πu(z)v(z) We integrate with respect to λ. 2π 0 u(z + re iθ )v(z)dθ. (2πu(z)v(z))dλ = = = 2π 0 2π 0 2π ( 0 ( ( u(z + re iθ )v(z)dθ)dλ u(z + re iθ )v(z)dλ)dθ u(z)v(z re iθ )dλ)dθ 2π u(z)( v(z re iθ )dθ)dλ 0 We can also rewrite the left side: (2πu(z)v(z))dλ = 2π u(z)( v(z)dθ)dλ 0 Hence we see that u(z)( 2π 0 (v(z re iθ ) v(z))dθ)dλ 0. We Taylor expand the intergrand v(z re iθ ) v(z). v(z re iθ ) v(z) = v x (z)r cos θ v y (z)r sin θ + 2 v xxr 2 cos 2 θ + 2 v yyr 2 sin 2 θ + v xy r 2 cos θ sin θ + o(r 2 ) We hence get an expression

11 SCV 2π u(z)( v x (z)r cos θ)dθ)dλ π u(z)( u(z)( u(z)( u(z)( u(z)( 0 2π 0 2π 0 2π 0 2π 0 v y (z)r sin θ)dθ)dλ 2 v xxr 2 cos 2 θ)dθ)dλ 2 v yyr 2 sin 2 θ)dθ)dλ v xy r 2 cos θ sin θ)dθ)dλ o(r 2 ))dθ)dλ Hence after carrying out the inner integrals we see that u(z)( 2 v xxπr v yyπr 2 + o(r 2 ))dλ 0. If we divide by πr2 2 and let r 0, we see that u(z) v(z) 0. Corollary.6.0a If u is a C 2 subharmonic function, then u 0. Proof. Let v 0 be a compactly supported C 2 function in the domain of u. By (.6.3), we have that u v 0. Integrating by parts twice, we see that v u 0. Since this is valid for all compactly supported nonnegative C 2 functions v, it follows that u 0. Theorem.6. Let u L loc () and assume that (.6.3) holds. Then there is one and only one subharmonic function U in which is equal to u almost everywhere. If ϕ is an integrable non-negative function of z with compact support and ϕ =, we have for every z (.6.4) U(z) = lim u(z δz )ϕ(z )dλ(z ). δ 0 We will divide the proof into lemmas. Lemma.6.a Assume u L loc (). Let ψ = ψ( z ) be a nonnegative C function with compact support in the unit disc and ψ =. Then the function u δ (z) := u(z δz )ψ(z )dλ(z ) = δ 2 u(w)ϕ( z w )dλ(w) δ

12 2 JOHN ERIK FORNÆSS is C in δ. If V, we have that u δ L (V δ ) u L (V ). Moreover u δ u L on compact subsets. Proof. The last equality is obtained by the change of variable, w = z δz. The fact that u C follows from differentiation under the integral sign in the last integral. The inequality in the L norm follows by integration with respect to z first. To show the last statement, write u = u + u 2 where u is continuous and u 2 has small L norm. The convergence for (u ) δ is obvious. And the L norm of (u 2 ) δ is as small as we wish. Lemma.6.b Suppose U is subharmonic. Then (.6.4) holds when we set u = U in the integral on the right side. In particular it follows that if two subharmonic functions are equal almost everywhere, they are identical. Proof. It follows by (.6.) that for small δ, U(z) U(z δz )ϕ(z )dλ(z ). By upper semicontinuity of U it follows that the upper limit of the right side when δ 0 is at U(z). Hence (.6.4) holds with u = U. Lemma.6.c Assume that u C 2 () and that u 0. Then u is subharmonic. Moreover u δ u. Proof. Fix z 0 δ. Let u z0 (w) = 2π 0 u(z 0 + e iθ w)dθ for w < δ. Then u z0 is C 2 and u z0 0. Moreover u z0 only depends on w. We calculate the Laplacian of u z0 at points w = x + iy, x 0, y = 0. We can write u z0 (x, y) = u z0 ( x 2 + y 2, 0) = g( x 2 + y 2 ). We get that g (x)+g (x)/x 0 for x > 0, so xg (x) + g (x) 0, x 0. It follows that xg (x) increases. The value at x = 0 is 0, so g (x) 0. So g(x) is increasing. By Theorem.6.3, it follows that u is subharmonic, and we also get that u δ decreases to u δ when δ 0. Lemma.6.d Assume u L loc satisfies (.6.3). Let ψ = ψ( z ) be a nonnegative C function with compact support in the unit disc and ψ =. Then the function u δ (z) := u(z δz )ψ(z )dλ(z ) is C and subharmonic in δ. Moreover u δ u L on compact subsets. Proof. Suppose that u L loc and that u(z) v(z) 0 for all functions that are C 2 with compact support and with v 0. Then it follows that also u δ has this property. Then, by Lemma.6.c it follows that u δ is

13 SCV 3 subharmonic. Also by Lemma.6.a we have convergence in L norm to u. Lemma.6.e Let u and ψ be as in Lemma.6.d. Then if δ, ϵ > 0, Let ϵ 0 : (u δ ) ϵ u δ Let δ 0 : (u ϵ ) δ u ϵ (u δ ) ϵ = (u ϵ ) δ Let ϵ 0 : u ϵ V for some subharmonic function V. Proof. We first show that (u δ ) ϵ u δ. By Lemma.6.d we have that u δ is C and subharmonic. Hence by Corollary.6.0.a it follows that u δ 0. Hence it follows by Lemma.6.c that (u δ ) ϵ u δ. The second limit holds for the same reason. To prove the following equality, we see that (u δ ) ϵ (z) = u δ (z ϵz )ψ(z )dλ(z ) = ( u(z ϵz δz )ψ(z )dλ(z ))ψ(z )dλ(z ) = (u ϵ ) δ (z) We show that u ϵ (z) u ϵ2 (z) if ϵ > ϵ 2. We have shown that for each δ > 0, (u δ ) ϵ (z) (u δ ) ϵ2 (z), hence (u ϵ ) δ (z) (u ϵ2 ) δ (z). Hence by Lemma.6.c, u ϵ (z) u ϵ2 (z). Finally, by Theorem.6.2, the limit of u ϵ is subharmonic. To finish the proof of Theorem.6., it suffices to note that by Lemma.6.d the function V obtained is equal to u a.e. Also by Lemma.6.b, we have that (.6.4) holds for V and so also for u. Now we turn to Chapter 2, Elementary properties of functions of several complex variables. 2. Preliminaries.

14 4 JOHN ERIK FORNÆSS We introduce coordinates in C n. Write z = (z, z 2, z n ) where each z j = x j + iy j. Let u be a C function. We introduce differential operators as in one variable. Then we get du = u = u u 2 x j 2 y j u = u + u 2 x j 2 y j n j= u dz j + n j= u dz j. We write for short the first sum on the right as u and the last sum as u We say that u is a form of type (0, ) and u a form of type (0, ). Definition 2.. A function u C () is said to be analytic or holomorphic if du is of type (, 0). Equivalently, u = 0 and also equivalently, the function satisfies the Cauchy-Riemann equations in each variable separately which is again equivalent to saying that u is analytic in each variable separately. The set is analytic functions on is called A(). If I = (i,..., i p ) is a multiindex of integers between and n we write dz I = dz i dz ip, and we write I = p. If J = (j,... j q ) is a multiindex of integers between and n we write dz J = dz i dz iq, and we write J = q. The following expression is called a (p, q) form f = I =p, J =q f I,Jdz I dz J. It is OK to think of this as an expression without meaning.

15 SCV 5 6. Hormander 2., 2.2, 2.5 Let f = I,J f I,Jdz I dz J be a (p, q) form. We use the usual convention that of two differentials, say dz i and dz j are switched, the form changes sign. We define the exterior differential df as the form df = I,J df I,Jdz I dz J. If we write f = I,J f I,J dz I dz J and f = I,J f I,J dz I dz J then df = f + f. This writes df as a sum of a (p+, q) form and a (p, q+) form. We write 0 = d 2 f = 2 f +( + )f + 2 f and these have type (p+2,q), (p+,q+) and (p,q+2) respectively so all three terms must vanish. 2 = 0, + = 0, 2 = 0. This implies that if f is a (p,q+) form and we want to solve the equation u = f then is is necessary that f = Applications of Cauchy s integral theorem in a polydisc. Let w = (w,..., w n ) be a point in C n and let r = (r,..., r n ) be positive numbers. We define the polydisc D with center w and polyradius r to be the set D = D n (w, r) = {z C n ; z j w j < r j, j =,..., n}. The set D 0 := { z j w j = r j, j =,..., n} is called the distinguished boundary of D. Theorem 2.2. Let D be an open polydisc and let u be in C (D). If u is an analytic function of z j D (w j, r j ) whenever the other variables are constant, z k w k r k, then (2.2.) u(z) = ( ) n u(ζ,..., ζ n )dζ... dζ n 2πi 0 D (ζ z ) (ζ n z n ) for all z D. Hence u is C in D.

16 6 JOHN ERIK FORNÆSS Proof. We use Cauchys integal formula inductively, i.e. Theorem.2.. We get then u(z) = u(z,..., z n, ζ n )dζ n 2πi ζ n w n =r n ζ n z n = 2πi ( ) 2 = 2πi ( ) 3 = 2πi = ( ) n = 2πi ζn wn =rn ( 2πi ζ j w j =r j,j=n,n ζ j w j =r j,j=n 2,n,n 0 D ζ n w n =r n u(z,...,z n 2,ζ n,ζ n)dζ n ζ n z n u(ζ,..., ζ n )dζ... dζ n (ζ z ) (ζ n z n ) ζ n z n u(z,..., z n 2, ζ n, ζ n )dζ n dζ n (ζ n z n )(ζ n z n ) )dζ n u(z,..., z n 3, ζ n 2, ζ n, ζ n )dζ n 2 dζ n dζ n (ζ n 2 z n 2 ) (ζ n z n ) The last statement follows by differentiation under the integral sign. Corollary If u A(), then u C () and all derivatives of u are also analytic. We use the following multiindex notation: We write α = (α,..., α n ) for nonnegative integers α j. We call α a multiorder. Set α! := α! α n!. We define α = ( z ) α ( z n ) α n, α = ( z ) α ( z n ) α n. Theorem If K is a compact subset of and K ω, then there exist contants C α for all multiorders α so that if u A() then (2.2.3) sup α u C α u L (ω). K Proof. Assume first that K = D n (w, r), ω = D n (w, r ) and = D n (w, r ) for multiradii r = (s,..., s), r = (s,..., s ), r = (s,..., s ), s < s < s. Let C j denote the constants in the one variable version of this theorem,

17 SCV 7 Theorem.2.4. Let z K. We get α u(z) C αn ( ) α ( ) α n u(z,..., z n, ζ n )dλ(ζ n ) ζ n w n <s z z n C αn C αn ζ n w n <s ( ) α ( ) α n 2 u(z,..., z n 2, ζ n, ζ n ) ζ n w n <s z z n 2 dλ(ζ n ) dλ(ζ n ) ΠC αj ω u(ζ) dλ(ζ). To complete the proof we cover the compact set in the Theorem by finitely many such polydiscs. Corollary If u k A() and u k u uniformly on compact subsets, then u A(). Proof. This follows from the Cauchy integral formula: ( ) n u k (ζ,..., ζ n )dζ... dζ n u k (z) = 2πi (ζ z ) (ζ n z n ) 0 D We take limits on both sides. Then it follows that u is analytic on D. Corollary If u k A() and is a uniformly bounded sequence on any compact subset of, then there is a subsequence u kj which converges uniformly on compact subsets to a limit u A(). Proof. By Theorem the first derivatives of the u k are uniformly bounded on compact subsets. Hence, by Ascoli, there is a subsequence u kj which converges uniformly on compact subsets to a limit u. By the previous corollary, u A(). Let a α (z) be holomorphic functions in. We say that α a α converges normally if α sup K α(z) converges for each compact subset of. In this case the sum α a α(z) is a holomorphic function on. Theorem Suppose that u is analytic in a polydisc D(0, r), r = (r,..., r n ). Then u(z) = α u(0) z α α! α for every z D and the convergence is normal.

18 8 JOHN ERIK FORNÆSS Proof. We use the formula for the sum of a geometric series for ζ 0 D and z D. (ζ z ) (ζ n z n ) = z α ζ α. ζ α ζ n The series converges normally in D. If u C (D) then we get from 2.2. that ( ) n ( u(z) = 2πi α ζ 0 D ) u(ζ) ζ α dζ dζ n z α. ζ ζ n We also obtain by differentiation of (2.2.) that ( ) n (2.2.3) α u(ζ) u(0) = α! 2πi ζ α dζ dζ n ζ ζ n ζ 0 D Hence the theorem follows if u C (D). To prove it in general, prove it for any strictly smaller polydisc. We obtain also from (2.2.3) the following Theorem (Cauchy s inequalities) If u is analytic on the polydisc D(0, r) and if u M, then α u(0) Mα!r α. For the proof apply (2.2.3) to any smaller polydisc Domains of holomorphy If C and p, then the function z p cannot be extended analytically from across p. We express this fact by calling a domain of holomorphy. In C n, n > it might be sometimes possible to extend holomorphic functions past the boundary. For example, let = C n \ D(0, r), r = (s,..., s). Then all holomorphic functions on extend to holomorphic functions on all of C n. To see this, use the Cauchy integral formula in the last variable: g(z,, z n, z n ) = 2πi ζ =S f(z,...,z n,ζ)dζ ζ z n. Here we use any number S > s. It is clear that if some z j > s for j > n, the function g = f. Also the integral defines an analytic function in C n. We see then that this defines an anlytic extension to all of C n. For this reason we will not call this domain a domain of holomorphy. The precise definition of domain of holomorphy is a little complicated. The reason is seen from the example f = z which is well defined in the complement of the set {x + i0, x 0}. This function can locally be extended across the boundary at any x + iy, x > 0. But these local extensions dont agree with the function as defined on the other side of the real axis. We now give the precise definition of domain of holomorphy.

19 SCV 9 Definition 2.5. An open set in C n is called a domain of holomorphy if there are no open sets and 2 in C n with the following properties: (a) = 2. (b) 2 is connected and not contained in. (c) For every u A() there is a function u 2 A( 2 ) such that u = u 2 in. Definition If K is a compact subset of, we define the A() hull ˆK of K by Exercises (2.5.) ˆK = {z ; f(z) sup f if f A()}. K ) Let be an open set in C. For K compact in show that ˆK is compact and that the distance of ˆK to the boundary of is the same as the distance of K to the boundary. 2) Let K C be a compact subset. Let U be a connected component of C\K. Show that U ˆK if and only if U is a bounded set and U. 3) Show that a polydisc in C n is a domain of holomorphy.

20 20 JOHN ERIK FORNÆSS 7. Hormander 2.5, 2.6 Lemma Let f A(). Let K be a compact subset of. Let d K denote the sup of the numbers r so that n (z, (r,, r)) for all z K. Then, if w ˆK then the power series of f around w converges normally in n (w, (r,, r)). Proof. Fix any 0 < r < d. Then f is uniformly bounded by some constant M on all polydiscs n (z, (r,..., r)) when z K. It follows from Theorem that for any z K and any multiindex α, α f(z) Mα!r α. Then these inequalties must also hold at every point w ˆK. But then this implies that the power series expansion of f around w converges normally in (w, (r,..., r)). Theorem Let be a domain of holomorphy. Let K be a compact subset. Then if d(l) denotes the sup of all radii r so that B(z, r) for all z L,, then d(k) = d( ˆK). Proof. If we instead measure boundary distance using polydiscs of multiradius (r,..., r) then the result holds by Lemma By scaling in each variable, we see that if D is any polydisc such that z + D for all z K, then also w + D for all w ˆK. Next we can choose any orthonormal basis for C n and define polydiscs in these coordinates. Then we see that the result also holds for such polydiscs. Next let B(r) be a ball such that z + B(r) for all z K, then since the ball is a union of polydiscs included rotated polydiscs, we see that w+b(r) for all w ˆK. Theorem If is an open set in C n, then the following conditions are equivalent: (i) is a domain of holomorphy. (ii) If K is a compact subset of, then ˆK is a compact subset of and d( ˆK) = d(k). (iii) If K is a compact subset of, then ˆK is also a compact subset of. (iv) There exists a function f A() such that it is not possible to find and 2 satisfying (a) and (b) in Definition 2.5. and f 2 A( 2 ) so that f = f 2 in. Proof. Notice that if K D for some polydisc, then ˆK is also contained in D. Suppose that (i) holds and that K is a compact subset of. Then by Theorem 2.5.4, ˆK is a closed set in C n. Since ˆK is also bounded, it follows that ˆK is a compact subset of. Moreover, it follows from Theorem that d( ˆK) = d(k). Hence (i) implies (ii). It is clear that (ii) implies (iii) and that (iv) implies (i). Hence it only remains to show that (iii) implies (iv).

21 SCV 2 Let D be a polydisc. For each ζ, let D ζ = ζ + rd denote the largest polydisc contained in. Let M be a countable dense subset of. We need a lemma: Lemma 2.5.5a Suppose that satisfies (iii). Then there exists a holomorphic function f on such that for each ζ M, there is no holomorphic function g defined on some neighborhood U of D ζ such that f = g on D ζ. Proof of Lemma 2.5.5a: Let ζ n be a list of the points in M such that each point in M is listed infinitely many times. Let K K n be compact subsets of such that each compact set in is contained in some K n. Since ˆK j is a compact subset of, there exists a point z j D ζj \ ˆK j. Hence there is a function f j A() so that f j (z j ) = and sup Kj f j < 2 j. We can choose f j so that f j is not identically in any connected component of. Let f = Π j=( f j ) j. Since the sum j2 j is convergent, this infinite product converges to an analytic function which does not vanish identically on any connected component of. All derivatives of f up to order j vanish at z j. Hence there can be no analytic function g defined on any neighborhood U of any D ζj agreeing with f on D ζj. This finishes the proof of Lemma 2.5.5a. We next continue with the proof that (iii) implies (iv). Let, 2 be any two open sets satisfying (a) and (b) in Definition We assume that there is a holomorphic function f 2 on 2 such that f 2 = f on. We will show that this leads to a contradiction. We can find a curve γ(t) 2, 0 t so that γ(0) and γ(), γ([0, ). By analytic continuation, f = f 2 on an open set containing γ([0, )). We then get a contradiction to the conclusion of Lemma 2.5.5a by chosing a point ζ j in this neighborhood very close to since 2 will contain a polydisc centered at γ(). 2.6 Pseudoconvexity and plurisubharmonicity. Definition 2.6. A function defined in an open set C n with values in [, ) is called plurisubharmonic if (a) u is upper semicontinuous. (b) For arbitrary z and w in C n, the function τ u(z + τw) is subharmonic in the part of C where it is defined. Theorem A function u C 2 () is plurisubharmonic if and only if (2.6.) n j,k= 2 u(z) z k w j w k 0, z, w C n.

22 22 JOHN ERIK FORNÆSS Proof. By Corollary.6.0a and Lemma.6.c, a C 2 function defined on an open set in C is subharmonic if and only if u 0. We calculate: u(z + τw) n u = w k τ ζ k 2 u(z + τw) τ τ = k= n n j= k= 2 u ζ j ζ k w j w k. The Theorem follows. Theorem Let 0 ϕ C0 (Cn ) be equal to 0 when z >. Let ϕ depend only on z,..., z n, and assume that ϕdλ = where dλ is the Lebesgue measure. If u is plurisubharmonic in, it follows that u ϵ (z) = u(z ϵζ)ϕ(ζ)dλ(ζ) is plurisubharmonic, that u ϵ C in ϵ, and that u ϵ u. ( We assume that u is not identically on any connected component of.) Proof. That u ϵ when ϵ 0 was proved in Lemma.6.e in the case n =. To show this when n >, choose first ϵ = (ϵ,..., ϵ n ) and define u ϵ (z) = u(z ϵ ζ,, z n ϵ n ζ n )ϕ(ζ)dλ(ζ) The one variable result implies that this expression decreases when we decrease only one of the ϵ i. Hence repeating this process n times show that u ϵ u ϵ 2 if ϵ ϵ 2. From Theorem.6.3 it follows that u ϵ (z) = u(z ϵζ)ϕ(ζ)dλ(ζ) = u(z ϵζ,, z n ϵζ n )ϕ(ζ)dλ(ζ )... dλ(ζ n ) u(z ϵζ,, z n ϵζ n, z n )ϕ(ζ)dλ(ζ )... dλ(ζ n ) uz ϵζ,, z n ϵζ n 2, z n, z n ))ϕ(ζ)dλ(ζ )... dλ(ζ n 2 )... u(z) By upper semicontinuity of u it follows that lim sup ϵ 0 u ϵ u. Hence u ϵ u when ϵ 0. To show that u ϵ is plurisubharmonic, we fix a complex line τ z + τw and show that u ϵ (z + τw) is subharmonic as a function of τ. By theorem.6.3 it suffices to find for each τ 0 such that z + τ 0 w

23 arbitrarily small r > 0 so that u ϵ (z + τ 0 w) 2π We calculate: 2π u ϵ (z + (τ 0 + re iθ )w)dθ 2π 0 = 2π ( 2π = 0 ϕ(ζ)( 2π SCV 23 2π 0 u ϵ (z + (τ 0 + re iθ )w)dθ. u(z + (τ 0 + re iθ )w ϵζ)ϕ(ζ)dλ(ζ))dθ 2π 0 ϕ(ζ)u(z + τ 0 w ϵζ)dλ(ζ)) u(z + (τ 0 + re iθ )w ϵζ))dθdλ(ζ)) = u ϵ (z + τ 0 w) Definition If K is a compact subset of C n, we define the P () hull ˆKP of K by ˆK P = {z ; u(z) sup u u P ()}. K Since f P () for all f A() we have that ˆK P ˆK. Theorem Any of the following two conditions imply that log d(z, c ) is plurisubharmonic and continuous: (a) is a domain of holomorphy. (b) ˆK P whenever K is a compact subset of.

24 24 JOHN ERIK FORNÆSS 8. Hormander 2.6, 4.-Banach spaces Proof. In the case (a) we will use Theorem which implies that ˆK is a compact subset of whenever K is a compact subset of. Pick a unit vector ξ. We define a distance in the ξ direction: For z, we set d ξ (z) := sup{t; z + τξ, τ C, τ < t}. We show that log(d ξ ) is plurisubharmonic. Assume not. Then there is a complex line L and a disc D L so that d ξ has value at the center strictly larger than the average value on the boundary. We can assume that L is the z axis and D is the unit disc. We can choose a holomorphic polynomial P (z) with h = R(P (z)) so that h > log d ξ on the boundary of the disc and h(0) < log d ξ (0). Consider the complex discs D t = D t (ζ) for t C, t and for ζ C, ζ, D t (ζ) = (ζ, 0,..., 0) + tξe P (ζ). If ζ is on the boundary of the unit disc, we have that log d ξ (ζ) < h(ζ) and hence te P (ζ) = t e h(ζ) < e log dξ(ζ) = d ξ (ζ). It follows that boundaries of the discs D t are all in. We will let K be the compact union of all these boundaries. Then ˆK P or ˆK is compact in. For t = 0 the interior of the disc is in. For those t for which the whole disc is in, we have then that the distance from the disc to the boundary of has a uniform lower bound. Hence no such disc gets too close to the boundary. So it follows that also for all t, t the disc D t is in. Let ζ = 0, t. Then D t (0) = tξe P (0). Hence e P (0) < d ξ (0), so log d ξ (0) < h(0), a contradiction. To finish the proof note that log d is the sup of all log d ξ and that this is continuous. Theorem If is a proper open subset of C n, the following conditions are equivalent: (i) log d(z, c ) is plurisubharmonic in. (ii) There exists a continuous plurisubharmonic function u in such that c = {z ; u(z)} for every c R. (iii) ˆK P for all compact sets K. Proof. To prove that (i) implies (ii), we define u(z) = z 2 log d(z, c ). Then if z n is a sequence in converging to, (including if is unbounded,) then u(z n ). (ii) obviously implies (iii). That (iii) implies (i) follows from Theorem Definition The domain is called pseudoconvex if = C n or if the equivalent conditions in Theorem are fulfilled.

25 SCV 25 We have shown that domains of holomorphy are pseudoconvex. The Levi problem asks if pseudoconvex domains are domains of holomorphy. This is solved in Chapter IV and is major result in this course. Theorem 2.6. Let be a pseudoconvex open set in C n, let K be a compact subset of, and ω an open neighborhood of ˆK P. Then there exists a function u C () such that (a) u is strictly plurisubharmonic, that is the hermitian form in (2.6.) is strictly positive definite for every z, i.e. c(z) w 2, c > 0. (b) u < 0 in K, but u > 0 in \ ω. (c) {z ; u(z) < c} for all c R. Lemma 2.6.a There exists a continuous plurisubharmonic function v(z) satisfying (b) and (c). We prove first the Lemma. By theorem (ii) there exists a continuous plurisubharmonic function u 0 on satisfying (c). We can assume that u 0 < 0 on K by subtracting a constant if necessary. Set K = {z ; u 0 (z) 2} and let L = {z \ ω; u 0 0}. If L is empty, we choose v = u 0. Then (b) is also satisfied. So we assume that L is nonempty. By continuity of u 0, the sets K and L K are compact. Hence K L δ for all small enough δ. Let p L. Then p ω c and hence p / ˆK P. Therefore there exists a plurisubharmonic function w p on such that w p (p) > 0 and w p < 0 on K. Using (w p ) ϵ as in Theorem 2.6.3, we get for small enough ϵ a smooth plurisubharmonic function in ϵ so that (w p ) ϵ < 0 on K and (w p ) ϵ > 0 on some open neighborhood U ϵ,p of p. By compactness we can cover L by finitely many such neighborhoods, U ϵj,p j. Let w = sup{(w pi ) ϵi on maxj ϵ j }. Then w is continuous and plurisubharmonic, w > 0 on L and w < 0 on K. In particular we can assume that w is plurisubharmonic in a neighborhood of K. Let C denote the maximum value of w on K. Since L K, C > 0. Note then that if < u 0 (z) < 2, then w C < Cu 0, hence Cu 0 = max{w, Cu 0 } on the set < u 0 < 2. Hence the function v(z) = max{w, Cu 0 } on {u 0 < 2} and v(z) = Cu 0 when u 0 > is well defined on all of and is locally plurisubharmonic, hence plurisubharmonic (see Corollary.6.5). We then have a continuous plurisubharmonic function on which satisfies (b) and (c). This finishes the proof of Lemma 2.6..a We now prove the Theorem. Proof. Let v be as in Lemma 2.6.a. For any c R, let c := {z ; v(z) < c}. We use the notation of Theorem Set v j (z) = v(ζ)ϕ((z ζ)/ϵ j )ϵ 2n j dλ(ζ) + ϵ j z 2, j = 0,,.... j+

26 26 JOHN ERIK FORNÆSS We choose ϵ j small enough that v 0, v < 0 on K and v j < v + in j. We also have that v j C (C n ) and we can arrange that v is strictly plurisubharmonic and > v in some neighborhood of j. Let χ denote a C convex increasing function on R such that χ(t) = 0, t 0 and χ (t) > 0, t > 0. Let j : On j \ j, v j + j > v + j (j ) + j = 0, where the first inequality holds on j and the second inequality holds in the complement of j. Hence χ(v j + j) is strictly plurisubharmonic in a neighborhood of j \ j. Let u m = v 0 + m a jχ(v j + j). Then u 0 is strongly plurisubharmonic on a neighborhood of 0. Since χ(v + ) is plurisubharmonic on and strongly plurisubharmonic on a neighborhood of \ 0, the function u will be strictly plurisubharmonic in a neighborhood of if we choose a large enough. We also choose a large enough so that u > v on. Similarly, we can next choose a 2 large enough that u 2 is strongly plurisubharmonic in a neighborhood of 2. Inductively we see that u m can be chosen strongly plurisubharmonic in a neighborhood of m and also larger than v there. Notice that u 0 < 0 on K. Also v + < 0 on K, so u = v 0 < 0 on K. Finally also observe that for j 2, v j + j < v + + j < (j 2) + 2 j = 0 on j 2. Therefore all u m = v 0 < 0 on K. Moreover on any j, if m > j+2, u m = u j, so the infinite sum is locally finite, so the limit exists and is strongly plurisubharmonic on.

27 SCV 27 We next move on to Chapter 4 in Hormander: L 2 estimates and existence theorems for the operator. We start with some functional analysis. Let G, G 2 denote two complex Banach spaces with norms, 2 respectively. Let E denote a complex subspace of G, not necessarily closed. We will consider linear maps T : E G 2. Let G 3 = G G 2 denote the product space with Banach norm (x, y) 2 3 = x 2 + y 2 2. We say that T is closed if the graph of T, G T = {(x, T x); x E} is a closed subspace of G 3. Example 4.0. G = G 2 = L 2 (0, ) and T x = x. Here the derivative is in the sense of distributions. So T x(ϕ) = xϕ. Here E consists of those L 2 functions x for which T x is an L 2 function. Then T is a closed operator: If (x n, T x n ) converge to (x, y) then for any test function ϕ we have T x n ϕ = x n ϕ so taking limit one gets yϕ = xϕ. Therefore y equals x in the sense of distributions. Let G i denote the dual Banach space of G i. So an element y G i is a continuous linear function (also called functional) from G i to C, x y(x). Moreover there is a constant C y so that y(x) C y x i. The smallest such constant C y is denoted by y i. An important theorem is the Hahn-Banach Theorem. Theorem 4.0.2, Hahn-Banach Let L G be a linear subspace of a Banach space. Suppose that ϕ : L C be a linear function with bounded norm, i.e. ϕ(x) C x. Then ϕ extends to a linear function ϕ : G C such that ϕ(x) = ϕ(x) for all x L and ϕ(x) C x for all x G. In particular the extension belongs to G. 9. Hormander 4.-Banach spaces We will next add the hypothesis that the operator T is densely defined, i.e. the subspace E is dense. We will define the adjoint operator T : D T G where D T G 2 is a linear subspace. We say that y belongs to D T if there exists a constant C so that y(t x) C x for all x E. Hence y y(t x) extends to a continuous linear functional z on E = G such that z G C. We set T (y) = z. If (y, y 2 ) G G 2 then this defines a continuous linear functional on G G 2 by (y, y 2 )(x, x 2 ) = y (x ) y 2 (x 2 ). Let G T G G 2 denote those (y, y 2 ) for which (y, y 2 )(x, x 2 ) = 0 for all (x, x 2 ) G T. Clearly G T is a closed subspace of G G 2. Lemma G T = G T.

28 28 JOHN ERIK FORNÆSS Proof. Suppose first that (y, y 2 ) G T. Then if x D T we have that y (x) y 2 (T x) = 0. This implies that y 2 (T x) = y (x) y x. Hence y 2 satisfies the requirement to be in D T. Moreover y satisfies the equirement to equal T (y 2 ). Hence (y, y 2 ) G T. Suppose next that (y, y 2 ) G T. Hence y = T (y 2 ). This implies that for any x D T, we have that y (x) = y 2 (T x). Hence (y, y 2 ) G T. Corollary The graph of T is closed. If G is a complex Banach space, we define G to be the dual of G. There is a natural isometric embedding ϕ of G into G : For x G, define ϕ(x)(y) = y(x) for y G. Then ϕ(x) G. Also ϕ(x) x and by the Hahn Banach theorem you have equality: Given x 0, choose a linear function ỹ on Cx by ỹ(x) = x and extend to G by Hahn-Banach. Call the extension y. Then y G and y =. Now ϕ(x)(y) = y(x) = x = x y so ϕ(x) x. Hence ϕ(x) = x. We say that G = G if this map is surjective. The Banach space is called reflexive in this case. In this case, we also have that G = G. Note that if G, G 2 are Banach spaces, then the dual of G G 2 equals G G 2. Namely, if ϕ is a continuous linear function on G G 2, then ϕ(x, x 2 ) = ϕ(x, 0) + ϕ(0, x 2 ) = y (x ) + y 2 (x 2 ). Lemma Let G be a reflexive Banach space, i.e. ϕ(g) = G, and let H be a closed subspace of G. Then (H ) = ϕ(h). Proof. Since ϕ is an isometry and H is closed, ϕ(h) is also closed. This follows because Banach spaces are complete. Suppose that x H and y H. Then y(x) = 0 and hence ϕ(x)(y) = y(x) = 0. Hence ϕ(x) H. Hence ϕ(h) H. Suppose next that z G \ϕ(h). Hence there exists an η G so that η(z) 0 while η vanishes on ϕ(h). By reflexivity, we have that there exists a y G so that z(y) = η(z) 0 while w(y) = η(w) = 0 for all w ϕ(h). Hence if x H and w = ϕ(x) ϕ(h), then y(x) = (ϕ(x))(y) = w(y) = 0. So, y H. Since z(y) 0, it follows that z cannot be in H, so H ϕ(h). Corollary If G, G 2 are reflexive and T : G G 2 is a densely defined closed linear operator, then T = T. Proof. Set G j = ϕ j(g j ). Then G T = (G T ) = G T = {(ϕ (x), ϕ 2 (T x)); (x, T x) G T }. We write this imprecisely as G T = G T, or T = T.

29 SCV 29 Lemma Assume that G, G 2 are reflexive Banach spaces. Then the operator T is densely defined. Proof. Suppose that there exists a y 0 G 2 \D T. Then there exists a z 0 G 2 so that z 0 (y 0 ) 0 while z 0 (y) = 0 for all y D T. So z 0 0. Hence for the point (0, z 0 ) G G 2 we have that 0(T y) z 0 (y) = 0 for all y D T. It follows that (0, z 0 ) G T = G T. We write (0, z 0) = (ϕ (0), ϕ 2 (x)) for some (0, x) G G 2. Then (0, x) G T. But T (0) = 0 and x cannot be zero since ϕ 2 (x) = z 0 0, a contradiction. We recall the uniform boundedness principle (Banach-Steinhaus theorem). Theorem Let F denote a family of continuous linear functionals on a Banach space B. Suppose that for every x B there exists a constant c x so that F (x) c x x B for all F F. Then there exists a constant C so that F C for all F F. The proof uses Baire category. For any number A the set of x where c x A is closed. (Closedness follows from continuity of the functionals.) Hence for some A the set has interior. Notation: Let y G and let H G be a linear subspace. We denote by y H G the norm of the linear functional y restricted to H. So y H G is the smallest c so that y(x) c x for all x H. We next give a more general version of Theorem 4.. in Hormander. Theorem 4.. Let G, G 2 be reflexive Banach spaces and let T : G G 2 be a densely defined closed linear operator. Let F G 2 be a closed subspace containing the range of T, R T. Then F = R T if and only if there exists a constant C > 0 such that (4..) y F G 2 C T y G y D T. If any of the two equivalent conditions are satisfied, then there exists for every z F an x D T with T x = z and x G C T x G2 = C z G2 for the same constant C. Proof. Suppose that R T = F. We will apply the Banach-Steinhaus theorem to a family of linear functionals on the Banach space F. Namely, let G denote the family of y D T for which T y G. Define for each such y a linear function L y F on F given by L y (x) = y(x). This is a continuous linear functional defined on F. For a given x F, pick some z D T for which x = T z. Then we have that for any L y F, that L y (x) = y(x) = y(t z) = (T (y))(z) T (y) G z G z G. Hence the family F is bounded uniformly on any given x F. Hence by the Banach-Steinhaus Theorem, there is a constant C so that (L y ) F G 2 C for any y D T with T y G. Then it follows that (L y ) F G 2

30 30 JOHN ERIK FORNÆSS C T y G y D T. Since (L y ) F = y(x), x F we get that y F G 2 C T y G for all y D T. We next suppose that (4..) is satisfied. Fix a z F. If y D T, w = T (y) R T, set ϕ(w) = y(z). Note that if w = T (y ) = T (y 2 ), then T (y y 2 ) = 0. Hence by (4..), (y y 2 )(z) = 0, so y (z) = y 2 (z) and therefore ϕ(w) is well defined. Also, by (4..), ϕ(w) = y(z) y F G 2 z G2 C T y G z G2 = C w G z G2, hence ϕ is a bounded linear functional with norm at most C z G2. This is then a bounded linear function on the Range of T in G with norm C z G2. We extend ϕ to all of G using the Hahn-Banach theorem. Then ϕ G with norm ϕ G C z G 2. Recall the definition of as it applies in this situation. We say that ϕ G belongs to D (T ) G if there exists a constant c so that ϕ(t y) c y G 2 for all y D T. Since ϕ(t y) = y(z) z G2 y G 2 and our z is fixed, it follows that ϕ D T G. By reflexivity there is an x G with norm C z such that u(x) = ϕ(u) for all u G. Moreover x D T. So whenever y D T, y(z) = ϕ(t y) = (T (y))(x) = y(t x). So we have shown that for any fixed z F, there is an x D T, x G2 C z G so that y(z) = y(t x) for all y D T. Since D T is dense in G 2, it follows that y(z T x) = 0 for all y G 2. By the Hahn-Banach theorem this implies that z T x = 0. Let N T denote the nullspace of T. Clearly N T is contained in D T and N T is a closed subspace. Next we study the case of three reflexive Banach spaces, G, G 2, G 3. Also we consider closed, densely defined linear operators, T : G G 2 and S : G 2 G 3 satisfying the condition that T (G ) N S. Hence S T is defined on D T and S T 0. We call this for short an S, T system. Definition 4..a An S, T system satisfies the Basic Estimate if there exists a constant C so that for every y D T and every u D S we have that (4..5) y(u) C( T (y) G u G2 + y G 2 S(u) G3 ). Theorem 4..b If we have an S, T system satisfying the basic estimate, then T (G ) = N S.

31 SCV 3 Proof. The space F = N S satisifes the condition of Theorem 4.., namely, F is a closed subspace of G 2 and it contains R T. We apply the Basic estimate to y D T and u F. Then y(u) C T (y) G u G2 since S(u) = 0. This is the estimate (4..) in Lemma 4... The theorem follows. 0. Hormander 4.-L p spaces We introduce some Banach spaces. Let be an open subset of C n. Let < r, s <, r + s =. If ϕ is a continuous real function on, we define L r (, ϕ) = {f : C, f r e ϕ dλ =: f r L r (,ϕ) < }. Here dλ is Lebesgue measure and f is assumed to be measurable and locally in L r, f L r loc (). We define similarly Ls (, ϕ). We know that the dual of L r (, ϕ) is L s (, ϕ) and vice versa. In particular these spaces are reflexive. We have for f L r (, ϕ) and g L s (, ϕ) that g(f) = fge ϕ. We write g(f) =< f, g > ϕ and get < f, g > ϕ = < g, f > ϕ. Also we have g(f) f L r (,ϕ) g L s (,ϕ). We can do the same for (p, q) forms. Let L r (p,q)(, ϕ) denote the space of forms of type (p, q) with coefficients in L r (, ϕ). f = f I,J dz I dz J. I =p J =q where refers to summing over strictly increasing multiindices. We set f r = I,J f I,J r and f r ϕ = f r e ϕ. Then the dual of L r (p,q)(, ϕ) is L s (p,q) (, ϕ) and we have < f, g > ϕ= I,J fi,j g IJ e ϕ. Set f g = I,J f I,Jg I,J for the pointwise product. Similarly we define L r (p,q) () loc. Let D() denote the space of C functions on with compact support in. Similarly we define D (p,q) (). We observe that D (p,q) () is dense in L r (p,q)(, ϕ). Let < r, s < with r + s =. Let ϕ, ϕ 2 be continuous functions on. Consider the operator. This gives rise to a linear densely defined closed operator T : L r (p,q) (, ϕ ) L r (p,q+) (, ϕ 2). An element u L r (p,q) (, ϕ ) is in D T if u, defined in the sense of distributions belongs to L r (p,q+) (, ϕ 2) and then we set T u = u. The operator is densely defined since it is defined on D (p,q) (). The closedness is as in Example 2.. Our goal is to show that the range of T consists of all those f for which f = 0 for some choices of ϕ j and r, r.

32 32 JOHN ERIK FORNÆSS Consider the case of 3 Banach spaces, G = L r (p,q) (, ϕ ), G 2 = L r 2 (p,q+) (, ϕ 2), G 3 = L r 3 (p,q+2) (, ϕ 3) with operators T : G G 2 and S : G 2 G 3. This is then an (S, T ) system and our goal is to prove the Basic Estimate under suitable conditions. We will assume that < r 3 r 2 r <. Lemma 4..3 a Let η ν be a sequence of C functions with compact support in. Suppose that 0 η ν and that on any given compact subset of we have η ν = for all large ν. Suppose that n (4..6) e ϕ 3 η ν / z k r 3 e r 3ϕ 2 /r 2. k= If r 3 < r 2 we add the extra condition that has a finite volume. Then for every f D S the sequence η ν f f in G 2. Moreover η ν f D S and S(η ν f) S(f) in G 3. Proof. The sequence η ν f f and η ν f converges pointwise to f, so ην f f r 2 e ϕ 2 dλ 0 by the Lebesgue dominated convergence theorem. We have that S(η ν f) = η ν S(f) + η ν f in the sense of distributions and S(f) L r 3 (ϕ 3 ) so to show that η ν f D S we need to show that η ν f L r 3 (ϕ 3 ). We have the pointwise estimate that η ν f r 3 e ϕ 3 f r 3 e r 3ϕ 2 /r 2. We show that the right side is an L function on. If so, η ν f L r 3 (ϕ 3 ) and the Lebesgue dominated convergence theorem implies that the integral converges to 0. If r 2 = r 3 the function is in L by the hypothesis that f G 2. Suppose that r 2 > r 3. We then get if r 3 r 2 + t =, ( ) r3 /r 2 f r 3 e r 3ϕ 2 /r 2 = ( f r 3 e r 3ϕ 2 /r 2 ) r 2/r 3 ( dλ) /t ( r3 /r 2 = f r 2 e 2) ϕ ( dλ) /t < Then the lemma follows since η ν S(f) S(f) in G 3. Lemma 4..3 b Let η ν be a sequence of C functions with compact support in. Suppose that 0 η ν and that on any given compact subset