6.2 Mean value theorem and maximum principles

Transcription

1 Hence g = u x iu y is analytic. Since is simply connected, R g(z) dz =forany closed path =) g has an integral function, G = g in. Call G = U + iv. For analytic functions d G = d G. Hence dz dx g = d dz G = U x + iv x CR = U x iu y =) U x = u x, U y = u y, i.e. r(u u) =andu = u+ constant. WededucethatV is a harmonic conjugate of u. Remark. Especially the conjugate exists in discs where u is harmonic =) locally the conjugate always exists. Let us then look at sets of uniqueness for harmonic functions. Theorem 6.3. Let u be harmonic in and let u =in an open subset of. Then u =in. Proof. Let be open with u =. Denoteg = u x iu y, whence g is analytic in and g = =) g in =) u is a constant. Theorem 6.4. Every harmonic function is C 1 -di erentiable. Proof. Immediate consequence of Theorem 6. applied locally. Theorem 6.5. Let f : 7! be analytic and let u : 7! R be harmonic. Then u f : 7! R is harmonic. Proof. Let z and assume z B, where B is a ball. Pick a harmonic conjugate v : B 7! R of u B. Then u+iv is analytic in B. Especially, g f = u f +iv f is analytic in f 1 (B) =) u f is harmonic in f 1 (B) =) claim. 6. Mean value theorem and maximum principles Theorem 6.6. (Mean value theorem) If u : 7! R is harmonic, then u satisfies the mean value property (MVP), i.e., for any B(z,r) u(z) = 1 u(z + re it )dt Proof. FT1 =) analytic functions satisfy MVT. Find a conjugate v of u in B B(z,r), B a ball. Apply MVT on u + iv =) Take real parts. u(z)+iv(z) = 1 (u + iv)(z + re it )dt 8

2 Proof. Another method, based directly on the definition of harmonic functions: Define a(r) = 1 u(z + re it )dt = 1 u(z + rx)ds x =1 d Enough to show that a is constant, i.e. a. Compute by the divergence theorem dr ( R A dn = R r A da = R V V a (r) = 1 ru(z + rx) x ds = 1 ru(x) n ds r x = 1 div(ru)da = 1 u da = r r B(z,r) Remember that div ru = r ru = u. B(z,r) Definition. Let u : 7! R. Then u satisfies the sub mean value property (SubMVP) is u is continuous in 8z 9 r = r (z) > sothat if r<r and B(z,r). u(z) apple 1 u(z + re it )dt Theorem 6.7. (A local maximum principle) maximum at z, then u is constant. Let u be harmonic in. Ifu has a local Proof. Assume that u(z) apple u(z )forz B(z, ). By the MVP (Theorem 6.6), for <r< we have = 1 u(z ) u(z + re it dt Since the integrand is continuous, we must have u(z + re it )=u(z ) 8 t, i.e. u = u(z ) As r< was arbitrary, u = u(z )inb(z, ), whence u = u(z )in by unique continuation. Definition. If C is a domain we 1 for the boundary with regard to C. In other 1 if is 1 [ {1} if is unbounded. Next we obtain a stronger maximum principle. 81

3 Theorem 6.8. (A strong maximum pronciple) Let u satisfy the SubMVP in the domain. Then 1. If u takes its supremum in (i.e. u(z )=sup z u(z) for some z ), then u is constant.. Assume that lim sup z! u(z) apple M<1 for 1. Then u apple M in and equality at one point implies that u is a constant 1. Proof. 1. Assume u(z 1 )=S := sup z u(z). Denote A = {z : u(z) =S}. Now A 6= ; and A is closed (in ) since A = u 1 ({S}). If we show that A is open, connectedness of implies that A =, whichiswhatwewanttoshow. Takez A, let r = r (z) > sothat by the SubMVP if < <r 1 u(z ) u(z + e it ) dt Again by continuity we deduce u = u(z )=S ), and since < <r is arbitrary we get B(z,r) A, whence A is open.. Observe that it is enough to check S apple M, then the claim follows from part 1. Pick a sequence (z n )withu(z n ) n!1! S. We may assume that z n! z 1. If z 1, thenu(z 1 )=S. By part 1. u is constant, whence trivially S apple M. If z we have directly from the assumption S = lim u(z n ) apple lim sup u(z) apple M n!1 z!z 1 Corollary 6.9. Let u : 7! R be harmonic and assume that lim sup u(z) apple M for all 1. Then u apple M in, and equality can occur only if u is a constant. Corollary 6.1. (The minimum principle) As in Corollary 6.9, let lim inf z! u(z) M for 1. Then u M in and equality at any point implies that u is constant. 1 Recall that lim sup z! u(z) :=lim r! (sup zb(,r)\ u(z)) if is finite, and lim sup z!1 u(z) := lim R!1 (sup z{ z >R}\ u(z)) 8

4 Corollary (A uniqueness theorem) If u and v are harmonic in and lim u(z) =lim v(z) z! z! (finite limits) for 1, then u = v in. Proof. Apply the minimum and maximum principle on u v. Example. By Corollary 6.11, if u vanishes on the boundary of the domain, then u vanishes in the whole domain. 6.3 The Poisson formula The following theorem is of fundamental importance (especially together with the inverse result, Theorem 6.14 below). Theorem 6.1. Assume that u is harmonic on D = B(, 1) (i.e. u is harmonic in a neighbourhood of ). Then, if z = re it D, we have 8 R 1 1 z u(w) dw z w >< R 1 1 z u(z) = u(e i )d z e i >: R 1 1 r 1+r r cos(t ) u(ei )d, where all the formulas are the same (just di erent ways to write the same integral). Remark. That the last two integrals are indentical is seen by the cosini theorem of elementary geometry, or just by computing z e i = re it e i = re it e i re it e it e =1+r i(t ) i( t) + e r =1+r r cos(t ) Proof. Fix z D. Consider the analytic function '(w) = z w 1 zw Clearly ' is analytic in a neighbourhood of D, and it is injective, with inverse ' 1 = '. One checks easily (see the computation in the beginning of proof of Theorem 4.6) that '(D) =D and '(@D) =@D. If w = '( ew), we may compute for that ' ( ew) = z z 1 (1 z ew) 83

5 whence dw = d ew ' ( ew) = d ew 1 z 1 zw = d ew 1 z, since ew =1. z ew Consider the function h(w) = u('(w)) on D. By Theorem 6.5 h is harmonic in a neighbourhood of D, so the mean value property yields that u(z) =h() = 1 h(w) dw = 1 u('(w)) dw = 1 ew =1 u( ew) 1 z d ew z ew Definition. The kernel in the above theorem is called the Poisson kernel, anditis denoted e.g. by K(z,w) = 1 1 z z w (z,w@d) or K(z, ) = 1 1 z e i w (z,e We collect the fundamental properties of the Poisson kernel in a lemma. Lemma Poisson kernel satisfies: (P1) K (P) R K(z,w) dw =18z D (P3) Fix and >. Then there is > such that if z D and with then K(z,w) <. z w < and w w > (P4) For fixed the function z 7! K(z,w) is harmonic in. Proof. 1. This is evident.. This follows by applying Theorem 6.1 on the constant function u Choose := 3 /(andassumethat < 1). Then if w w > and z w <, first of all z w < / sothatthetriangleinequalityyields w z > w w w z > 84

6 Moreover Put together 1 z =(1+ z )(1 z ) apple K(z,w) = 1 1 z z w apple ( ) apple <. 4. Observe that Re w + z w z =Re(w + z)( w z) w z =Re apple 1 z wz zw + = 1 z w z w z w z since wz zw =iim wz. Thus we have shown that K(z,w) = 1 Re w + z w z Especially, z 7! K(z,w) is harmonic as the real part of the analytic function 1 w + z w z. We are now ready to prove an important converse to Theorem 6.1: Theorem Assume that f 7! R is continuous. Define u in u(z) = f(w)k(z,w)dw as Then 1. u is harmonic in D. lim z! u(z) =f( ) for 3. u extends continuously to D with = f Proof. By continuity, f apple M Fix Let " > begiven. Then,by uniform continuity of f, there is > sothat apple " and f(w ) f(w) < " if w w < ) (6.) Apply (P3) to pick so that (P3) is valid with above and chosen. Assume that z w < so that the conclusion of (P3) is valid. In order to estimate u(z) f(w ) 85

7 compute u(z) f(w ) (P ) = (f(w) f(w )) K(z,w)dw (P 1) apple w w + f(w) f(w) f(w ) K(z,w) dw f(w ) K(z,w) dw w w < (6.) apple M K(z,w) dw + " K(z,w) dw w w w w < (P 3),(P ) apple M + " apple (4 M +1)" This proves the. claim since " > wasarbitrary. ByaneasyExercise,. implies 3. Finally, by smoothness we may di erentiate under the integral, which yields u = in D by (P4). Remark. Word by word the same proof yields a Poisson formula for C + = {z : y>}, or R n+1 +. One just verifies the properties (P1)-(P4) for the corresponding kernel. Remark. In a general case disc B(a, R) thepoissonformulaisobtainedbyalinear change of variables (z 7! a + R ). We get u(z) = 1 R w a =R R z a u(w) dw = 1 w z R z a Re i z + a u(a + Rei )d As a corollary we obtain a general existence and uniqueness result for the solution of the Dirichlet problem in a disc: Theorem Let B = B(a, r) be a disc and let f C(@B). Then there is a unique harmonic function u on B such that u extends continuously to B with = f Proof. Existence is immediate from Theorem 6.14 (see also above remark). If there are two solutions u 1 and u, then lim z! u 1 (z) =lim z! u (z) =f( ) foreachboundary whence u 1 = u by Corollary Consequences of the Poisson formula Recall the SubMVP from page

8 Definition. If u is continuous in the domain we say that u satisfies the mean value property (MVP) if u(z) = 1 u(z + re i )d whenever B(z,r). If this holds only for small enough radii r < r (z), where r (z) > foreachz, thenwesaythatusatisfiesthe weak MVP. We know that harmonic function satisfy the MVP. The converse is also true, and allows to define harmonicity by just assuming u to be continuous and have the (weak) MVP: Theorem Let u :! R be continuous. Then u is harmonic if and only if it satisfies the MVP in (or equivalently, it satisfies the weak MVP in ). Proof. Assume that u satisfies the weak MVP in. TakeB(z,r) with r<r (z) and define v(z) = u(z + rw)k((z z )/r, w) dw for z B = B(z,r), and v(z) =u(z) on@b. Then v is harmonic in B. Moreover, v u and u v satisfy the weak MVP, whence the SubMVP, and they thus obey the strong maximum principle and minimum principle (Theorem 6.8) in B. Especially, since v = u we deduce that v = u in B, which shows that u is harmonic in B. Since z was arbitrary, the claim follows. The following result deals with inessential singularities of harmonic functions. The idea of the proof is important by itself! Theorem Let u be harmonic and bounded in the punctured disc B(z,r) \{z }. Then u extends to a harmonic function in B(z,r). Proof. Choose <R<r. Call B = B(z,R). Use Poisson formula (6.1) to define a continuous function v on B such that = and v is harmonic inside B. The Theorem follows as soon as we show that u = v in B. Choose " > andconsiderthefunction z z h = u v + " log R Then h is harmonic in B(z,R) \{z } and lim sup z! h( ) = ( z = R 1 = z By the strong maximum principle (Corollary 6.9) we have h inb(z,r) \{z }. Especially, for all " > u(z) apple v(z) log z R Letting! weobtainuapplev. A symmetric argument yields v apple u. 87

9 6.5 Harnack type results Theorem 6.18 (Harnack s inequality). Let u be harmonic and u Then, if r<rwe have on B(z,R). R r R + r u(z ) apple u(z) apple R + r R r u(z ) (Harnack s inequality) for z B(z,r). Proof. By a linear change of variables z 7! z + Rz we may assume that R =1,z =. Observe that the Poisson kernel satisfies 1 1 z max K(z,w) apple ( w z ) = 1 min K(z,w) 1 1 z (1 + z ) = 1 1 z (1 z ) = 1+ z 1 z 1 z 1+ z Hence, by the Poisson formula we may compute if u is harmonic in D: u(z) = u(w)k(z,w) dw (6.3) apple 1+ z u(w) dw 1 z = 1+ z 1 z u() 1 (6.3) (6.4) If u is not harmonic in! 1., but just in D, apply the above to u( z) ( < 1) and let The left hand side inequality is proven in a similar manner using (6.3). Remark. The inequalities in 6.18 are optimal (exercise). Theorem 6.19 (Harnack s principle). Let u n be a increasing sequence of harmonic functions in a domain, whence the limit u(z) = limu(z) exists everywhere. Then n!1 one of these applies: 1. u = 1 identically.. u is harmonic in and u n! u locally uniformly. Before proving the Theorem, let us observe a useful fact. Lemma 6.. If u n is harmonic in for each n, u n! u locally uniformly as n!1, then u is also harmonic in. Proof. The MVP is clearly preserved in local uniform limits, and the claim follows from Theorem Recall that local uniform convergence implies uniform convergence over any compact set. 88

10 Proof. (Theorem 6.19) Wemayassumeu n u 1 ). Denote u n (otherwiseconsiderthesequence A = {z : u(z) =1}, A 1 = {z : u(z) < 1} By connectedness it is enough to show that both A and A 1 are open. Let z A. Choose R>sothatB(z,R). Then if z B(z, R ). Harnack s inequality yields u n (z) R z z R + z z u n(z ) R R/ R + R/ u n(z )= 1 3 u n(z ) Letting n!1we obtain z A, hence A is open. Similarly one check that A 1 is open. By connectedness one of these sets is whole. Assume that u<1. In order to prove local uniform convergence, pick any z and select again B(z,R). Since u n (z ) n!1! u(z ), if > wemaychoosen so that Then by Harnack s inequality in B(z, R ) apple u m (z ) u n (z ) < if m>n n u m (z) u n (z) apple R + R R R (u n (z ) u n (z )) apple 3 since u m u n inb(z, R ). Letting n!1we obtain apple u(z) u n (z) apple 3 for n n and z B which yields the local uniform convergence. Finally, the limit function is harmonic by Lemma Reflexion principle z, R Definition. C is symmetric with regard to R if z () z. Wedenote + = {z, Im z>}, = {z, Im z<} =) = + [ [ ( \ R) (disjoint union) Because is connected, a symmetric domain always intersects R (why?). Lemma 6.1. If u C( ) and u satisfies the weak MVP, i.e. then u is harmonic. u(z) = 1 u(z + re it )dt, 89 r < r (z)

11 Proof. The proof of Theorem 6.16 applies as such. Theorem 6. (Schwarz reflexion principle). Assume that is symmetric with regard to R. If u is harmonic in + and lim z! u(z) =8z \ R, then u extends to a harmonic function in by the formula 8 >< u(z), z + h(z) =, z \ R >: u( z), z Proof. Clearly h is continuous. Obviously, h satisfies the weak MVP at each point z + [ (choose r (z) sosmallthatthecirclebelongto \ R). If z R \, choose first r > sothatb(z,r ). Thenif<r<r, we have by the symmetry in the definition that =h(z )= 1 h(z + re it )dt The harmonicity of h now follows from Lemma Applications to analytic functions Theorem 6.3. Let be a symmetric domain with regard to R. Let f = u + iv : + 7! C be analytic and assume that lim z! v(z) =for all \ R. Then there is an analytic extension of f to such that f( z) =f(z) 8 z. Proof. The imaginary part v is harmonic and satisfies the conditions of Theorem 6.. Thus we may extend v to whole (the extension is still denotes by v) so that v( z) = v(z) 8 z. Let us first show that for any z R \ and a disc B := B(z,r) there is a local extension of f from B + to B. Let U be a conjugate function of v in B = B(z,r). Then U = u + c in B + (why?), where c is a constant. Thus, by adjusting U by a constant and by denoting g = U + iv, we have that g is analytic in B with g B + = f. Moreover, as Im g = v B =onb \ R we see that the function g(z) g( z) isanalyticinb and it vanishes on B \ R. Hence g(z) = g( z) in B, so that g provides the local extension to B with the desired symmetry property. Choose for each a \ R r = r(a) sothatb(a, r(a)) and let g a denote the extension to B(a, r(a)) constructed above. Define 8 >< f(z), z + g(z) = lim w!z >: f(z), z R \ f( z), z Then g is well-defined and analytic at any a. Namely, for a/ R this is evident, and if a R \, weconsiderb := B(a, r(a)) and the extension g a defined above, and obtain by definition that g B = g a and hence g is analytic at a. 9

12 Remark. Reflexion with regard to circles can be analogously defined (exercise). This time one sets z f = f(z) z f(z) in case of the unit circle. If reflexion point in the circle is denoted by z? = have f(z? )=f(z)?, and of course the needed assumption of f is that f(z)! 1as z! 1in +. z z,we As out last application to analytic functions, let us discuss the following surprising rigidity for conformal maps of non-simply connected domains. Theorem 6.4. Denote A(r, R) ={R : r< z <R}. A(r,R ) are conformally equivalent if and only if i.e. the annuli are similar to each other! R 1 r 1 = R r The annuli A(r 1,R 1 ) and Proof. We may clearly assume r 1 = r = 1. Assume that f : A(1,R 1 ) 7! A(1,R )isa conformal bijection. Denote S p R ). Then f 1 (S) iscompactina(1,r 1 ), so for small enough > A(1, 1+ ) \ f 1 (S) =;. Hence f(a(1, 1+ )) A(1, p R ) [ A( p R,R ), and since f(a(1, 1+ )) is connected we must have that either f(a(1, 1+ )) A(1, p R )orf(a(1, 1+ )) A( p R,R ). We may assume that f(a(1, 1+ )) A(1, p R ) (6.5) since if this is not the case we may replace f(z) by R f(z). If z n! 1 +, we must have f(z n )! 1or f(z n )! R since f(z n ) cannot have any limit point in A(1, p R ), and by (6.5) we actually have f(z n )! 1. Similarly, f(z n )! R if z n! R + 1 (Exercise: why cannot now f(z n )! 1if z n! R + 1?). Consider the harmonic function (recall Theorem 6.5) u(z) =log f(z) a log z where a = log R log R 1. By what we just proved, u(z) z!! 1 ). Corollary 6.11 yields that u ina(1,r 1 ). We obtain that f(z) = z a in A(1,R 1 ) Pick any z A(1,R 1 ), and denote by z a some branch of exp(a log z) inasmall neighbourhood of z. Then f(z) =1locally =) f(z) =cz a in that neighbourhood z a 91

13 ( c =1). Butweknowthatz a has an analytic branch in the annulus A(1,R 1 )ifand only if a. Thus f(z) =cz a for some a. This is injective only if a = ±1. Since the inner circles correspond to each others, a =1, c =1. Butthenf(z) =e i z,and R = R 1. Remark. The proof shows that the conformal self-maps of the annulus are f(z) =cz or f(z) =c R 1 z with c =1. Another proof could be obtained by extending f to a conformal self-map of C (=) f(z) =az + b) byrepeatedreflexionovertheboundarycircle(seeremark on page 91). 9

14 Chapter 7 Dirichlet s problem 7.1 Subharmonic functions Definition. Let C be a domain. A function u : 7! R is subharmonic in if it satisfies the SubMVP in, i.e.itiscontinuousand u(z) apple 1 for <r<r = r (z), r (z) <d(z,@ ). u(z + re it )dt Remark. Instead of continuity, one could only require a weaker upper semicontinuity property. This is important in the general theory of subharmonicity. However, for our purposes continuous subharmonic functions su ce. Definition. u is superharmonic if u is subharmonic. Theorem 7.1. Assume that u and v are subharmonic in. Then also are subharmonic on if a, b. au + bv, max(u, v) Proof. That one may multiply by positive constants is evident. Assume that both u and v satisfy the SubMVP at z for radii r<r,andz. Bysymmetry,mayassume that u(z) apple v(z). Then, for r<r max(u, v)(z) =v(z) apple 1 v(z + re it )dt apple M max(u, v)(z + re it )dt Theorem 7. (Global maximum principle). If u is subharmonic in and u attains its supremum at z, then u is constant. 93

15 Theorem 7.3 (Strong maximum principle). If u is subharmonic in and lim sup u(z) apple M<1 z! 1 then u apple M in and equality at one point implies that u is constant in. Proof. Both Theorems 7. and 7.3 are direct consequences of Theorem 6.8. Remark. Recall that 1 is the boundary of with regard to C. The local maximum principle is not valid: a subharmonic function may be constant locally without reducing to a constant. E.g. since linear function is harmonic, u is subharmonic, where (use Theorem 7.1). u(x, y) =max( 1 x, 1,x 1) The following is an important characterisation of subharmonic functions: Theorem 7.4. Let u : 7! R be continuous. Then u is subharmonic if and only if the following condition holds: whenever is a subdomain, v : 7! R is harmonic, then the di erence u v obtains its supremum in if and only if u v is constant. (other words, subharmonic functions are those who obey the (global)maximum principle even after you add modify them by adding an arbitrary harmonic function.) Proof. If u is subharmonic in, then for any harmonic v in, v is harmonic, whence by Thm 7.1 u v is subharmonic in. The validity of the condition follows now from Theorem 7.. Conversely, assume u is continuous and the condition holds. Assume B(z,r). Let v be the harmonic function in B(z,r) that is continuous in B(z,r)andsatisfies = We claim that u apple v in B(z,r). If not, then u v takes its maximum on B(z,r)at an inside point B(z,r), whence the condition of the theorem (applied to v on B(z,r)) implies that v u is constant, i.e. v = u in B(z,r). Thus in any case u apple v. Especially, we have u(z ) apple v(z )= 1 Hence u is subharmonic. v(z + re it )dt = 1 94 u(z + re it )dt

16 Corollary 7.5. If u : 7! R is subharmonic, then u(z) apple 1 u(z + re it )dt for all balls B(z,r) Proof. This follows from the latter part of the proof of Theorem 7.4. Corollary 7.6. Let be a domain, u subharmonic in and v harmonic in. Assume that lim sup u(z) apple lim inf v(z) 1 z! z! Then u apple v in. Proof. Now u v is subharmonic. Use Theorem 7.3. Theorem 7.7. Let, C be domains. Assume that f :! is a conformal bijection and u is subharmonic in. Then u f is subharmonic in. Proof. Exercise. Theorem 7.8. Assume that u C ( ). Then u is subharmonic if and only if u in. Proof. Denote for fixed z, ) a(r) = 1 Previously we computed that a (r) = 1 r u(z + re it )dt z z <r u da. If u, then this shows that a is increasing. Especially, u() = lim r! + a(r) apple a(r), so that u is subharmonic. In turn, if u(z ) < forsomez, thenbycontinuity u<inasmallneighbourhoodofz, whence d a<forsmallr, andu() >a(r) dr for small enough r, andu is not subharmonic. Example. If u is C and convex in C, then (because u is convex in all directions) u xx and u yy =) u =) u is subharmonic. Actually, the mere convexity (with no C assumptions) implies that u is subharmonic. z is subharmonic for all R in C\{}! (Compute u) It is not at all convex if (, 1). It is also subharmonic in whole C if >. 1}, con- max(log z, ) is subharmonic in whole C (log z is harmonic in { z stant is always subharmonic, then use Theorem 7.1). 95

17 7. The Poisson modification Let u be subharmonic in, B = B(z,r). We may solve the Dirichlet problem in B with the boundary values by applying the Poisson formula. Thus, let h be harmonic in B, continuous in B and = Definition. In the above situation, the function ( u(z), z \ B u B (z) = h(z), z B is the Poisson modification of u in the disc B. Theorem 7.9. The Poisson modification u B (of a subharmonic function u) is subharmonic and u B u. Proof. Since h is continuous in the closed set B, andu in \ B, it is clear that u B is continuous. Moreover, it is clearly subharmonic at any point z \ B (since u is) or z B (as h is harmonic). In addition, by Corollary 7.6 h u in B, so u B u. Then, if we have (let r<d(z,@ )) u B (z) =u(z) apple 1 so u B is also subharmonic at any u(z + re it )dt apple 1 u B (z + re it )dt Theorem 7.1. Locally uniform limits of subharmonic functions are subharmonic. Proof. Exercise. 7.3 Perron families Theorem Let A 6= ; be a family of harmonic functions in the domain. Assume that A fulfills the condition: If u, v A, then there is w A so that w max(u, v) Then the function h(z) =sup ua u(z), z is either harmonic or it is identically 1 in. Proof. Let us fix z. Wemaypickasequence(u n )froma so that h(z )= lim n!1 u n (z ) By the assumed property of A we may define inductively a sequence ũ n from A so that 96