Open mapping theorem

1 SOLUTIONS TO COMPLEX PRELIM PROBLEMSC IHAN BAHRANIn this document, I will collect my solutions to some (not all, probably not even half)of the Complex Prelim problems that were asked in the past in theorems get used a lot. I will copy some of them from David C. Ullrich swonderful bookComplex Made mapping theoremThis is very useful in general. It s easy to forget the connectedness assumption, so Iwill state it (V) denotes the set of analytic maps from an open mapping open andf H(V). Also letWbe an open andconnected set contained inV. Thenf(W)is either a singleton (that is,fis constantonW) or open 2011, 7 and Fall 2010, that there is no one-to-one conformal mapof the punctured disk{z C: 0<|z|<1}onto the annulus{z C: 1<|z|<2}.

2 LetDdenote the unit open disk andAthe described annulus. Suppose, to the contrary,that there existsf H(D {0}) which is injective andf(D {0}) =A. Firstly, since|f(z)|<2 for allz D {0}, we havelimz 0zf(z) = 0 is a removable singularity off. Thus g H(D) such thatg|D {0}=f. As 0 is alimit point ofD {0},g(0) is a limit point ofg(D {0}) =A. Thusg(D) A={z C: 1 |z| 2}.Moreover, sincefis injective,gis not constant. Thus by the open mapping theorem ,we getg(D) Int(A) = =g(0) A. Sincef(D {0}) =A, there existsz D {0}such thatg(z) =f(z) = . Sincez6= 0, there exist disjoint open disksV,W Dsuch thatz Vand 0 W. Again by the open mapping theorem ,g(V) andg(W) are open subsetsofA.

3 Thusg(V) g(W) is open. However, sincefis a bijection andV,W {0}are1 SOLUTIONS TO COMPLEX PRELIM PROBLEMS2contained inD {0}, we haveg(V) g(W) =g(V) g({0}) g W {0} =f(V) { } f W {0} = f(V) { } f(V) f W {0} =f(V) { }={ }.This is a 2011, a holomorphic function on a connected open that ifh(z)2=h(z) for allz Vthenhis constant onV. Find all possiblevalues for of all,Vis nonempty1. Letf=h3 H(V). So for allz V,f(z) =h(z)2h(z) =h(z)h(z) =|h(z)|2 R. That is,f(V) R. Supposefis not constant. Then the openmapping theorem yieldsf(V) Int(R) = (we are consideringRas a subspace ofChere), a contradiction. Thus for allz V,h(z)3=f(z) =afor somea C.

4 Thereforeif we let =e2 i3, we haveh(V) {|a|13, |a|13, 2|a|13},soh(V) is finite. Nonempty open sets inCare infinite, thush(V) is not open. Againby the open mapping theorem we deduce thathis constant. The rest is the kind ofstuff Calc 1 students fail to do correctly quite often: basic algebra. For somex,y R,h(z) =x+iyfor allz V. So(x+iy)2=x+iyx2 y2+i(2xy) =x iy .Thusx2 y2=xand 2xy= y. The second equation givesy= 0 orx= 12. Going tothe first equation, the first case yieldsx2=xand the second case yields14 y2= the solutions are 0,1, 12+i 32, 12 i fractional transformationsI think the most useful linear fractional transformations for the prelims are the onesthat map a half plane to the unit disk.

5 And, thanks to Ullrich s book, I know thatthere is a way to do this which is really cool and impossible to forget. LetUbe theupper half plane andDbe the open unit disk. The neat geometric observation is that1 Why? Because the correct definition of connectedness excludes the empty space. Otherwise wewouldn t get a unique decomposition of every topological space into its connected components. Thesame reason as why we don t count 1 as a prime number: to have unique TO COMPLEX PRELIM PROBLEMS3z Uif and only ifzis closer toithan it is to i. This yieldsz U |z i|<|z+i| z iz+i <1 z iz+i the LFTz iz+imapsUontoD.

6 In exactly the same way,z 1z+ 1maps the righthalf plane ontoD. Let s roll:Fall 2012, that a M obius transformation maps a straight line or circleonto a straight line or takes work! I d like to know if there is a shorter argument. We start with a lemma:Lemma an LFT. ThenT 1(R)is a circle inC (that is, a line or a circleinC). (z) =az+bcz+dwherea,b,c,d Csuch thatad bc6= 0. Note that T 1(R) T(w) R a +bc +d=T( ) =T( ) =a +bcw+d ac| |2+ad +bc +bd=ac| |2+ad +bc +bd (ac ac)| |2+ (ad bc) + (bc ad) +bd bd= 0(?)We have two cases: ac ac= 0. Then (?) becomes(ad bc) + (bc ad) +bd bd= 0(ad bc) +(bc ad) +bd bd= if we writeA=ad bc, andB=bd, we getA A =B B2iIm(A ) = 2iIm(B)Im(A ) = Im(B).

7 ( )We observe thatA6= 0. Suppose not. Then we havead=bcand soacd=acd=bccc(ad bc) = bc6= 0, we getc= 0. But thenad= 0 soad= 0 and hencead bc= 0, a contradiction. SinceA6= 0, ( ) describes a TO COMPLEX PRELIM PROBLEMS4 ac ac6= 0. Then we can write (?) as| |2+ad bcac ac +bc adac ac +bd bdac ac= bcac ac, soA=ad bcac ac=bc adac acand writeB=bd bdac ac. Thus weget| |2+A +A +B= 0| |2+A +A +AA=AA B| +A|2=AA B( )Note thatAA B=ad bcac ac bc adac ac bd bdac ac=(ad bc)(bc ad) (ac ac)(bd bd)(ac ac)2=abcd+abcd aadd bbcc[2i Im(ac)]2=ad(bc ad) bc(ad bc) 4 [Im(ac)]2=(ad bc)(ad bc)4 [Im(ac)]2=|ad bc|24 [Im(ac)] if we letr=|ad bc|2 Im(ac)>0 (ad bc6= 0), ( ) becomes| +A|2=r2,which describes a circle.

8 Definition ,z2,z3be three distinct points inC . We define an LFT (,z1,z2,z3)by(,z1,z2,z3)(z) = (z,z1,z2,z3) = (z z1)(z2 z3)(z z3)(z2 z1)if none ofz1,z2,z3are z2 z3z z3ifz1= z z1z z3ifz2= z z1z2 z1ifz3= Note that (,z1,z2,z3) sendsz1,z2,z3to 0,1, , s sends circles inC to circles inC . an LFT andCbe a circle inC . Pick three distinct points 1, 2, 3onC. Letzi=T( i) fori= 1,2,3 and letS= (,z1,z2,z3). Note that by constructionz1,z2,z3 S 1(R), which is a circle inC by Lemma 2. But the three distinct pointsSOLUTIONS TO COMPLEX PRELIM PROBLEMS5z1,z2,z3determine a unique circleC inC ; thusS 1(R) =C . In a similar fashion,we get (S T) 1(R) =Csince 1, 2, 3uniquely determineC.

9 Thus we haveT(C) =T((S T 1)(R)) = (T (S T) 1)(R) =S 1(R) =C . Fall 2012, (z) is analytic on the punctured unit diskD {0}, andthe real part off(z) is positive. Prove thatfhas a removable singularity at {z C: Re(z)>0}. We may writef:D {0} H. Also by the abovediscussion, we have an invertible analytic map :H Dz7 z 1z+ 1with the inverse 1(z) =z+ 1 z+ 1=z+ 11 z. Note thatg:= fis analytic and mapsD {0}toD. In particular,gis bounded. Thuslimz 0zg(z) = 0and hence 0 is a removable singularity ofg. So there exists an analytic functionh:D Csuch thath|D {0}=g. Sinceh(D {0}) D, by continuity we geth(D) D. Thereare two cases by the open mapping theorem :(1)his constant.

10 Then picka D {0}, soh(0) =h(a) =g(a) D.(2)h(D) Int(D) = any case, we may writeh:D D. So forming the composition 1 his legal, and( 1 h)|D {0}= 1 g=f. We effectively removed the singularity offat 2009, (z) be an analytic function onCwhich takes value in the upperhalf planeU. Show thatfis deal. (z) =z iz+imapsUconformally toD, thus fis an entire functionwhich takes values inD. Thus by Liouville s theorem , fis constant. is invertible,sofis also 2009, an explicit conformal equivalence which maps the open setbounded by z 12i =12and|z i|= 1 onto the upper half am too lazy to learn drawing circles in LaTeX now, so the reader should draw {z C: z 12i <12}andD2={z C:|z i|<1}.