Math 185 Homework Problems IV Solutions

Math 185 Homework Problems IV Solutions Instructor: Andrés Caicedo July 31, 22 12 Suppose that Ω is a domain which is not simply connected Show that the polynomials are not dense in H(Ω) PROOF As mentioned in lecture, since Ω is not simply connected, there is a cycle Γ in Ω and a point α / Ω such that Ind α (Γ) Fix such Γ and α Then for at least one of the loops γ in Γ we must have Ind α (γ) Assume moreover that γ is smooth; we will discard this extra assumption at the end 1 By Cauchy s formula, Ind α (γ) = 1 2πi γ dz z α Let g(z) = 1 Then g H(Ω) We claim that g is not limit of polynomials z α (in the sense of the metric d of Problem 1) Were the polynomials dense in H(Ω), g would be limit of a sequence of them, so our claim solves the problem To see the claim, recall that convergence with respect to d is equivalent to uniform convergence on compact subsets of Ω Suppose g is limit in H(Ω) of a sequence (p n ) n of polynomials in z The set K = γ [, 1] is compact, so p n g uniformly on K Since uniform convergence allows us to exchange the order on limits and integrals, p n g But γ γ γ p = for any polynomial p (since each polynomial is the derivative of an analytic function, namely another polynomial) This is a contradiction Hence, all that remains is to show that we can replace any given (continuous) loop γ in Ω with a smooth one γ 1 such that Ind α (γ) = Ind α (γ 1 ) If we manage to do this, the argument above applies to γ 1 instead of γ, and we would be done 1 If we take as part of the definition of cycle that the loops of which it consists are smooth, then we don t need this extra step 1

By a simple change of variables, we may assume γ : [ π, π] C = C \ {} and α = Let P : [ π, π] C be a trigonometric polynomial sufficiently close to γ Explicitly, this means that for some δ >, 1 P (x) γ(x) < δ/4 for all x [, 1], and 2 δ < γ(x) for all x [, 1] In lecture it was shown that such functions P exist For example, letting S n (x) = n k= n we can take P (x) = σ n (f, x) := 1 n + 1 e ikx 1 π γ(t)e ikt dt, 2π π n S l (x) for n sufficiently large l= Given such a trigonometric polynomial P, we know that Ind(P ) = Ind(γ) ie, we defined Ind(γ) this way, and showed it does not depend on the choice of P and coincides with the old meaning of Ind(γ) if γ itself is smooth Hence, we can take γ 1 = P, and this will complete the argument, provided that γ 1 [ π, π] Ω But γ [ π, π] is a compact subset of Ω, so for some ρ >, for any z C and t [ π, π], if z γ(t) < ρ then z Ω Hence, it suffices to require in addition to 1 and 2 above that δ 4ρ This completes the proof 13 Show that the official definition of index of a loop given in lecture coincides with the definition using liftings, ie, if γ : [a, b] C is a continuous loop, / γ [a, b] and c : [a, b] C is a continuous lifting of γ, so e c(t) = γ(t) for all t [a, b], then c(b) c(a) Ind(γ) = 2πi PROOF If γ is smooth, ie, γ C 1( [, 1] ), this was shown in lecture Suppose now γ is only continuous By a change of variables, we may assume γ : [, 1] C = C \ {} Let P : [, 1] C be a trigonometric polynomial sufficiently close to γ As in Problem 12, this means that for some δ >, 1 P (x) γ(x) < δ/4 for all x [, 1], and 2 δ < γ(x) for all x [, 1] For example, letting s n (x) = n k= n e ik(2πx π) 1 π ( t γ 2π π 2π + 1 ) e ikt dt, 2 2

we can take P (x) = σ n (f, x) := 1 n + 1 n s l (x) for n sufficiently large l= (In slightly more detail: In lecture we showed that if g : [ π, π] C is continuous, then σ n (g, x) = 1 n l e ikx 1 π g(t)e ikt dt g(x) as n + 1 2π l= k= l n, the convergence being uniform on [ π, π] The formula given above comes from this one; simply reparametrize [, 1] as [ π, π] ie, consider the change of variables x [, 1] 2πx π [ π, π] and its inverse t [ π, π] t 2π + 1 [, 1]) 2 We have that for any such P, Ind(P ) = Ind(γ) Choose P such that, moreover, P (x) γ(x) < δ/8 for all x [, 1], and let Q(t) = P (t) P () + γ() Then Q(t) γ(t) < δ/4, so Ind(Q) = Ind(γ) Also, Q() = γ() and, since P and γ are loops, Q(1) = γ(1) We claim that Q and γ are fixed endpoints homotopic in C Were that the case, we are done: Let c : [, 1] C be a (continuous) lifting of γ, and let d : [, 1] C be the unique lifting of Q such that d() = c() Then (as shown in lecture) c and d are fixed endpoints homotopic in C, so d(1) = c(1), and Ind(γ) = Ind(Q) = d(1) d() 2πi = π c(1) c(), 2πi as we wanted to prove Hence, all that remains is to show that γ, Q are fixed endpoints homotopic in C For this, we explicitly define a homotopy witnessing this fact: Let Then 1 H is continuous H(t, s) = (1 s)γ(t) + sq(t), (t, s) [, 1] [, 1] 2 H(t, ) = γ(t) for all t [, 1] 3 H(t, 1) = Q(t) for all t [, 1] 4 Let z = γ() = Q() = γ(1) = Q(1) Then H(, s) = (1 s)γ() + sq() = (1 s)z + sz = z for all s [, 1] 5 In the same way, H(1, s) = z for all s [, 1] 6 Finally, H(t, s) for all (t, s) [, 1] [, 1] For suppose H(t, s) = Then s ( γ(t) Q(t) ) = γ(t), but s ( γ(t) Q(t) ) γ(t) Q(t) < δ/4 < δ < γ(t) This is a contradiction 3

We are done 14 Let D = B(, 1) Let f : D D be continuous The goal of this problem is to prove Brower s fixed-point theorem: There is z such that f(z) = z 1 Suppose otherwise Define g : D D as follows: Let g(z), for z D, be the unique point where the ray that starts at f(z) and passes through z meets D Show that g(z) = z + t z ( z f(z) ) for some real t z Show that z t z is continuous (explicitly compute it as the root of some quadratic, be careful about the choice of sign in the radical) and conclude that g is continuous 2 Show that g(ζ) = ζ for any ζ D Use results from lecture to obtain a contradiction PROOF If z f(z) for all z D, then the ray joining f(z) to z is uniquely determined, and the function g is well defined The direction of the line l containing this ray is that of the vector z f(z), so the line l = { t ( z f(z) ) : t R } is parallel to l Hence, if a translation of the plane makes the image of l meet l, in fact this image coincides with l It follows that l = { z + t ( z f(z) ) : t R }, from which it is clear that the form of g is as described If the map z t z is continuous, then g is composition of continuous functions, hence continuous We have 1 = g(z) 2 = ( ( )) z + t z z f(z) ))( z + tz ( z f(z) = t 2 ( z z f(z) 2 + 2t z z 2 R ( z f(z) )) + z 2, which shows t z is the root of a quadratic equation Explicitly, t z = 2( z 2 R ( z f(z) )) ± 4 ( z 2 R ( z f(z) )) 2 4 z f(z) 2 ( z 2 1) 2 z f(z) 2 ( z 2 = ± R ( z f(z) )) 2 + z f(z) 2 (1 z 2 ) ( z 2 R ( )) z f(z) z f(z) 2 Let l(t) = z+t ( z f(z) ), so l is continuous and injective, l( 1) = f(z), l() = z and l(t z ) = g(z) Since z is a point in the segment [g(z), f(z)], it follows from connectedness of R that t z Therefore, in the expression for t z we found above, the sign in front of the radical must be positive (the expression inside the radical is in absolute value bigger than or equal to the term outside the radical, 4

since 1 z 2 ) This gives t z as a continuous function of z, as we needed to show In lecture it was shown that if h : D D is continuous, then there is ζ D such that h(ζ) = ζ In particular, this must be true for g But if z D, then by definition g(z) = z, since z and g(z) are in the same ray beginning at f(z) D, and both are on D This is a contradiction, and completes the proof: The function g cannot exist, hence for some z D it must be the case that f(z ) = z 15 Let P and Q be polynomials in z Let D = B(, 1) Suppose that a deg P deg Q b P (z) Q(z) for all z D c All zeroes of Q lie in D 1 Show that there is an analytic function f defined on a neighborhood of C \ D such that whenever Q(z), then f(z) = P (z) Q(z) 2 Show that f(z) 1 for all z C \ D 3 Fix λ with λ > 1 and let g(z) = P (z) λq(z) Show that if g(ζ) = Q(ζ), then ζ D 4 Use Gauss theorem from Problem 1 to conclude Bernstein s theorem: With P, Q as above, it must be the case that P (z) Q (z) for all z D 5 Let P be a polynomial in z of degree at most n such that P (z) 1 for z 1 Show that P (z) n for z 1 PROOF (1) Let f(z) = P (z) for z not a zero of Q Let ε > be sufficiently small Q(z) to ensure that if a zero of Q lies in { ζ : 1 ε < ζ 1 }, then in fact it lies in D Let V = { z D: Q(z) = } Then f is analytic in {ζ : 1 ε < ζ } \ V It suffices to show that if z D and Q(z ) = then lim z z f(z) exists Letting f(z ) = lim z z f(z) for z V, the function so defined extends f to a continuous function in Ω = {ζ : 1 ε < ζ } Since V is finite, results from lecture show that f so defined is in fact analytic in Ω, and (1) follows 5

Suppose, then, that Q(z ) = and z = 1, say Q(z) = (z z ) n H(z) where n > and H(z) is a polynomial such that H(z ) Since P (z) Q(z) for all z D, we have P (z ) =, so P (z) = (z z ) m I(z) for some m > and polynomial I(z) such that I(z ) We are done if m n, because then lim f(z) = z z if m > n, I(z ) H(z ) if m = n Toward a contradiction, suppose m < n Then lim z z f(z) = But this contradicts that P (z) Q(z) for z D, since this inequality implies that f(z) 1 for all z D \ V ( ) 1 (2) Let h(ζ) = f for ζ < 1, ζ, h() = lim z f(z) (This limit ζ exists since deg P deg Q) Then h H ( 1 B(, 1 ε )) because h H ( 1 B(, 1 ε )\ {} ) and h C ( B(, 1 1 ε )) Hence, max ζ 1 h(ζ) = max ζ =1 h(ζ), by the maximum principle, ie, max z 1 f(z) = max z =1 f(z) 1 ( by (1) ), as we were trying to show (3) If g(ζ) = Q(ζ) then P (ζ) = λ If ζ / D, then Q(ζ) λ = P (ζ) Q(ζ) = f(ζ) 1, a contradiction Hence, ζ D (4) Let z D We claim that P (z ) Q (z ) Assume first that Q (z ) = By Gauss theorem, z lies in the convex hull of the zeroes of Q Then, Q(z ) = as well Otherwise, if l denotes the line tangent to D at z, there is by continuity a line l parallel to l, but so close to it that it divides D into two regions, the smaller of which contains no zeroes of Q But then the larger region of D determined by l is a convex set containing all these zeroes Hence, it must contain their convex hull, so it contains all the zeroes of Q, including z This is a contradiction 6

C l z z 3 z 2 4 z 1 z z z 2 3 z1 O zn-1 zn-1 z n Since Q(z ) = = Q (z ), z is a double zero of Q The argument of (1) shows that in this case we must have that z is a double zero of P as well, hence P (z ) = and in particular P (z ) Q (z ) holds Suppose now that Q (z ) and, toward a contradiction, that P (z ) > Q (z ) Let λ = P (z ) Q (z ) Then λ > 1 and P (z ) λq (z ) = Let g(ζ) = P (ζ) λq(ζ), so g is as in (3) and g (z ) = The roots of g lie in D, since by (3) g(ζ) = implies that ζ D or else Q(ζ) =, in which case by hypothesis we must have ζ = 1 Hence, the roots of g lie in D as well, and the same argument as in two paragraphs above shows that if g (ζ) = and ζ = 1, then g(ζ) = as well, and therefore Q(ζ) = = P (ζ) In particular, since g (z ) = and Q (z ), we see that P (z) f(z ) = lim f(z) = lim z z z z Q(z) = lim z z P (z) P (z ) z z Q(z) Q(z ) z z = P (z ) Q (z ) = λ But then f(z ) = λ > 1, a contradiction, and we are done (5) This is a corollary of Bernstein s theorem: Simply take Q(z) = z n Then P (z) 1 = Q(z) whenever z D By Bernstein s theorem, P (z) Q (z) = n for z D Then P (z) n for all z D, by the maximum principle 7

16 Let D = B(, 1) Suppose f = u + iv H(D), f(z) = n a nz n 1 Let < r < 1 and n > Show that a n = 1 πr n u(re iθ )e inθ dθ = i πr n v(re iθ )e inθ dθ 2 Use the first identity to show that if f() = 1 and u, then a n 2 for all n 1 3 Conclude that f(z) 1 + z 1 z 4 Show that f(z) for any z D 5 Conclude that f(z) 1 z 1 + z PROOF (1) Let < r < 1 Fix n N Then Cauchy s formula for the derivatives gives ie, a n = f (n) () = 1 n! 2πi = a n = 1 2 1 2πr n ( 1 πr n z =r f(re iθ )e inθ dθ, f(z) 1 2π f(re iθ ) dz = zn+1 2πi r n+1 e i(n+1)θ ireiθ dθ u(re iθ )e inθ dθ + Hence, we are done if we show that for n >, since this gives us that i πr n f(re iθ )e inθ dθ =, u(re iθ )e inθ dθ = i ) v(re iθ )e inθ dθ v(re iθ )e inθ dθ The series f(z) = n= a nz n converges absolutely and uniformly in B(, r) Thus, we are allowed to exchange the integral and the series in the following identity: f(re iθ )e inθ dθ = ā k r k e i(n+k)θ dθ k= { Let m Z Since e imθ if m =, dθ =, the series above vanishes 2π otherwise unless n =, in which case it equals 2πā This shows (1) 8

(2) Let < r < 1 Then a n = 1 πr n u(re iθ )e inθ dθ 1 πr n As shown in (1), a = 1 2π ( Since a = f() = 1, it follows that 1, so u(re iθ ) dθ + i a n 2 r n (2) follows from this, letting r 1 u(re iθ ) dθ ) v(re iθ ) dθ v(re iθ ) dθ = and 1 2π u(re iθ ) dθ = (3) If a n 2 for all n 1, then for any z D, f(z) a n z n a + 2 z n = 1 + 2 z 1 z = 1 + z 1 z, as required n= n=1 (4) If z D and f(z) =, then f(z) is not an interior point of f D, since f D { ζ : Rζ } because u By the open mapping theorem, it must then be the case that f is the function identically equal to But f() = 1, contradiction (5) Let g(z) = 1 f(z) = u(z) f(z) 2 i v(z) f(z) 2 g is well defined by (4), so g H(D), Rg and g() = 1 Applying (3) to g instead of f we find 1 1 + z = g(z) f(z) 1 z, which is equivalent to the inequality we want References 9

1 C Berenstein and R Gay, Complex Variables An Introduction, Springer- Verlag, 1991 2 M Marden, Geometry of Polynomials, American Mathematical Society, 1966 3 W Rudin, Principles of Mathematical Analysis, McGraw-Hill, 1976 1