4.5 The Open and Inverse Mapping Theorem

Transcription

1 4.5 The Open and Inverse Mapping Theorem Theorem [Open Mapping Theorem] Let f be analytic in an open set U such that f is not constant in any open disc. Then f(u) = {f() : U} is open. Lemma If is a ero of f of order m N, then there are r, > such that for every w D(, ), the function f w has m eros (c.m.) in D(, r). Proof of Open Mapping Theorem assuming the lemma. Let w f(u), we will show that there is > such that D(w, ) f(u). Let U be such that f( ) = w. Then is a ero of f w. Since f is not constant near, we have ord (f w ) = m for some m N. Applying the lemma, we see that there is > such that for every w D(, ), the function (f w ) w = f (w + w) has at least a ero in U, i.e., w + w f(u). Thus, D(w, ) f(u). Remarks. If U is a domain, then the condition in the theorem is equivalent to that f is not constant. We have seen before that if a holomorphic function f defined on a domain U satisfies that f is constant or 3 Re f + Im f =, then f is constant. Now we see such kind of results all follow easily from the open mapping theorem. Proof of the lemma. For some r >, we have f() = a n ( ) n, < r. Let g() = a m ( ) m and h() = n=m f() g() ( ) m = n=m+1 a n ( ) n m. Then h is analytic at, and h() =. Pick r (, r ) such that h =r < am. This is possible because lim h() = and a m. Let = am rm >. Fi any w D(, ). Let f w () = f() w and g w () = g() w. Note that g w () = iff ( ) m = w a m. If w =, g w has one ero:, whose order is m; if w, then g has m eros, each of which is simple because g w() = a m ( ) m 1. Thus, g w has m eros (c.m.), whose distance from are all equal to ( w a m )1/m, which is less than ( a m )1/m < r. Thus, g w has m eros in { < r}. For { = r}, and g w () = a m ( ) m w a m ( ) m w > a m r m a m r m, f w () g w () = f() g() = r m h() r m h =r < a m r m From Rouché s theorem, f w also has m eros (c.m.) in D(, r). 66 < g w ().

2 Definition An analytic function f defined on U is called an analytic isomorphism if there is an analytic function g defined on f(u) such that g(f()) = for every U. If f is an analytic isomorphism defined on U such that f(u) = U, then we say that f is an analytic automorphism of U. We say that f is a local analytic isomorphism at a point if there eists an open set U containing such that f is an analytic isomorphism on U. It is clear that an analytic isomorphism f must be injective. The following theorem states that the converse is also true. Theorem Let f be holomorphic and injective on an open set U. Then f is an analytic isomorphism. Proof. Let V = f(u). Then V is an open set by the open mapping theorem. Define g = f 1 on V. Since f is injective, g is well defined. It suffices to show that g is holomorphic. Since f is injective, it is not constant in any open disc. From the open mapping theorem, for any open set O U, g 1 (O) = f(o) is open. This shows that g is continuous. Let Z denote the set of eros of f. Let U \ Z and w = f( ). Since g and g 1 = f are both continuous, we see that w w iff g(w) g(w ). Thus, g(w) g(w ) g(w) g(w ) lim = lim w w w w g(w) g(w ) f(g(w)) f(g(w )) = 1 f (g(w )), where in the last step we used that f (g(w )) = f ( ). Thus, g (w ) eists. So g is holomorphic on V \ f(z). Note that every w f(z) is a singularity of g V \f(z). Since g is continuous on f(u), it is bounded near every w f(z). So these singularities are all removable. This means that if we suitablly redefine g at points in f(z), then the new g is holomorphic on V. However, the new g and the old g are both continuous on V, and may differ only at f(z). The two functions have to be the same. Thus, the g without modification is holomorphic on U. If f and g are both holomorphic, and g = f 1, then from Chain rule, g (f())f () = 1 for U, which implies that f () for U. Thus, if f is a local analytic isomorphism at, then f ( ). On the other hand, f () for U does not imply that f is an analytic isomorphism. One eample is f() = e. Note that (e ) = e for all C, but e is not injective on C. Theorem [Inverse Function Theorem] If f is holomorphic at, and f ( ), then f is a local analytic isomorphism at. Proof. Let w = f( ). Since f ( ), is a ero of f w of order 1. Applying Lemma to f w, we find r, > such that for any w D(w, ), f w has eactly one ero in D(, r). This means that, for every w D(w, ), there is eactly one D(, r) such that f() = w. Let U = f 1 (D(w, )) D(, r). Then f is injective on U, which is open and contains. From the previous theorem, f U is an analytic isomorphism. 67

3 Homework Chapter V, 3: Suppose f 1 is an analytic isomorphism defined on U, and f is an analytic isomorphism defined on f 1 (U). Prove that f 1 1 and f f 1 are also analytic isomorphisms. 4.6 Evaluation of Definite Integrals We are going to apply the Residue theorem to compute definite integral f()d := lim f()d + lim R R R f()d. Suppose that f is holomorphic on the closed upper half plane {Im } ecept for finitely many isolated singularities: k, 1 k n, in the open upper half plane {Im > }. Choose R > such that k < R for 1 k n. Note that [,R] f = 1 f( + t(r ()))(R ())dt = R f()d. Let denote the semicircle with radius R: γ(t) = Re it, t. Then Γ := [, R] is a positively oriented Jordan curve. From Residue formula, we get R f()d + f = Γ If we know that lim R f =, then we get R lim R f()d = i f = i n Res k f. k=1 n Res k f. (4.1) R The limit lim R f()d is called the principal value integral, and is denoted by P. V. f()d. k=1 It is slightly different from f()d. Their relations are: 1. If f()d eists, then P. V. f()d eists and is equal to f()d.. If P. V. f()d eists, f()d may or may not eist. f() =. Since R d = for all R, P. V. d =. But since d = + and d =. One countereample is d does not eist 68

4 3. If f() is nonnegative or an even function on R, then the eistence of P. V. f()d implies the eistence of f()d, which must be equal. Moreover, if f is even, and f()d eists, then f()d = f()d = 1 f()d. Eample. We want to calculate d 1. Let f() =. Then f is meromorphic in C with 4 poles: 1 = 1 + i, = 1 + i, 3 = 1 i, 4 = 1 i, among which 1 and lie in {Im > }. Since ( 4 + 1) = 4 3 does not equal to at j, 1 j 4, we get Res j f = j = 1 4 j 4 j = j 4. To estimate f, we note that f() 1 R 4 1 on since = R 4 1 > on, which implies that Since f is even, we get f L( ) f SR R R 4, R. 1 f()d = i(res 1 f + Res f) = i ( 1 + ) =. Homework Chapter VI :. Eample. Compute sin d. Although we know that sin etends to an entire function, this does not help use compute the integral. In fact, we will integrate the function f() := ei. Note that f has eactly one pole in C, which is. The pole lies on R, so the above method does not apply. Suppose R > >. Let be as in the last eample. We consider a contour (closed curve): Γ = [, R] [, ] S. Note that f is holomorphic on and inside Γ. From Cauchy s Theorem, = Γ f = f f + S R f()d + f()d. Note that R f()d = R cos R d + i sin d; 69

5 Since cos is odd and sin f()d = is even, we get cos d + i sin d. R Thus, ( R lim lim R We will show that lim R First, we have So f()d + f()d + R sin f()d = i d. ) sin f()d = i d. f =, and evaluate lim S f. f = f e ireit Re it ireit dt = ie ireit dt. e sin t dt = The last equality holds because sin( t) = sin t. Lemma sin t t for t /. / e sin t dt. Proof. Since sin (t) = sin t on [, ], sin t is a concave function on [, t ]. Note that y = is the equation of a straight line that passes through (, ) and (, 1). Since the curve y = sin t also passes through these two points, the concaveness of sin on [, t ] implies that sin t for t /. From the lemma / e sin t dt / e t/ dt e t/ dt = R, which tends to as R. e Now we evaluate lim i S. For > and θ (, ], let γ,θ denote the curve γ (t) = e it, t θ. Note that if θ =, we get the circle { = }, and θ = corresponds to the semicircle S. Lemma Suppose f has a simple pole at. Then lim + γ,θ f = iθ Res f. 7

6 Proof. Let a = Res f. We may write f() = a + h() such that h is holomorphic at. We compute γ,θ a d = θ a θ e it ieit dt = iadt = iθa. Since h is bounded near, and L(γ,θ ) = θ as, we get γ,θ h as +. Thus, lim + γ f = iθa = iθ Res f. Since Res e i = 1, from the lemma we get lim e S i d = i. Thus, sin d = 1 i lim e S i d =. Eample. Compute cos(a) d for a R. 1+ First we assume that a. Let f() = eia. Let R > 1. Let Γ be the contour [, R] S 1+ R. Using Residue formula, we get R f()d + SR f = i Res i f = i eiai i = e a. Note that R R cos(a) R f()d = 1 + d + i sin(a) R 1 + = cos(a) 1 + d. The second term vanishes because sin(a) is odd. 1+ For, Im, so Re(ia). Thus, e ia = e Re(ia) 1 on. So f SR 1 which implies that f f SR L( ) R R 1, as R. Thus, letting R and using the fact that cos(a) 1+ is even, we conclude that cos(a) 1 + d = e a. R 1, Now we consider the case a <. The above argument does not work because e ia 1 on. You may work on a lower semi circle, and repeat the above argument. There is a simple way to do this. Using the fact that cos is even, we get cos(a) 1 + d = cos( a) 1 + d = e a, where the second equality holds because a >. Thus, we get Homework Chapter VI : 8(a), 9. e ia 1 + d = e a, a R. 71