4 Uniform convergence

Transcription

1 4 Uniform convergence In the last few sections we have seen several functions which have been defined via series or integrals. We now want to develop tools that will allow us to show that these functions are analytic. Recall that in general, it is not enough to know that the sum fx lim n f n x converges everywhere and that each f n x is differentiable to deduce that f is differentiable, or that if f is differentiable, that f x lim n f k x. For example, if f nx sinn2 x n then f n x fx : 0 for all x R, but f nx f x 0. We ll see that things behave a little more nicely for analytic functions because of the convergence properties of power series. Recall that the above is asking whether lim lim f n z + δz f n z δz 0 n δz f n z + δz f n z lim lim n δz 0 δz so what we are trying to do is swap the order of two limits. Similarly, asking whether lim n f n z dz lim f nz dz n is a question of whether two limits can be swapped, as the integral is also a kind of limit. In order to do this sort of limit swapping you need some extra hypotheses! Definition: Let f, f 2, f 3,... be a sequence of complex-valued functions defined on a common domain S C. Then f n f uniformly on S or f n converges to f uniformly on S if for all ǫ > 0 there is an integer N such that for every n N, f n z fz < ǫ whenever z S, that is, such that sup f n z fz ǫ. z S This is to be constrasted with the weaker notion of pointwise convergence: f n f pointwise on S means that for all z S, lim n f n z fz, or in full, ǫ > 0 z S N n N f n z fz < ǫ., In this definition N Nǫ, z, that is, N depends on both the choice of ǫ and z, whilst in the definition of uniform convergence N Nǫ, that is, N depends on ǫ alone. 8

2 Recall: Uniform convergence is a metric space convergence, using the metric d f, g f g sup fz gz. z S Of course one must limit this to functions where this quantity is defined, such as the set of all bounded functions on S. The definition of uniform convergence doesn t require that the functions are bounded, but one has to take great care if this is not the case! Pointwise convergence is not in general a metric space convergence, which is partly why it is less well-behaved. The following result shows that uniform limits and contour integration commute. Lemma. For n, 2, 3,... let f n be a continuous function on a region Ω and suppose that f n f uniformly on Ω. Then f is continuous on Ω and, for any contour lying in Ω, f n z dz fz dz. lim n Proof. The continuity of f follows from the standard ǫ/3 argument that you saw in earlier courses, using the fact that for any z, w Ω and any n, fz fw fz f n z + f n z f n w + f n w fw. Furthermore standard integral bounds tell us that f n z dz fz dz f n z fz dz f n f length. and so as f n f 0. f n z dz fz dz Example: The fact that a contour has finite length is important here! Let f n z e z n2 / for n, 2, 3,... and for < Im z <. Then f n fz 0 uniformly on the n strip. However π f n z dz 2 fz dz. The reason the above proof doesn t work in this case is that length. As we saw above, in the real setting, even the taking of uniform limits and differentiation do not commute. However, in the complex setting, the previous lemma plus Morera s theorem gives the following result. 9

3 Theorem 2. For n, 2, 3,... suppose f n is analytic in a disk z z 0 < R and suppose f n f uniformly in each closed disk z z 0 R δ, where δ > 0. Then f is analytic in z z 0 < R, and, moreover, f n f uniformly in each disk z z 0 R δ. Proof. Given Morera s theorem it is enough to show that fz dz 0 whenever is, say, a triangular contour in z z 0 < R. For such a contour choose δ > 0 small enough so that is contained in z z 0 R δ. Then fz dz lim f n z dz 0 n where the first equality follows by the previous lemma and the second equality follows from Cauchy s theorem. We have proved that f is analytic in z z 0 < R. Next assume z z 0 R δ and let C denote the circle z z 0 R δ/2. By Cauchy s formula for deriviatives we have f z f n z 2πi Now given ǫ > 0 we can choose N so that, for all ζ C, C fζ fn ζ dζ ζ z 2 fζ f n ζ < ǫδ2 4R whenever n N. Therefore, for n N, f z f nz 2π ǫδ 2 4R 2πR δ/2 2 since ζ z δ 2 for ζ C. In other words f z f nz < ǫ for n N. Since z is an arbitrary point satisfying z z 0 R δ and N does not depend on z we have proved the result. The restriction of the region to a disk can be relaxed in the above result. We will frequently be interested in the uniform convergence of series of functions f n on an more general set. Roughly speaking if the series converges uniformly on some compact set, then so does its series of deriviatives assuming the functions are analytic f n, and moreover f n d f n. dz Recall: f n is defined to be the limit lim N s N where s N f +f 2 + +f N is the Nth partial sum of the series of functions. 20

4 Theorem 3. Suppose that {f n } is a sequence of analytic functions defined on a region Ω C. Suppose that f n converges uniformly on each compact subset of Ω with limit function f. Then f is analytic on Ω and moreover f n converges uniformly to f on each compact subset of Ω. Proof. This result follows very directly from the previous theorem. Suppose that K is a compact subset of Ω. For each z K, there exists δz > 0 such that the disk Dz, 2δz {w : z w < 2δz} Ω. Clearly {Dz, δz : z K} forms an open cover of K, so by compactness, one can choose a finite subcover: K Dz, δz Dz k, δz k. Let D j denote the closure of Dz j, δz j, j,...,k. The series converges uniformly to f on each of these closed disks, so by the previous theorem, f is analytic on each of the open disks, and hence on K. Furthermore s N f uniformly on each of the closed disks D j Since there are only finitely many of these disks covering K, one may therefore deduce that s N f uniformly on K. The most commonly involved test to guarantee uniform convergence is the Weierstrass M- test. Theorem 4. Weierstrass M-test Suppose that S is a nonempty subset of C and that {f n } is a sequence of functions defined on S. Suppose that for each n, there exists a constant M n such that f n M n. If M n < then the series f n converges uniformly on S. Recall: This theorem just a version of the Banach space theorem that if X is a Banach space and {x n } is a sequence in X with x n <, then x n convergences in the Banach space. Exercise: Check you could prove this! Recall: A series of complex numbers c n converges absolutely if c n converges. As in the real numbers, absolute convergence implies convergence. A series of complex functions fn converges absolutely on S if f n z converges absolutely for every z S Exercise: i Show that under the hypotheses of the Weierstrass M-test, the series f n converges absolutely as well as uniformly. 2

5 ii Give an example of a series of analytic functions on a set S which converges uniformly, but not absolutely. iii Give an example of a series of analytic functions on a set S which converges absolutely, but not uniformly. Exercise: Let f n z sin n2 z, n, 2, 3,.... Then f n f 0 uniformly on the real line. n Furthermore, each f n entire, as is the limit function. So why doesn t f n f on the real line? Theorem 5. If a nz n has radius of convergence ρ > 0 meaning that the series converges, indeed converges absolutely for z < ρ then a nz n converges uniformly in any disk z R where R < ρ. Proof. Assume ξ satisfies R < ξ < ρ then a n ξ n converges so that there is a constant M such that But then, for all z satisfying z R, we have a n M ξ n n, 2,.... a n z n M ξ nrn, neither M nor ξ depends on z. lim n a nξ n 0. Hence But R/ ξ < so that the geometric series R/ ξ n converges. Hence, by the Weierstrass M-test a n z n converges uniformly for z < R. Remark: Note that we need not have uniform convergence on z < ρ. For example, / z z n, z < pointwise but not uniformly on z <. Nevertheless we can conclude, by Theorem 3, that d a n z n dz Example: Show that n, n 0 na n z n z < ρ. z + n n is uniformly and absolutely convergent on each compact subset K of the punctured plane C\{n : n Z}. 22

6 Solution. Since K is compact, K is bounded and so there is a constant M such that z M for all z K. Therefore z + n n z nz + n M + z/n Now for n > 2M, + z/n > /2 whilst inf + z/n δ > 0, z K, n 2M for all z K. since K is compact, K is closed which means C\K is open. Therefore there exists a constant A such that But P A n 0 z + n n A for all n Z and z K. n2 is convergent. Hence the result follows by the Weierstrass M-test. Since each function a n z z+n n n, n 0 is analytic on C\{n : n Z} it follows that z + n n is analytic on C\{n : n Z} and moreover, by Theorem 3, that d dz z + n n z +. n, n 0 n, n 0 We have shown that we can differentiate this sum by differentiating each term. 23

7 5 Infinite products 5. Interpolation A common and classical problem in calculus is to find a function that takes specified values at certain specified points. For example, it is easy to find a polynomial p for which you have specified N values pz j c j, j,...,n. Indeed you can choose the Lagrange interpolating polynomial, N i j pz c z z i j i j z j z i. j Written another way, this expresses pz as c j l j z where l j is a polynomial for which l j z i δ ij. If you wish to find an analytic function that satisfies infinitely many conditions, things are much more complicated. We have already seen, for example, that it is impossible to find an entire function f such that f/j { 0, if j is even /j, if j is odd. It is of course easy to construct a continuous function with these constraints! A natural question to ask is whether one can ever construct anything like the Lagrange interpolating polynomial with infinitely many terms. This would require us to write some form of infinite product of terms z z j. This obviously requires some care as such an infinite product can easily be zero or unbounded! One of the amazing truths about analytic functions is that essentially all entire functions can be written as infinite products. This was discovered but not proved, as was so much else, by Euler in about 750. We shall begin this section by studying one example: sin z2. This example introduces most of the key ideas. We shall then look more carefully at the theory of infinite products of numbers, such as 2 π 4 24

8 and infinite products of functions. Later we shall apply the ideas to two further examples, the gamma function and the zeta function. 5.2 A nontrivial example The basis of our example is the following partial fraction series representation of cot. π cot z + 2z z 2 for z 0, ±, ±2,.... To prove this we consider the contour C N which is the boundary of the rectangle bounded by the lines y N, y N, x N + 2 and x N + 2. y x N + 2 y N x N N N + x y N We first show that on C N there is a bound for cot which is independent of N, that is, there is a B such that, for all N, 2, 3,... cot B whenever z C N. Note that cos cosπx cosh πy i sin πx sinh πy sin sin πx cosh πy + i cosπx sinh πy. Therefore, on x ± N + /2, cos sin sinhπy cosh πy sinh πy + sinh 2 πy 25

9 whilst on y ±N, cos sin So in fact we can take B 2. cos2 πx cosh 2 πn + si πx sinh 2 πn si πx cosh 2 πn + cos 2 πx sinh 2 πn cos2 πx + sinh 2 πn si πx + sinh 2 πn + sinh 2 πn sinh 2 πn + sinh πn + πn < 2. Suppose now that z / Z and make sure that N > z. We will now evaluate I N,z π cotπw 2πi C N w 2 z dw. 2 Note that from the above bounds, and using the fact that if w C N, then w 2 N 2 and hence w 2 z 2 N 2 z 2, I N,z π cotπw 2π C N w 2 z 2 dw which tends to 0 as N. C N N 2 z dw 24N +, 2 N 2 z 2 On the other hand, for any N, the integral is just the sum of the residues at the singularities inside C N. The singularities occur where sin πw 0 and where w 2 z 2 0, that is, at integers w N, N,..., N, N and w ±z. π cot πw Let fw. Then each integer is a simple zero of sin πw and thus is a simple w 2 z2 pole of f and the Laurent expansion is of the form fw c w n + c 0 + c w n + c 2 w + Calculating the residue is easy at such a point: πw n cosπw Resf, n limw nfw lim w n w n w 2 z 2 sin πw z 2. A similar argument shows that the residues at z and z are Thus the Residue Theorem gives I N,z Resf, z + Resf, z + π cot 2z N n N and Resf, n π cot 2z π cot 2z + π cot 2z + N n N z 2 26

10 Thus, with z fixed, and taking a limit as N, lim I N,z 0 π cot + π cot + N 2z 2z or, using symmetry and pulling out the n 0 term, n z 2 0 π cot z z z 2. Therefore, multiplying through by z, as claimed. π cot z 2z z 2 Clearly each term on the right hand side has an antiderivative related to logz 2 and the left hand side has an antiderivative related to logsin log. If no-one is looking you might write that sin logsin log log or sin logz 2 log z 2 z 2...before you realize that you have no idea what any of this means! Our task then is to turn this into something that does make sense to make this rigorous. sin Let us begin by considering lsinz on the punctured disk, 0 < z <. Lemma 6. lsinz is analytic on the punctured disk, 0 < z <, where, as usual, is the principal branch of log. Proof. fz is analytic in any region where the analytic function fz is neither zero nor takes a negative real value, since this is the cut for the function. First note that sin /z can only vanish where sin 0, and none of the zeros is inside the punctured disk. Next suppose that the function takes a negative real value, k, or sin k where k > 0. Equating real parts gives sinπx cosh πy kπx or simply sin πx k πx 27

11 where k > 0. y fx sin πx gx k πx, k > 0 2 x From the graphs this can only occur if x 0 or x >. So we can assume that x 0 and equate the imaginary parts to obtain sinh πy kπy. Since k > 0 the only solution is y 0 and therefore the function lsin is analytic on the punctured disk. In fact, the function sin / clearly has a removable singularity at z 0 and so we can make lsin analytic at z 0, provided we give sin / the value there. Consequently lsin is analytic on the disk z <. Moreover, for 0 < z < we have d dz lsin z d sin dz d dz sin z π π cot z. From this we deduce that the deriviative of lsin at z 0 must equal lim z 0 π cot z 0. Note also that Thus, for z <, d dz z2 2z z2 2z z 2. d dz lsin z 2z z 2 d dz z2 d z2. dz 28

12 Here we could interchange d/dz and since the series of s converges uniformly on any compact set not containing an integer. One easy way to prove this is to observe that w 2 w when w /2. Therefore z2 2 z 2 2 for z < and n 2. But 2 n2 < and so we have the result by the Weierstrass M-test we can clearly forget a finite number of terms in the series if we want to. Therefore, for z <, sin z2 + C for some constant C. Substituting z 0 shows that C 0. That is, sin z2 lim N N z2 At this point we would like to take the sum inside the logaripi*xthm, but we need to remember that a + b is not necessarily ab. Rather. a + b ln a + i Arg a + lnb + i Arg b ln ab + iarg a + Arg b If Arg a + Arg b π, π then everything does work OK. Exercise: a Show that Arg z2 n sin z 2. 2 b Use the fact that t. sin t for t [0, 0.25] to show that for z < and n 2, z sin 2 <. n. 2 c Deduce that for any N, N Arg z2 Pi*x In light of this then, sin < π 2 +. N n2 z2 n π π < 2.3 < π. lim N N z2 lim N N z2 29

13 Now recalling that exp w w for all w 0, and, of course, that exp is continuous, sin sin exp N exp lim z2 We shall write this as lim N exp lim N sin N N z2 N z2. z2 for z <. We will soon see that this equation holds for all z. If we substitute z /2 we obtain so that π 2 4 π 2 2 2n 2n + 2n 2n + 2 or π which is traditionally and rather unsatisfactorily written as Wallis product π Our immediate aim is to provide a general theoretical framework for what we have done, to regularize the example and to allow us to treat other examples. 5.3 Products of constants Definition: If {p n } is a sequence of non-zero complex numbers we say that the infinite product p n converges to P if the sequence of partial products P N p p 2 p N converges to a non-zero limit P. If the infinite products converge to zero or infinity then the product is said to diverge. Infinite products which do not converge are said to diverge. Remark: Intuitively, for the infinite product to converge we d expect to need that p n converges to 0, which means that p n. As we ll see shortly, this turns out to be correct, 30

14 so it is common to consider infinite products of the form + α n where α n for all but a finite number of n. Clearly we then must have α n 0 for the product to exist. Definition: More generally we agree to say that an infinite product p n exists if. at most a finite number of factors are zero; and 2. the product of the non-vanishing terms exists in the above sense. Hence a convergent infinite product has the value 0 if and only if one or more of its factors is 0. Example: a Consider so that the infinite product exists if and only if lim 2 n n 2 3 n exists and is non-zero. But the value is lim n 0. Therefore the infinite product n does not exist. n It diverges to 0. Ignoring the initial 0, this corresponds to the fact that diverges to. N n log n + log n + Example: b Consider + n n Here P N /2 if N is odd and N+2 2N+ if N is even. Clearly then lim N P N /2 so that the infinite product exists and equals /2. Note that if 0 were prefixed as the first factor in example b then the product would still exist and would equal 0. We say that it converges to 0. Example: c It is tempting to think that if P p n exists then P p n. This need not be the case however. Let p n exp iπ/2 n, so that the partial products are given by P N N p n exp iπ /2 N. In this case P N P, so we have p n πi while P πi. 3

15 Lemma 7. If P p n exists then p n. Proof. If P exists, but is zero, then we need only look at those terms past the last p n that is zero, so without loss of generality, let s assume that P 0. With the notation as above then, lim p P n n lim P n n P n P. Theorem 8. P p n exists and is nonzero if and only if p n converges. Proof. As we saw in the concrete example we did earlier, the main issue here is that we might not have p n p n. Recall however that for any N N N Arg p n Arg p n + 2k N π, some k N Z, and so Thus P N N p n N p n + 2k N πi. N N N N p n exp p n exp p n + 2k N πi exp p n. One direction is now easy! If p n converges then P N exp p n as N since exp is continuous and so P N is convergent with a non-zero limit. For the converse, assume that P lim P N exists and is nonzero. Let us first assume that P, 0. Then is continuous at P and hence P lim N P N lim N P N exists. As above, for all N, c N : P N N p n + 2k N πi for some integer k N. Since the sequence {c N } converges it is Cauchy, and in particular c N c N p N + 2πik N k N 0, as N. Thanks to Joel for googling the fix to this proof! 32

16 But we know that p N, and hence that p N 0. This implies that 2πik N k N 0 too, which can obviously only happen if the sequence {k N } is eventually constant. Thus, there exists an integer k such that for all sufficiently large N, N p n P N 2kπi. Since the right-hand side converges to P 2kπi, the left-hand side converges too. The remaining situation is if P, 0. If every p n were real and positive, then P would be too, so there must exist at least one term, say p m which is not real and positive. In this case P n mp n converges to a nonzero point which is not on the negative real axis. The previous case would then imply that n m p n converges, and hence that p n converges too. Remark: It is worth noting that we actually showed that if p n is non-zero then the sum p n can only differ from P by an integer multiple of 2πi. The following result provides us with a very simple test. Theorem 9. p n converges absolutely if and only if p n converges absolutely. Therefore if p n converges absolutely then p n converges. Proof. Let us write p n + α n. Then, for α n <, the Taylor series for gives + αn α n α 2 n 2 + α3 n 3 α4 n 4 + α n 2 + α n 3 + α n α n αn + α n 2 + α n If we further assume α n /2 then the geometric series in the last line has sum at most /2 and so, for α n /2, + α n α n 2 α n or, equivalently, 2 p n pn 3 2 p n. The result now follows by the comparison test since clearly for n large enough, we can assume that α n /2. 33

17 Corollary 20. If α n < then Exercise: Show that + α n converges. α n < if and only if + α n converges. If either of these above conditions hold then we say that +α n converges absolutely. It therefore follows that if + α n converges absolutely then + α n converges. This is, of course, because the similar result for series is true. To complete the discussion we shall see, by two examples, that the convergence of α n is neither necessary nor sufficient for the convergence of the product + α k. Example: a Let α 2n n /2 and α 2n n /2 + n. Then α n is divergent being the harmonic series. Note that n n n n 3/2 so + α n + n 3/2 is convergent. Example: b On the other hand if α n n /n 2 then 2 α n is clearly convergent, whilst 2 + α n is divergent. Exercise: Prove this last statement! Hint: Prove and use the fact that + α 2n + α 2n + < 2n 2n. 34