UNIFORM CONVERGENCE Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 Suppose f n : Ω R or f n : Ω C is sequence of rel or complex functions, nd f n f s n in some sense. Furthermore, suppose we know tht f n ll hve certin properties. Wht cn we sy bout properties of f? It turns out, tht, t lest for certin notions of convergence, we cnnot relly sy nything bout f. In order for limits of functions to preserve useful properties, like continuity or integrbility, we need certin type of convergence of functions known s uniform convergence. 1. Uniform Convergence If f n : Ω R (everything works the sme for C) is sequence of functions, we sy tht f n converges to f pointwise on Ω if, for ech x Ω, f n (x) f(x). In other words, once we fix x, the sequence f n (x) hs limit, nd we declre f(x) to equl tht limit. This is probbly the most nive nd obvious definition of wht it mens for sequence of functions to converge to limit. Unfortuntely, this notion of convergence is wek, in the sense tht the limiting procedure fils to preserve vriety of useful properties: Suppose ll the f n re continuous on Ω. If f n f pointwise, then f my not be continuous. For exmple, consider f n (x) = x n, Ω = [0, 1]. Then f n pointwise converges to the function f(x) where f(x) = 0, 0 x < 1, f(1) = 1. Obviously this function is not continuous t x = 1. If the fct tht this discontinuity is t n endpoint of closed intervl bothers you, it is esy to modify this exmple so the discontinuity occurs t n interior point (just symmetriclly extend f n to [0, 2]). The bove exmple lso mkes it cler tht if the f n re differentible, then its pointwise limit my not be differentible. A homework problem provides n explicit exmple of sequence of functions f n defined on closed intervl I which pointwise converges to f on I, but for which lim f n dx f dx. n I I In other words, in generl it is not possible to interchnge limit nd n integrl sign. Actully, it is not even cler tht the integrl on the right exists (nmely, pointwise limit of Riemnn-integrble functions my not be Riemnn-integrble). 1
2 UNIFORM CONVERGENCE Becuse pointwise convergence fils to preserve vriety of useful properties, we will define stronger notion of convergence, nd then prove tht some of these properties re preserved under uniform convergence. However, certin properties still will not be preserved under uniform convergence; evidently limit processes cn destroy lots of useful properties! Suppose f n : Ω R is sequence of functions nd f : Ω R is nother function such tht for ll ε > 0, there exists n N = N(ε) such tht for ll n > N, f n (x) f(x) < ε for ll x Ω. Then we sy tht f n converges uniformly to f on Ω. When we write N = N(ε), we emphsize the fct tht while N is llowed to depend on ε, it does not depend on x. Wht s the difference between this definition nd pointwise convergence? We cn rewrite the definition of pointwise convergence s sying tht for ech x Ω nd ε > 0, there exists some N = N(ε, x) such tht for ll n > N, f n (x) f(x) < ε. The difference is tht in the uniform convergence definition, the N in tht definition only depends on ε nd not on x, while in the pointwise definition, N is llowed to depend on both ε nd x. Although this might seem like minor difference, it ctully mkes huge difference. In mthemtics, uniformity usully refers to the property of certin prmeter not depending on nother prmeter in this cse, in the definition of uniform convergence, the prmeter N does not depend on x. In mny situtions, uniformity is good property to hve, lthough its specific consequences vry from sitution to sitution. The following more geometric interprettion of uniform convergence might be useful. The property tht f n (x) f(x) < ε for ll x Ω cn be geometriclly represented by sying tht the grph of f n (x) is lwys within bnd of width ε in either verticl direction of the function f(x). Exmples. If f n converges uniformly to f, then f n lso converges pointwise to f. This is n exercise in the homework. In prticulr, to test whether f n converges uniformly to some function, it suffices to first check if this sequence converges pointwise to f, nd then to check whether f is the uniform limit of f n. Let us directly check tht the sequence of functions f n (x) = x n does not converge uniformly to its pointwise limit f(x). (Recll tht f(x) = 0 if 0 x < 1, f(1) = 1.) Select ny ε < 1, like ε = 1/10, sy. We clim there is no N such tht f n (x) f(x) < 1/10 for ll x [0, 1]. Indeed, if there were, then f n (x) < 1/10 for ll x [0, 1), since f(x) = 0 for x [0, 1). But clerly this is flse, since f n (x) is continuous function on [0, 1], nd f n (1) = 1, so tht for x close to 1, f n (x) >.1. Are there ny convenient criteri for determining whether sequence of functions uniformly converges? There re few, but we will primrily be interested in one criterion, which we will pply to power series inside its disc of convergence. We sy tht series f n uniformly converges to f if its prtil sums uniformly converge to f. Theorem 1 (Weierstrss M-test). Let f n : Ω C be sequence of functions. Suppose there exist sequence of non-negtive rel numbers M n such tht f n (z) M n
UNIFORM CONVERGENCE 3 for ll z Ω nd M n converges. Then f n uniformly converges on Ω to function f(z). Proof. First we check tht f n (z) pointwise converges. Indeed, for fixed z, we hve f n (z) M n < converges, so tht f n (z) converges bsolutely, nd hence converges to some limit, which we cll f(z). Then f n (z) pointwise converges to f(z). We now check tht f n (z) uniformly converges to f(z). Let ε > 0 be rbitrry, nd let s n (z) = n k=0 f k(z) be the nth prtil sum of f n (z). We wnt to show tht there exists some N such tht for ll n > N, f(z) s n (z) < ε for ny z Ω. Notice tht f(z) s n (z) = f k (z) f k (z) M k. Recll tht k=1 M k converges to some limit, so this mens tht for n sufficiently lrge, M k < ε. So select N to be bound for this to be true. Exmple. One of the most useful pplictions of the Weierstrss M-test is tht it shows power series uniformly converges in ny closed disc contined in its disc of convergence. Suppose n z n is power series with rdius of convergence R, nd let Ω be ny closed disc z < R, where R < R. Recll tht in the proof of the theorem which shows the existence of the rdius of convergence, we sw tht n R n is bounded from bove by convergent geometric series r n. (The exct rgument involved using properties of lim sup.) In prticulr, in the course of tht proof we showed tht n z n < cr n whenever z R, where r < 1 is some rel number (which does depend on R, but not on z ) nd c some rel constnt depending on R. Therefore we cn tke M n = cr n in the Weierstrss M-test. 2. Properties of uniform convergence Now tht we ve defined wht uniform convergence mens, nd sw few exmples of sequences/series of functions which did nd did not uniformly converge, let s prove some of the bsic properties of uniform convergence. First, we will show tht the uniform limit of sequence of continuous functions is still continuous (unlike the f n (x) = x n exmple): Proposition 1. Suppose f n converges uniformly to f on Ω. Then f is continuous on Ω. Proof. The method of proof is simpler version of the proof tht power series cn be differentited term-by-term. We wnt to show tht f is continuous on Ω, so this mens given ny Ω, we wnt to show tht for ny ε > 0, there exists some δ > 0 such tht if x < δ (nd x Ω), then f(x) f() < ε. The wy we will prove this is by writing f(x) f() in slightly funny wy, pply the tringle inequlity, nd then use the hypotheses to individully bound ech of the remining pieces. (If you look t the proof of the fct tht power series cn be differentited term-by-term, the strtegy is exctly the sme.) More precisely, we write
4 UNIFORM CONVERGENCE f(x) f() = (f(x) f n (x)) + (f n (x) f n ()) + (f n () f()) f(x) f n (x) + f n (x) f n () + f n () f(). First, becuse f(x) is the uniform limit of f n (x), we know tht there exists n N such tht for ll n > N, f(x) f n (x) < ε/3 for ll x Ω. But we lso know tht ech f n is continuous, so there exists some δ > 0 such tht for ll x < δ, f n (x) f n () < ε/3. (It is possible tht δ depends on n.) So, in summry, given ε > 0 nd x Ω, select n lrge enough so tht both f(x) f n (x) < ε/3, f() f n () < ε/3. This is possible becuse of uniform convergence. Next, select δ > 0 such tht x < δ implies f n (x) f n () < ε/3. Then f(x) f() < ε if x < δ, by dding up these three bounds on the inequlity bove. The second result we prove is tht interchnge of uniform limit of functions with n integrl over finite intervl (versus n improper integrl, like n integrl over R) is llowed. A homework exercise sks you to generlize this to contour integrls. Proposition 2. Suppose f n : [, b] R is sequence of continuous functions which converges uniformly to f on [, b]. Then lim n f n (x) dx = f(x) dx. (Actully, the theorem is still true for f n just Riemnn-integrble, but we use this more restrictive version becuse the proof hs fewer technicl detils in it, nd the continuous version will be generl enough for our purposes.) Proof. We lredy hve shown tht f is continuous, hence integrble. Notice tht f(x) f n (x) dx f(x) f n (x) dx. However, we lso know tht given ε > 0, there exists N such tht f(x) f n (x) < ε for n > N. Therefore, when n > N, f(x) f n (x) dx f(x) f n (x) dx (b )ε. Since b is finite, this implies tht lim n f(x) f n(x) dx = 0, which is wht we wnted to prove. The contour integrl version of this proposition is the primry reson why we re interested in uniform convergence: in one of the next mjor theorems we prove we will wnt to be ble to justify switching limit nd integrl opertion. As mtter of fct, you cn check tht on the question on lst week s HW which required justifying limit nd integrl, tht setup is just specil cse of the contour integrl version of this proposition: in other words, knowing this theorem would hve sved substntil mount of work! (On the other hnd, if you did crefully
UNIFORM CONVERGENCE 5 justify why switching limit nd integrl ws legl, you bsiclly did this proof for tht prticulr exmple.) Finlly, one might sk whether uniform limits preserves differentibility, nd if the limit of derivtives is the derivtive of the limit. Unfortuntely, in generl the nswer is no, s the following exmple shows: Exmple. Let f n = 1 sin nx. Then f n n uniformly converges to f(x) = 0 on x [0, 2π], sy: indeed, notice tht f n (x) 0 < 1/n for ll x R. However, f n(x) = cos(nx), which does not pointwise converge to ny function, let lone converge to f (x) = 0. There is slightly more restrictive cse where uniform limits does preserve differentibility, however. We mention the theorem more out of completeness thn nything else: Proposition 3. Suppose f n uniformly converges to f on (, b), nd f n converges uniformly to g on (, b). Then f is differentible, nd f = g. The reson the lst exmple does not contrdict this proposition is becuse the derivtives of f n did not uniformly converge (let lone pointwise converge) to ny function. Actully, nother reson for mentioning this proposition is tht we will see the sitution with holomorphic functions is much simpler: uniform limit of holomorphic functions is lwys holomorphic, nd we cn interchnge the limit nd differentibility opertions. This is one exmple of how holomorphic functions re more well-behved thn their rel differentible counterprts.