Uniform convergence and its consequences

Transcription

1 Uniform convergence nd its consequences The following issue is centrl in mthemtics: On some domin D, we hve sequence of functions {f n }. This mens tht we relly hve n uncountble set of ordinry sequences, since for ech x D, we hve the sequence {f n (x)}. Pointwise convergence Definition: Suppose tht sequence of functions f n : D R is given nd tht, for ech x D, lim f n(x) exists. We write f(x) for this limit, nd sy tht the sequence of functions {f n } converges pointwise to f on the domin D. Exmple : On (0, ), let Then f n (x) = nx nx. lim f /(nx) n(x) = lim =, x > 0, so f n converges pointwise to the constnt function f(x) =. Exmple 2: On (, ), let f n (x) = + x + x x n. Then, s we know, f n converges pointwise to the limit /( x). Of course, in this cse, we write the limit s the sum of n infinite series: x n =, for x <. x The issue t hnd is the behvior of the limit function f. For instnce, suppose ech of the functions f n is continuous; is it true tht f is continuous? More generlly, which properties of the sequence {f n } survive the limiting process? As nother exmple, suppose tht n x n converges to some f(x) on D. Here the sequence of functions is the sequence of prtil sums of the series: n f n (x) = k x k. k=0

2 Ech member f n of the sequence is polynomil; it cn be differentited, for instnce: f n(x) = n k k x k. k= Wht cn we sy bout the sequence {f n}. Does it converge? If so, does it converge to f, the derivtive of the originl limit? For the geometric series bove, pointwise convergence of the sequence of derivtives is equivlent to the convergence of the infinite series nx n, n= which cn be demonstrted using the rtio test: ( ) b n+ n + b n = x x s n, n so the sequence of derivtives of the geometric series converges on the sme intervl s the originl series. But wht s the limit? We suspect tht nx n = n= ( x) 2, but if true, this must be shown seprtely - it doesn t follow from the bove. Exercise: () Use Tylor s theorem to show tht the bove series converges to /( x) 2. Tht is, show tht the reminder term in Tylor s theorem goes to 0 s n provided tht x <. Nturlly, things do not lwys work out the wy we imgine. Here re some importnt counterexmples to ber in mind:. f n (x) = x n on the intervl [0, ]. Here, for 0 x <, we hve x n 0. On the other hnd, for x =, x n. So here is cse of sequence of continuous functions converging pointwise on closed intervl to limit function which is not continuous. 2. Define g n : [0, ] R by 0 : x = 0 g n = n : 0 < x n 0 : < x n For ech x (0, ], there exists n n N such tht < x. And therefore g n n(x) 0 s n. And for x = 0, g n (0) = 0, n. We therefore hve pointwise convergence of the sequence g n (x) to the limit function g(x) = 0. On the other hnd, we hve 0 g n (x)dx =, n, while 0 g(x)dx = 0. 2

3 Here the sequence of functions converges pointwise to continuous limit, but the integrls of the functions do not converge to the integrl of the limit. The specific functions g n re not continuous, but tht s not the source of the problem: insted of step functions, we could use tent functions which re continuous, nd chieve the sme unplesnt result. Exercise: (2) Construct sequence of continuous functions on [0, ] which converges pointwise to the the zero function nd whose integrls behve s bove. 3. Let f n (x) = x n /n on the intervl [0, ]. This sequence converges pointwise everywhere to the function f(x) = 0. But the sequence f n(x) is the sme s the sequence in the first exmple. So here is n exmple of sequence of functions converging to continuous limit, but the derivtives of the sequence do not converge pointwise to the derivtive of the limit (there s problem t x =, s bove). 4. Let s n (x) = n k+ sin kx ( ). k k= We hve rgued (not rigorously) tht s n (x) f(x) = x/2, nd in prticulr, tht s n (π/2) π/4. Over wht intervl re these considertions vlid? Wht bout the sequence of derivtives, or the sequence of integrls? 2 Uniform convergence The concept we need to mke sense of this is clled uniform convergence. Definition: The sequence f n of functions defined on the domin D is sid to converge uniformly to the function f on D if for ny ɛ > 0, n N such tht for ll x D, f n (x) f(x) < ɛ. This mens tht () every one of the sequences {f n (x)} converges to f(x), nd tht (b) the number n in the definition works for every x D. In some sense, every single one of these sequences converges t the sme rte. Geometriclly, if f n f uniformly on [, b], nd ɛ > 0 is given, drw ribbon extending distnce ɛ in either direction from the grph of f on [, b]. Then there exists n N such tht the grph of f n lies inside the ribbon for ll n N. Definition: The infinite series n (x) is sid to be uniformly convergent in D if the sequence of prtil sums is uniformly convergent sequence of functions in D. Exmples: () The sequence f n (x) = x n converges, but the convergence is not uniform: suppose tht 0 < ɛ <, 0 < x <. Then f n (x) f(x) < ɛ x n < ɛ n log x < 3

4 log ɛ log ɛ = log(/ɛ) < n log(/x), or finlly log(/ɛ) log(/x) < n. For fixed ɛ, s x gets closer to, the frction on the left pproches +, so for ny given n we cn lwys find n x tht violtes the inequlity. So the convergence is not uniform. (2) On the other hnd, for f n (x) = x n /n the convergence to the limit function f(x) = 0 is uniform, since x n n n < ɛ n > ɛ. Exercise: (3) Show tht the sequence of functions you constructed in exercise 2 bove is not uniformly convergent. 3 Consequences of uniform convergence One of the min tools used to demonstrte uniform convergence is clled the Weierstrss M-test; it s comprison test: Theorem: Let the terms of the series n (x) be defined on some domin D, nd let M n be convergent series of positive constnts. If the inequlity n (x) M n holds for ll n nd for ll x D, then the series n (x) is uniformly convergent on D. Proof: We will show tht Cuchy s convergence criterion is stisfied: if s n (x) is the n th prtil sum of the series, then, if m < n, s n (x) s m (x) = m+ (x) + + n (x), nd so s n (x) s m (x) M m+ + + M n, since k (x) M k. Since the series M n is convergent, if ɛ > 0 is given, there exists N N such tht if N m < n, M m+ + + M n < ɛ, nd so, independent of x, we hve s n (x) s m (x) < ɛ, which estblishes the uniform convergence of n (x) in D. Exmple: The series n= sin nx n 2 is uniformly convergent on R. Here we my tke s comprison series n=, since sin nx, n2 4

5 nd the series (/n 2 ) is known to converge. Theorem: continuous. If the functions f n re continuous in D, nd f n f uniformly in D, then f is Proof: Let ɛ > 0 be given. We need to show tht, for ny x 0 D, we cn find δ > 0 such tht x x 0 < δ f(x) f(x 0 ) < ɛ. We hve f(x) f(x 0 = f(x) f n (x) + f n (x) f n (x 0 ) + f n (x 0 ) f(x 0 ) f(x) f n (x) + f n (x) f n (x 0 ) + f n (x 0 ) f(x 0 ) Since f n converges uniformly to f, we cn find n N such tht n N f n (x) f(x) < ɛ/3 for ll x D. From the lst inequlity, we then hve f(x) f(x 0 ) f n (x) f n (x 0 ) + 2ɛ/3, n N. Now fix ny such n; since f n is continuous, there s δ > 0 such tht x x 0 < δ f n (x) f n (x 0 ) < ɛ/3. So x x 0 < δ f(x) f(x 0 ) < ɛ, which is wht we hd to show. Corollry: If the terms of n infinite series re continuous on D nd if the series converges uniformly on D, then the sum of the series is continuous function. 4 Integrtion nd differentition of uniformly convergent sequences We ll need the following Proposition: If f : [, b] R is continuous, then so is f, nd f(x)dx f(x) dx Proof (exercise): Hint: show tht the inequlity holds for ll Riemnn sums. Theorem: Suppose the functions f n re continuous on [, b] nd converge uniformly to f on [, b]. Then f(x)dx = lim f n (x)dx. (For n infinite series, this is equivlent to the sttement tht if the series n (x) = f(x) is uniformly convergent on [, b], nd if ech n (x) is continuous on [, b], then f(x)dx = n n (x)dx. 5

6 To mke sure you understnd this, write out exctly why this second sttement is true.) Proof: We need to show tht, given n ɛ > 0, we cn mke the quntity f n (x)dx f(x)dx less thn ɛ for n sufficiently lrge. By the proposition bove, f n (x)dx f(x)dx f n (x) f(x) dx. And since the sequence is uniformly convergent, there exists n n 0 such tht But then so we re done. Exmple: We know tht n n 0 f n (x) f(x) < f n (x) f(x) dx < ɛ, x [, b]. b ɛ dx = ɛ, b x = x n if x <. So choose ny r such tht 0 < r <. From the theorem, we know tht r r x dx = = r r r r x n dx x 2n dx (n odd gives 0) Evluting the integrls gives log + r ) (r r = 2 + r3 3 + r5 5 + Theorem: Suppose f(x) = lim f n (x), where the convergence is uniform on [, b]. Suppose lso tht the derivtives f n(x) re continuous nd converge uniformly on [, b]. Then f(x) is differentible, nd f (x) = lim f n(x). Proof: Since f n is uniformly convergent sequence of continuous functions, we my write g(x) = lim f n(x), 6

7 where the limit function g is continuous on [, b]. We need to show tht g(x) = f (x). Using the previous result on term-by-term integrtion, we my write, for ny x [, b], But then x x g(t)dt = lim x f n(t)dt. g(t)dt = lim [f n (x) f n ()] = f(x) f(), where in the second line, we ve used the uniform convergence of the originl sequence f n. Tking the derivtives of both sides, we hve g(x) = f (x), nd the theorem is proven. Exercise: Corollry: If the series n (x) converges uniformly to f(x) on [, b], nd if the derivtives n(x) re continuous nd lso converge uniformly on the intervl, then f (x) = n(x), n= so the series my be differentited term-by-term. Exmples: Consider the following expressions: ( x) 2 = 2 ( x) 3 = nx n = + 2x + 3x n= n(n )x n 2 = 2 + 6x + 2x n=2 These re obtined by differentiting both sides of the expression for the geometric series x = x n. Are the two expressions vlid? The nswer is positive if it cn be shown tht the series on the right hnd sides converges uniformly. If so, then the corollry sserts the equlity of the two sides. We consider the first series. We know tht the originl series converges for x <. Choose ny number r such tht 0 < r <. Then we hve, for ll x r, kx k kr k =: M k. But the series kr k = k= 7 k= M k

8 converges by the rtio test since M k+ lim k M k = lim k k + k r = r <. So, by the Weierstrss M-test, the series of derivtives converges uniformly for x < r. Since r is n rbitrry positive number less thn, the series of derivtives converges uniformly on the intervl x <, nd the first expression is vlid. Exercise: Show tht the second expression is vlid s well. 8