10. Sequences and series of functions: uniform convergence

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1 10. Sequences and series of functions: uniform convergence Pointwise and uniform convergence We have said a good deal about sequences of numbers. It is natural also to consider a sequence of functions (f 1, f 2,...). A simple example is: f n (x) = x n for each n. What might one mean by the limit of a sequence of functions? There are different possible answers to this question. The simplest one is as follows. Let E be some real interval (possibly the whole real line, or the positive half line). Definition. We say that (f n ) converges to f pointwise on E if, for each x E, we have f n (x) f(x) as n. For each fixed x, [f n (x)] is, of course, a sequence of numbers, so this is just saying that each such sequence of numbers tends to a limit; it is not really a new idea at all. Note that, for the statement to have any meaning, it is essential to specify the interval E. Example 1. Let f n (x) = x n, and let E = [0, 1]. By 1.13, f n f pointwise on E, where f(x) = In this example, each f n is continuous, but f is discontinuous at 1. So a pointwise limit of continuous functions can fail to be continuous. We now introduce a second type of convergence that will turn out to overcome this defect. What would one mean by one function being a good approximation to another (on a stated interval E)? A natural answer is as follows. We say that G ε-approximates f on E if g(x) f(x) ε for all x E. This means that g lies between the dotted lines in the diagram. Definition. We say that (f n ) converges to f (as n ) uniformly on E if, given any ε > 0, there exists n 0 such that for all n n 0, f n ε-approximates f on E. That is, f n (x) f(x) ε for all x E If (f n ) converges to f uniformly on E, then it converges to f pointwise on E. Proof. This follows at once from the definition with x fixed. 1

2 Another trivial observation is: if f n f uniformly on E, and E 1 E, then f n f uniformly on E 1. How do we determine whether convergence is uniform in particular cases? In most cases, this is done along the lines of one of the two following remarks. (1) Suppose we can find numbers M n, tending to 0 as n, such that f n (x) f(x) M n for all x E. Then f n f uniformly on E; this statement just rewrites the definition very slightly. It does not matter whether M n is the exact greatest value of f n (x) f(x) on E. (2) Suppose that there exists c > 0 such that, for each n, we can find a point x n E with f n (x n ) f(x n ) c. Then the definition fails with ε = c/2, so (f n ) does not tend to f uniformly on E. Example 2. Again, let f n (x) = x n, with f as in Example 1. This shows that the converse of 10.1 is false: pointwise convergence does not imply uniform convergence. Example 3. Let f n (x) = x n. However, (f n ) does tend to 0 uniformly on any bounded interval [ M, M], since f n (x) M/n on this interval, and M/n 0 as n. nx Example 4. Let f n (x) = 1 + n 2 x. Show that f 2 n 0 pointwise on [0, ). Does it tend to 0 uniformly (a) on [0, 1], (b) on [1, )? 2

3 Example 5. Let f n (x) = x + n x 2 + n 2. Show that f n 0 uniformly on R. Consequences of uniform convergence 10.2 PROPOSITION. Let E be a real interval. Suppose that (f n ) is a sequence of functions, each continuous on E, and that f n f uniformly on E. Then f is continuous on E. Proof. Choose x 0 E (for the moment, not an end point) and ε > 0. Of course, we already saw in Example 1 that pointwise convergence is not sufficient for this conclusion. The result for integration is very easy: 10.3 PROPOSITION. Suppose that (f n ) is a sequence of functions, each continuous on the interval [a, b], and that f n f uniformly on [a, b]. Then b a f = lim n b a f n. 3

4 Proof. Note. Under the weaker assumption that each f n is Riemann-integrable on [a, b], one can show that f is Riemann-integrable and the stated limit holds. (Exercise!) The next example shows that pointwise convergence is again not sufficient. Example 6. Let f n be as shown. Actually, pointwise convergence is sufficient in 10.3 under the additional condition that there exists M such that f n (x) M for all n and all x E. This is a major theorem, and we will not prove it here. Using both versions of the fundamental theorem of calculus (7.13 and 7.15), we can derive a result for differentiation. The statement is not quite as straightforward as the previous two results PROPOSITION. Suppose that (f n ) is a sequence of functions, each having a continuous derivative on [a, b], and that: f n f pointwise on [a, b], f n g uniformly on [a, b]. Then f is differentiable on (a, b), and f (x) = g(x) for a < x < b. (Similarly for one-sided derivatives at the end points.) Proof. 4

5 A simple example shows that the hypothesis f n g cannot be removed in 10.4, even if f n f uniformly. Example 7. Let f n (x) = 1 sin nx. Then f n n(x) 1/n for all x, so f n 0 uniformly on R. However, f n(x) = cos nx, which (for x 0) does not tend to any limit, so (f n) is not even pointwise convergent. Series of functions Let f 1, f 2,... be functions, and let s n = f f n. If (s n ) tends to a function f uniformly on E, we say that the series n=1 f n converges to f uniformly on E. A pleasantly simple sufficient condition is given by the next result PROPOSITION (the M-test ). Suppose that, for each n, we have f n (x) M n for all x E, and that n=1 M n is convergent. Then the series n=1 f n is uniformly convergent on E. Proof. By the comparison test, for each x E, the series n=1 f n(x) is convergent: let its sum be f(x). Note that if n=1 f n is uniformly convergent on E (with sum f), and each f n is continuous on E, then, by 10.2, so is f. Example 8. If n=1 a n is convergent, then n=1 a n cos nx is uniformly convergent on R, since Example 9. n=0 xn /n 2 is uniformly convergent on [ 1, 1], since Example 10. Recall that the Riemann zeta function is defined for x > 1 by: ζ(x) = n=1 1/nx. This series is uniformly convergent on [1 + δ, ) for any δ > 0, since 1/n x 1/n 1+δ on this interval. It follows that the zeta function is continuous for all x > 1. Example 11. Recall that the power series x x3 3 + x5 5 converges to tan 1 x for x 5

6 in [ 1, 1]. In fact, we showed in 8.2 that if s n (x) is the sum of n terms, then s n (x) tan 1 x 1/(2n + 1) on this interval. So the series is uniformly convergent to tan 1 x on [ 1, 1], but not by the M-test! 10.6 PROPOSITION. Let n=0 a nx n be a power series. Suppose that n=0 a n r n is convergent for some r > 0. Then the power series is uniformly convergent on [ r, r]. In particular, this holds for any r < R, where R is the radius of convergence. Proof. Note that this result says that, for any ε > 0, the sum of the series can be ε- approximated on [ r, r] by a polynomial, the sum of a suitable number of terms. simplest possible example shows that we can not say this for the open interval ( R, R): Example 12. The geometric series n=0 xn. For this series, R = 1, and the sum of the series is 1/(1 x), which is unbounded on ( 1, 1), so certainly cannot be ε-approximated by a polynomial there. By 10.6 and 10.2, we have re-proved continuity of power series functions (3.11), seemingly with no effort! Without very much effort, we can also re-prove the harder theorem on differentiation of power series (4.13): 10.7 THEOREM. Suppose that the power series n=0 a nx n converges for x < R, with sum f(x). Then f is differentiable on ( R, R), with f (x) = n=1 na nx n 1. Proof. By Lemma 4.11, the series n=1 na nx n 1 also converges for x < R: let its sum be g(x). For each n, let s n (x) = a 0 + a 1 x + + a n x n, The so that s n(x) = a 1 + 2a 2 x + + na n x n 1. Choose r < R. By 10.6, s n (x) f(x) uniformly on [ r, r]. Also, since the series for g(x) is also a power series, s n(x) g(x) uniformly on [ r, r]. By 10.4, it follows that f is differentiable on ( r, r), with f (x) = g(x) for x < r. But this holds for every r < R, so in fact it holds for all x with x < R. 6

7 A continuous function that is nowhere differentiable We have presented a number of consequences of uniform convergence that might seem fairly routine. We now describe one that most people find quite amazing! It is a fractal construction, first thought of by Weierstrass in the 1880 s. For each n 0, define f n (x) = inf{ x k/4 n : k Z}. To clarify what this means, write d n = 1/4 n. Then f n (x) = x on [0, 1 2 d n] and f n (x) = d n x on [ 1 2 d n, d n ]. These values are then repeated on each interval [kd n, (k + 1)d n ]. The function f n is continuous, since it comprises straight-line pieces, and 0 f n (x) 1 2 d n for all x. By the M-test, the series n=0 f n(x) is uniformly convergent on R, defining a continuous function f(x). Let s n = f 0 + f f n. The diagram shows s 0, s 1, s 2. We show that f is not differentiable at any point. Choose any x. For each r, let a r = 4 r [f(x + d r ) f(x)], b r = 4 r [f(x) f(x d r )]. Note that for all n r, we have f n (x ± d r ) = f n (x). Let A r be the union of all intervals [4kd r, (4k + 1)d r ] and [(4k + 2)d r, (4k + 3)d r ], and let B r = R \ A r. A bit of thought shows that if x A r, then x and x + d r lie under a single line segment of the graph of f r 1, and of f n for 0 n r 1 (the diagram helps!). Each such line segment has gradient ±1, so for 0 n r 1, we have 4 r [f n (x + d r ) f n (x)] = ±1. Now a r is the sum of r such terms, so it is an integer, even if r is even and odd if r is odd. In the same way, if x B r, then b r is even if r is even, odd if r is odd. Now define c r to be a r if x A r and b r if x B r. Then c r is even if r is even and odd if r is odd, so does not tend to a limit as r. But if f were differentiable at x, then (a r ) and (b r ), hence also (c r ), would converge to f (x). There is no hope of adequately drawing the graph of f! It has a strict local minimum at every point of the form k/4 r. Any function you can draw is really piecewise differentiable. 7