Uniform Convergence Where is it prove that one obtains the derivative of an infinite series by taking derivative of each term ? - - -Niels Henrik Abel We have already studied sequences and series of (constant) real numbers. In most problems, however, it is desirable to approximate functions by more elementary ones that are easier to in- vestigate. We have already done this on a few occasions. For example, we looked at the uniform approximation of continuous functions by step, piecewise linear, and polynomial functions. Also, we proved that each bounded continuous function on a closed bounded interval is a uniform limit of regulated functions. Now all these approximations involve estimates on the distance between the given continuous function and the elementary functions that approximate it. This in turn sug- gests the introduction of sequences (and hence also series) whose terms are functions defined, in most cases, on the same interval. 1 Sequence and Series of Functions (i) For a set E ⊂ R, let us denote by F(E; R) the set of all functions from E to R. We are interested in sequences and series in the sets F(E; R). For each sequence hfni 2 F(E; R)N and each x 2 E, the numerical sequence hfn(x)i 2 RN may or may not converge. Let x fn : [0; 1] ! R be a function, given by fn(x) = n ; x 2 [0; 1], then hfni is a sequence of functions on [0; 1]. (ii) Let E ⊂ R and let huni 2 F(E; R)N. Then the formal sum 1 X u1 + u2 + ··· + un + ··· = un n=1 is called an infinite series of functions with general term un. For each x 2 E, we have a numerical series 1 X u1(x) + u2(x) + ··· + un(x) + ··· = un(x) n=1 For each n 2 N, we can then define the partial sum n X sn(x) = u1(x) + u2(x) + ··· + uk(x) = uk(x) k=1 This defines a sequence hsni 2 F(E; R)N. 1 2 Small Overview On Uniform Convergence 2 Pointwise Convergence Definition 1. [Pointwise Convergent Sequence of functions ]: For each hfni 2 F(E; R)N; let E0 ⊂ E be the set of all points x 2 E such that the numerical sequence hfn(x)i 2 RN converges and let f(x) = lim fn(x); 8x 2 E0 (1) n−!1 which, in detail, means that, corresponding to an " > 0, 9N = N(x; ") 2 N, depends on both x and ", such that jfn(x) − f(x)j < "; whenever n ≥ N (2) The sequence hfni is then said to be pointwise convergent (or simply convergent) on E0 and the function f 2 F(E0; R); defined by (1) is called the pointwise limit (or simply limit) of hfni on E0. For example, let X = f1; 2; 3g and let fn(k) ≡ n(modk); k = 1; 2; 3 where n(modk) is the remainder when n is divided by k. Let a = 1, then fn(1) = 0 for n 2 N and hence fn(1) ! 0. On the other hand hfn(2)i = h1; 0; 1; 0; · · · i and hence the sequence is not convergent. Hence the sequence hfni is not pointwise convergent on X. We now look at a few examples and examine their pointwise convergence. Pay attention to the graphs of these functions to get an idea of what is going on. As far as possible, we shall investigate whether the given sequence is pointwise convergent and if so, we shall determine the limit function. EXAMPLE 1. (A discontinuous limit of continuous functions) Consider, the sequence hfn(x)in, n where, fn(x) = x , for all x 2 [0; 1] as y 6 depicted in the Fig. 1. For x 2 [0; 1). 1 We then have lim fn(x) = 0. On n−!1 the other hand, lim fn(1) = 1. The n−!1 n =n 1= 2 sequence is therefore pointwise conver- n = 5 gent to a function f(x), on [0; 1], where - x ( O ! 0; if 0 ≤ x < 1; f(x) = Figure 1: fn(x) for n = 1; 2; 3; 4; 5. 1; if x = 1 Note that each function fn(x) of the sequence is continuous on [0; 1], but the limit function is not continuous on [0; 1], it has a jump discontinuity at the point x = 1. nx n=2 EXAMPLE 2. Consider, the sequence hf (x)i , where, f (x) = 1 − ; n ≥ 1 for n n n n + 1 all x 2 (−∞; 1]. We then have lim fn(x) = 0, for 0 < x < 1 and lim fn(x) = 1, for n−!1 n−!1 x < 0. On the other hand, lim fn(0) = 1. Thus, the sequence hfn(x)in converges pointwise on n−!1 E0 = [0; 1] to the limit function f defined by ( 0; if 0 ≤ x ≤ 1; f(x) = 1; if x = 0 Dr. Prasun Kumar Nayak Home Study Materiel 3 Small Overview On Uniform Convergence n −nx EXAMPLE 3. Consider, the sequence hfn(x)in, where, fn(x) = x e ; n ≥ 1 and x ≥ 0, as y y 6 6 y = n y = e−n y = fn(x) 2 n −nx − − 1 - y = fn(x) = x e n n 1 2 e O n n x e - ee y = −n O (a) 1 x q (b) n −nx Figure 2: (a) fn(x) = x e (b) fn(x). 0 n−1 −nx depicted in the Fig. 2(a). Now, fn(x) = nx e (1 − x) = 0, gives x = 1 and the maximum −n −n value of fn(x) on [0; 1) is e . Therefore jfn(x)j ≤ e , and so lim fn(x) = 0 for all x ≥ 0. n!1 The limit function in this case is identically zero on [0; 1). EXAMPLE 4. Consider, the sequence hfn(x)in, where, for n ≥, 8 0; x < − 2 > for n > 2 1 > −n(2 + nx); for − n ≤ x < − n < 2 1 1 fn(x) = n x; for − n ≤ x < n > n(2 − nx); for 1 ≤ x < 2 > n n :> 2 0; for x ≥ n defined on (−∞; 1), as depicted in the Fig. 2(b). Here lim fn(0) = 0, for all n and lim fn(x) = n!1 n!1 2 0 if n ≥ jxj . Therefore f(x) = lim fn(x) = 0; −∞ < x < 1; n!1 so, the limit function in this case is identically zero on (−∞; 1). EXAMPLE 5. (Uniform limits of differentiable functions need not be differentiable): Consider, y 6 the sequence hfn(x)in, where fn : r f (x) 1 1 ! defined by f (x) = x2 + , R R n n f2(x) n 2 N, as depicted in the Figure 3, for all x 2 R. Here we clearly have y = jxj lim fn(x) = jxj for all x 2 . Thus, n-−!1 R O thex sequence is pointwise convergent on R. We also observe that fn is dif- Figure 3: Uniform limits of differen- ferentiable on R for all n 2 N with tiable functions need not be differen- 0 x fn(x) = q . tiable 2 1 x + n On the other hand, the limit function f(x) = jxj is not differentiable at x = 0. sin(n2x) EXAMPLE 6. Consider, the sequence hf (x)i , where, f (x) = for all x 2 and n n n n R n 2 N. Dr. Prasun Kumar Nayak Home Study Materiel 4 Small Overview On Uniform Convergence as depicted in the Figure 6. Since j sin αj ≤ 1 for all α 2 y 6 sin x R, we have obvious estimate f1(x) = 1 1 jfn(x)j ≤ n for any x 2 R. Here the limit function, f; O π is the (identically) zero func- - −π x sin(n2x) tion. Indeed, ≤ n 1 holds for all x 2 and n R 1 n 2 N and lim = 0. n−!1 n Figure 4: Graph of fn(x) of Example 6 0 Therefore, f(x) = f (x) = 0 for all x 2 R. On the other hand, cos(n2x) f 0 (x) = n2 · = n cos(n2x) n n does not converge to 0. In fact, lim f 0 (0) = lim (n) = 1: n−!1 n n−!1 2 EXAMPLE 7. Consider, the sequence hfn(x)in, where, fn(x) = [cos (n!πx)], for all x 2 [0; 1] p where, for each t 2 R, [t] denotes the greatest integer ≤ t. If x = q with (relatively prime) positive integers p and q; then n!x is an integer for all n ≥ q and hence cos2(n!πx) = 1. On the 2 other hand, if x 2 Q, then cos (n!πx) 2 (0; 1). It follows that the (pointwise) limit function, f, is given by ( 0; if x 2 \ [0; 1]; f(x) = Q 1; if x 2 [0; 1]=Q In other words, f is the Dirichlet function which is nowhere continuous on [0; 1]. In particular, f is not Riemann integrable. On the other hand, each fn has only a finite (in fact n! + 1) number of discontinuity points and hence is Riemann integrable. 2 n EXAMPLE 8. Consider the functions fn(x) = nx(1 − x ) , for all x 2 [0; 1]. The (pointwise) limit, f; is the identically zero function: f(x) = 0; 8x 2 [0; 1]. This is obvious for x = 0 and x = 1, and for x 2 (0; 1) it follows from the fact that lim nαn = 0, for all α 2 (0; 1). Now n−!1 Z 1 nhx(1 − x2)n+1 i1 n fn(x)dx = − = 0 2 n + 1 0 2(n + 1) Z 1 1 R 1 It follows that lim fn(x)dx = ; and yet 0 f(x)dx = 0.