Uniform Convergence

Home , Divergent series, Pointwise convergence, 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 investigate. 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 ∈ F(E; R)N and each x ∈ E, the numerical sequence hfn(x)i ∈ RN may or may not converge. Let x fn : [0, 1] → R be a function, given by fn(x) = n ; x ∈ [0, 1], then hfni is a sequence of functions on [0, 1].

(ii) Let E ⊂ R and let huni ∈ F(E; R)N. Then the formal sum ∞ X u1 + u2 + ··· + un + ··· = un n=1

is called an inﬁnite series of functions with general term un. For each x ∈ E, we have a numerical series ∞ X u1(x) + u2(x) + ··· + un(x) + ··· = un(x) n=1

For each n ∈ 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 ∈ F(E; R)N.

1 2 Small Overview On Uniform Convergence

2 Pointwise Convergence

Deﬁnition 1. [Pointwise Convergent Sequence of functions ]: For each hfni ∈ F(E; R)N; let E0 ⊂ E be the set of all points x ∈ E such that the numerical sequence hfn(x)i ∈ RN converges and let

f(x) = lim fn(x); ∀x ∈ E0 (1) n−→∞ which, in detail, means that, corresponding to an ε > 0, ∃N = N(x; ε) ∈ N, depends on both x and ε, such that

|fn(x) − f(x)| < ε; whenever n ≥ N (2)

The sequence hfni is then said to be pointwise convergent (or simply convergent) on E0 and the function f ∈ F(E0; R); deﬁned by (1) is called the pointwise limit (or simply limit) of hfni on E0.

For example, let X = {1, 2, 3} 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 ∈ 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 ∈ [0; 1] as y 6 depicted in the Fig. 1. For x ∈ [0, 1). 1 We then have lim fn(x) = 0. On n−→∞ the other hand, lim fn(1) = 1. The n−→∞ 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 ∈ (−∞, 1]. We then have lim fn(x) = 0, for 0 < x < 1 and lim fn(x) = ∞, for n−→∞ n−→∞ x < 0. On the other hand, lim fn(0) = 1. Thus, the sequence hfn(x)in converges pointwise on n−→∞ 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, ∞) is e . Therefore |fn(x)| ≤ e , and so lim fn(x) = 0 for all x ≥ 0. n→∞ The limit function in this case is identically zero on [0, ∞).

EXAMPLE 4. Consider, the sequence hfn(x)in, where, for n ≥,

 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 (−∞, ∞), as depicted in the Fig. 2(b). Here lim fn(0) = 0, for all n and lim fn(x) = n→∞ n→∞ 2 0 if n ≥ |x| . Therefore

f(x) = lim fn(x) = 0; −∞ < x < ∞, n→∞ so, the limit function in this case is identically zero on (−∞, ∞).

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 ∈ N, as depicted in the Figure 3, for all x ∈ R. Here we clearly have y = |x| lim fn(x) = |x| for all x ∈ . Thus, n-−→∞ 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 ∈ 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) = |x| is not differentiable at x = 0. sin(n2x) EXAMPLE 6. Consider, the sequence hf (x)i , where, f (x) = for all x ∈ and n n n n R n ∈ N.

Dr. Prasun Kumar Nayak Home Study Materiel 4 Small Overview On Uniform Convergence

as depicted in the Figure 6. Since | sin α| ≤ 1 for all α ∈ y 6 sin x R, we have obvious estimate f1(x) = 1 1 |fn(x)| ≤ n for any x ∈ 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 ∈ and n R 1 n ∈ N and lim = 0. n−→∞ n Figure 4: Graph of fn(x) of Example 6

0 Therefore, f(x) = f (x) = 0 for all x ∈ 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) = ∞. n−→∞ n n−→∞ 2 EXAMPLE 7. Consider, the sequence hfn(x)in, where, fn(x) = [cos (n!πx)], for all x ∈ [0, 1] p where, for each t ∈ 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 ∈ Q, then cos (n!πx) ∈ (0, 1). It follows that the (pointwise) limit function, f, is given by ( 0; if x ∈ ∩ [0, 1], f(x) = Q 1; if x ∈ [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 ∈ [0, 1]. The (pointwise) limit, f; is the identically zero function: f(x) = 0; ∀x ∈ [0, 1]. This is obvious for x = 0 and x = 1, and for x ∈ (0, 1) it follows from the fact that lim nαn = 0, for all α ∈ (0, 1). Now n−→∞ 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. n−→∞ 0 2 EXAMPLE 9. For each positive integer n ∈ N, let En be the set of numbers of the form x = p q , where p and q are integers with no common factors and 1 ≤ q ≤ n. Define fn(x) = ( 1; x ∈ En . If x is irrational, then x 6∈ En, for any n, so fn(x) = 0, n ≥ 1. If x is 0; x 6∈ En rational, then x ∈ En and fn(x) = 1 for all sufficiently large n. Therefore ( 1; x ∈ Q f(x) = lim fn(x) = c n→∞ 0; x 6∈ Q = R/Q

Dr. Prasun Kumar Nayak Home Study Materiel 5 Small Overview On Uniform Convergence

EXAMPLE 10. Let ql, q2,... be an enumeration of the rationals Q∩[0, 1] in the interval I = [0, 1]. Consider the functions fn[0, 1] → R deﬁned by: ( 1; if x ∈ {q , q , . . . , q } f (x) = l 2 n n 0; otherwise

Then the functions f3 converge pointwise to the f which is equal to 1 on the rationals and 0 on the irrationals. Each fn is integrable because it is discontinuous at only a ﬁnite number of points. But ( 1; if x ∈ {q : k ∈ } the pointwise limit is the Dirichlet function f(x) = k N This function is 0; otherwise not integrable on [0, 1]. x2n EXAMPLE 11. Let hf i be deﬁned by f (x) = , x ∈ and n ∈ . Therefore the limit n n 1 + x2n R N function is given by  2n  0; |x| < 1 x  1 f(x) = lim fn(x) = lim = ; |x| = 1 n→∞ n→∞ 1 + x2n 2  1; |x| > 1

Note that each fn(x) is continuous on R but f is not continuous at ±1. ∞ P Deﬁnition 2. [Pointwise Convergent Series of Functions] The series un(x) is said to be point- n=1 wise convergent (or simply convergent) on E0 ⊂ E with sum s ∈ F(E0; R)N if the sequence hsni ∈ F(E; R)N of partial sums converges (pointwise) to the function s on E0. In other words, if

s(x) = lim sn(x); ∀x ∈ E0 n→∞

∞ P n EXAMPLE 12. Consider, the series un(x), where, for each n ∈ N, un(x) = x for all x ∈ n=1 ∞ P (−1, 1) and u0 = 1. Then the series un(x) is (pointwise) convergent on (−1, 1) with sum n=1

∞ ∞ X X 1 s(x) = u (x) = xn = n 1 − x n=1 n=1 x x EXAMPLE 13. Consider the series + + ··· ; x ≥ 0. Here x + 1 (x + 1)(2x + 1) 1 1 1 u (x) = − ; S (x) = 1 − n (n − 1)x + 1 nx + 1 n nx + 1

Thus, when x > 0, lim Sn(x) = 1 and when x = 0, lim Sn(x) = 0 as Sn(0) = 0. Let n→∞ n→∞ 1 1 y = Sn(x), then (y − 1)(x + n ) = − n . The curve y = s(x), when x ≥ 0, As Sn(x) is certainly continuous, when the terms of the series are continuous, the approximation curves will always differ very materially from the curve y = s(x), when the sum of the series is discontinuous.

Dr. Prasun Kumar Nayak Home Study Materiel 6 Small Overview On Uniform Convergence

consists of the part of the line y = 1 for which x > 0 and the origin. As 6Sn(x) n increases, this rectangular hyperbola (Fig. 5) approaches more and more y = 1 closely to the lines y = 1, x = 0. If we reasoned from the shape of the ap- - proximate curves, we should expect to O x ﬁnd that part of the axis of y for which 0 < y < 1 appearing as a portion of the curve y = s(x) when x ≥ 0. Figure 5: Example 13. nx In this case, S (x) = and n 1 + n2x2 Sn(x6) lim Sn(x) = 0 for all values of x. n→∞ 0.5 The sum function s(x) of the series is continuous for all values of x, but we n = 5 shall see that the approximation curves n = 10 differ very materilly from the curve y = s(x) in the neighbourhood of the -x origin. The curve y = Sn(x) has a O maximum Figure 6: Example 14.

∞ X nx (n − 1)x EXAMPLE 14. Consider the series u (x), where, u (x) = − . n n 1 + n2x2 1 + (n − 1)2x2 n=1 1 1 1 1 at ( n , 2 ) and the minimum at (− n , − 2 ) as depicted in the Fig. 6. The points on the axis of x just below the maximum and minimum move in towards the origin as n increases. And if we reasoned from the shape of the curves y = Sn(x), we should expect to ﬁnd the part of the axis of 1 1 y from − 2 to 2 appearing as a portion of the curve y = s(x). ∞ X n EXAMPLE 15. Find the sum function of (cos x) on (0, π). n=1 ∞ P Solution: Let hSn(x)i be the sequence of partial sums of the series un(x), where, un(x) = n=1 (cos x)n. Thus cos x S (x) = cos x + cos2 x + ··· + cosn x = 1 − cosn x n 1 − cos x cos x n cos x lim Sn(x) = lim 1 − cos x = ; as | cos x| < 1 for 0 < x < π ∴ n→∞ n→∞ 1 − cos x 1 − cos x

cos x Therefore, the sum function is given by S(x) = lim Sn(x) = ; for 0 < x < π. n→∞ 1−cos x

3 Uniform Convergence

The pointwise limit of a sequence of functions may differ radically from the functions in the sequence.

Dr. Prasun Kumar Nayak Home Study Materiel 7 Small Overview On Uniform Convergence

(i) In Examples 1 and 2, each fn is continuous on (−∞, 1], but the limit function f is not continuous.

(ii) In Example 4, the graph of each fn has two triangular spikes with heights that tend to ∞ as n → ∞, while the graph of f(x) (the x-axis) has none.

(iii) In Examples 5 and 6, each fn is differentiable at x0, while the limit function f is not differ- 0 0 0 entiable at x0 or even if f (x0) exists, the hfn(x0)i exists need not converge to f (x0).

(iv) In Examples 8 and 9, each fn is integrable, while the limit function f is non integrable in any compact interval.

There is nothing in Deﬁnitions 1 and 2 to preclude these apparent anomalies.

Deﬁnition 3. [Uniformly Convergent Sequence of functions ]: Let E ⊂ R. We say that a sequence hfni ∈ F(E; R)N converges uniformly on E0 ⊂ E to a function f : E0 → R if, corresponding to an ε > 0, ∃N = N(ε) ∈ N, depends on ε only, such that

|fn(x) − f(x)| < ε; whenever n ≥ N and ∀x ∈ E0 (3)

We interpret the uniform convergence in a geometric way. Draw the graphs of fn and f.

Put a band of width ε around y 6 the graph of f. Condition (3) f(x) + ε states that if ε is any posi- V tive number, then for n > N ] the graph of y = fn(x) lies entirely below the graph of f(x) + ε and entirely above f(x) fn(x) the graph of f(x) − ε as de- f(x) − ε - a b picted on the Fig. 7. Thus O x draw a tube V of vertical ra- Figure 7: Geometrical signiﬁcance of uniform dius ε around the graph f. convergence.

For n large, the graph of fn is contained in the ε-tube V around the graph of f. Notice that the special feature of uniform convergence is that the rate at which fn(x) converges is independent of x ∈ E. For example,

EXAMPLE 16. Consider, the sequence hfn(x)in as in Example 1). The sequence hfn(x)in q 1 n 1 1 Indeed, if for ε = 10 , the point xn = 2 is send by fn to 2 and thus not all points x satisfying Eq. nm (3), when n is large. The graph of fn, as depicted in the ﬁgure 8, fails to lie in the ε-tube V . Here, fn(x) is converging very rapidly to zero for x near zero but arbitrarily slowly to zero for x near 1.

EXAMPLE 17. Prove that for the sequence hfni, fn → f pointwise on a ﬁnite set E(⊂ R) the convergence is uniform.

Dr. Prasun Kumar Nayak Home Study Materiel 8 Small Overview On Uniform Convergence

of functions, where, f : (0, 1) → is y 6 (1, 1) n R given by f (x) = xn for all x ∈ [0, 1], r n as depicted in the Figure 1. For each f1 f2 x = 0 x ∈ (0, 1) it is clear that fn(x) → 0. x = 1 The (pointwise) limit function, f, is - ( O 0; if 0 ≤ x < 1, V x f(x) = 1; if x = 1 Figure 8: Non uniform, pointwise con- Here the convergence is not uniform. vergence.

Solution: Let E = {x1, x2, ··· , xn} be a ﬁnite set in R. Since the sequence hfni converges pointwise to f, so ∀ε > 0, ∃Nk = Nk(xk; ε) such that

|fn(xk) − f(xk)| < ε; ∀n ≥ Nk; k = 1(1)n.

Let for a pre-assigned ε > 0, max Nk(xk; ε) = N(ε). Therefore, xk∈E

|fn(x) − f(x)| < ε; ∀n ≥ N(ε) and ∀x ∈ E

As N(ε) depends on ε and not on x, hfni converges uniformly to f on E. x EXAMPLE 18. Let f (x) = for x ∈ as depicted in the Fig.. The sequence hf (0)i is a n n R n a constant sequence h0i. Hence the limit function is f = 0. More generally, if a ∈ R, we get h n i as the pointwise sequence.

n where, fn(x) = x , for all x ∈ [0; 1] as depicted in the Fig. 9. For x ∈ [0, 1). y 6 We then have lim fn(x) = 0. On the 1 n−→∞ other hand, lim fn(1) = 1. The se- n−→∞ quence is therefore point wise conver- n = 1 n = 2 gent on [0, 1]. We note, however, that n = 5 the limit function - x ( O ! 0; if 0 ≤ x < 1, f(x) = Figure 9: fn(x) for n = 1, 2, 3, 4, 5. 1; if x = 1

ln(1 + n2x2) EXAMPLE 19. Show that hf 0 (x)i is uniformly convergent on [0, 1], where f (x) = n n n for x ∈ [0, 1].

Solution: Consider the sequence of functions hfni , deﬁned by fn : [0, 1] → R, where, fn(x) = log(1 + n2x2) ; x ∈ [−k, k], k > 0, as in the Fig. 10(a). Then hf i converges uniformly to n n 2nx f(x) = 0 on [0, 1]. Now, f 0 (x) = , for x ∈ [0, 1], as depicted in the Fig. 10(b). Clearly, n 1 + n2x2 fn(0) → 0, as fn(0) = 0 for all n ≥ 1. Let x 6= 0. For a given ε > 0, we see that 2nx 2 |f 0 (x) − 0| = ≤ < ε n 1 + n2x2 n|x|

Dr. Prasun Kumar Nayak Home Study Materiel 9 Small Overview On Uniform Convergence

y 6 6 0 f2 0 f2 1 f1

f1 f 0 j 3 f50 -x x- O 1 O 1 (a) (b) 0 Figure 10: Figures of fn and fn

2 2 whenever n > . Let us take N = [ ] + 1 ∈ , then ε|x| ε|x| N

0 |fn(x) − 0| < ε for all n ≥ N

0 0 Therefore, hfn(x)i converges everywhere to the function f (x) = 0 (Fig. 10(b)). Observe that N depends on both x and ε. Let

0 0 2x Mn = sup |fn(x) − f (x)| = sup 2 2 = sup g(x). x∈[0,1] x∈[0,1] 1 + n x x∈[0,1]

0 2(1−n2x2) 0 1 Then g (x) = (1+n2x2)2 , and g (x) = 0 implies x = n ∈ [0, 1]. Thus

0 0 2x 0 1 1 Mn = sup |fn(x) − f (x)| = sup 2 2 = fn( ) = . x∈[0,1] x∈[0,1] 1 + n x n n

0 Hence, Mn → 0 as n → ∞, and consequently hfni is uniformly convergent on [0, 1]. ∞ P Deﬁnition 4. [Uniformly Convergent Series of Functions ]: The series un(x) is said to be n=1 uniformly convergent on E0 if the sequence hsni of partial sums is uniformly convergent on E0, i.e., corresponding to a > 0, ∃ N() ∈ N such that

sn(x) − s(x) < , whenever n ≥ N and ∀x ∈ E0 (4)

∞ th P where sn(x) is the n partial sum of the series un(x). n=1

THEOREM 1. If hfni converges uniformly to f on E, the hfni converges pointwise to f on E. The converse is not always true, i.e., pointwise convergence does not imply uniform convergence .

Proof:

RESULT 1. John Kelley refers to the growing steeple: Consider, the sequence hfn(x)in, So there is no question of uniform convergence. Even if the function have compact domain of deﬁnition, and are uniformly bounded and uniformly continuous, pointwise convergence does not imply uniform convergence.

Dr. Prasun Kumar Nayak Home Study Materiel 10 Small Overview On Uniform Convergence

where, fn : [0, 1] → R defined as, for n ≥ 1,  1 y 6 2 f4  n x; for 0 ≤ x ≤  n  2 1 2 fn(x) = 2n − n x; for ≤ x ≤ n n   2  0; for ≤ x ≤ 1 n defined on (−∞, ∞), as depicted in the Fig. 11. f1 Here lim fn(x) = 0, for all x and fn converges n→∞ - pointwise to the function f = 0. The graph of O V x f , as depicted in the figure 11, fails to lie in the n Figure 11: Graphical representation of ε-tube V . fn(x)

3.1 Test for Uniform Convergence

THEOREM 2 (Weiestrass Mn test). A sequence hfni ∈ F(E; R)N, where E ⊂ R; converges uniformly on E0 ⊂ E if and only if, corresponding to an ε > 0, ∃N = N(ε) ∈ N, depends on ε only, such that n o Mn = sup |fn(x) − f(x)| : x ∈ E0 < ε; whenever n ≥ N

Proof: Necessary part : Let hfn(x)in converges to f(x) uniformly on E0. Then corresponding to an > 0, ∃N ∈ N such that

fn(x) − f(x) < , for n ≥ N, ∀x ∈ E0 n o ⇒ Mn = sup |fn(x) − f(x)| : x ∈ E0 < ε, for n ≥ N

⇒ Mn → 0 as n → ∞.

Sufﬁcient part : Let Mn → 0 as n → ∞, so corresponding to an > 0, ∃N1 ∈ N such that

Mn − 0 < , for n ≥ N1 ⇒ Mn < , for n ≥ N1 n o ⇒ sup |fn(x) − f(x)| : x ∈ E0 < ε, for n ≥ N1

⇒ fn(x) − f(x) < for n ≥ N1; ∀x ∈ E0

Therefore, fn(x) → f(x) uniformly on E0. 2 x EXAMPLE 20. Consider, the sequence hf (x)i , where, f (x) = ; x ∈ [a, b]. For any n n n 1 + nx2 x ∈ [a, b], x f(x) = lim fn(x) = lim = 0 n−→∞ n−→∞ 1 + nx2

Therefore, hfn(x)in converges pointwise to zero on [a, b]. Now, x x fn(x) − f(x) = − 0 = = g(x)(say). 1 + nx2 1 + nx2

Dr. Prasun Kumar Nayak Home Study Materiel 11 Small Overview On Uniform Convergence

For, x > 0, using A.M. ≥ G.M., we have

1 + nx r 1 x 1 x ≥ nx ⇒ ≤ √ 2 x 1 + nx2 2 n

√1 √1 Therefore, fn(x) = 2 n , when x = n ∈ [a, b]. Thus, n x o 1 M = sup − 0 : x ∈ [0, ∞) = √ → 0 as n → ∞ n 1 + nx2 2 n

Therefore hfn(x)i is uniformly convergent on [a, b].

n EXAMPLE 21. Consider, the sequence hfn(x)in, where, fn(x) = x for all x ∈ [0, 1]. The ( 0; if 0 ≤ x < 1, (pointwise) limit function, f, is f(x) = Now 1; if x = 1

n n o lim Mn = lim sup |x − f(x)| : x ∈ [0, 1] = lim 1 = 1 6= 0 n→∞ n→∞ n→∞ Therefore, the convergence is not uniform.

EXAMPLE 22. Consider, the sequence hfn(x)in, where fn : R → R r y 6 2 1 deﬁned by fn(x) = x + , n f1(x) n ∈ N, as depicted in the Figure 12, for all x ∈ R. Here we clearly f2(x) have lim fn(x) = |x| for allx = −1 x = 1 n−→∞ y = |x| x ∈ R. So, the sequence hfn(x)in - has pointwise limit f(x) = |x|, O x for all x ∈ R. Now Figure 12: Graph of fn as in example 22

n r 1 o n 1/n o 1/n 2 Mn = sup x + − |x| : x ∈ R = sup q : x ∈ R = √ n 2 1 1/ n x + n + |x| r n 2 1 o 1 lim Mn = lim sup x + − |x| : x ∈ R = lim √ = 0 ∴ n→∞ n→∞ n n→∞ n Thus the convergence is uniform. We draw an ε-band around the limit function. Indeed, the graph of fn, as depicted in the ﬁgure 12, lie in the ε-tube V . This also ensures geometrically that the convergence is uniform. sin(n2x) EXAMPLE 23. The sequence hf (x)i , where, f (x) = for all x ∈ and n ∈ has n n n n R N sin(n2x) 1 pointwise limit f(x) = 0, for all x ∈ . Since ≤ , for all x ∈ and n ∈ , so R n n R N n sin(n2x) o 1 lim Mn = lim sup − 0 : x ∈ R = lim = 0 n→∞ n→∞ n n→∞ n Thus the convergence is uniform.

Dr. Prasun Kumar Nayak Home Study Materiel 12 Small Overview On Uniform Convergence

x EXAMPLE 24. Consider, the sequence hf (x)i , where, f (x) = for all x ∈ [0, 1]. For a n n n nx + 1 given x ∈ [0, 1], x f(x) = lim fn(x) = lim = 0 n−→∞ n−→∞ nx + 1

Therefore hfn(x)i converges pointwise to zero on [0, 1]. Now, x x fn(x) − f(x) = − 0 = = g(x)(say). nx + 1 nx + 1 0 1 Then, g (x) = (nx+1)2 > 0, ∀x ∈ [0, 1], so, g(x) is strictly increasing function on [0, 1]. Thus g(x) attains its maximum value at x = 1. Therefore, x 1 Mn = sup fn(x) − f(x) = sup = x∈[0,1] x∈[0,1] nx + 1 n + 1

Now , Mn → 0 as n → ∞. Therefore hfn(x)i converges uniformly to 0 on [0, 1]. nx EXAMPLE 25. Consider, the sequence of functions hf (x)i , where, f (x) = for all n n n 1 + n2x2 y 6 1 n = 1 n = 2 n = 3 n = 4 n = 5

- ) 1 x

Figure 13: fn(x) for n = 1, 2, 3, 4, 5. x ∈ [a, b], containing 0. The graphs of fn for n = 1, 2, 3, 4 are shown in the Fig. 13. For any x ∈ [a, b] containing zero, nx f(x) = lim fn(x) = lim = 0 n−→∞ n−→∞ 1 + n2x2

Therefore, hfn(x)in is converges pointwise to zero on [a, b]. Now, nx nx fn(x) − f(x) = − 0 = = g(x)(say). 1 + n2x2 1 + n2x2 n(1 − n2x2) Then, g0(x) = . Thus g0(x) = 0 ⇒ x = ± 1 . Also (1 + n2x2)2 n h(1 + n2x2)2 · (−2n2x) + 2(1 − n2x2) · 2(1 + n2x2)i g00(x) = n (1 + n2x2)4 2 2 2 00 3 3 − n x n or, g (x) = −2n x · | 1 = − < 0 1 2 2 4 x= n x= n (1 + n x ) 4 Thus g(x) attains its maximum value at x = 1/n. Therefore, n nx o 1 M = sup − 0 : x ∈ [a, b] = n 1 + n2x2 2

Dr. Prasun Kumar Nayak Home Study Materiel 13 Small Overview On Uniform Convergence

Now , Mn 9 0 as n → ∞. Therefore hfn(x)i is not uniformly convergent on any interval containing 0. Also, The graph of fn, as depicted in the ﬁgure 13, fails to lie in the ε-tube V , about f = 0, so that non uniform convergence is veriﬁed.

−nx2 EXAMPLE 26. Consider, the sequence hfn(x)in, where, fn(x) = fn(x) = nxe , x ∈ [0, ∞). For any x ∈ [0, ∞), nx x f(x) = lim fn(x) = lim = lim = 0 n−→∞ n−→∞ enx2 n−→∞ x2enx2

Therefore, hfn(x)in is converges pointwise to zero on [0, ∞). Now, nx nx

fn(x) − f(x) = 2 − 0 = 2 = g(x)(say). enx enx n − 2n2x2 1 Then, g0(x) = . Thus g0(x) = 0 ⇒ n − 2n2x2 = 0 ⇒ x = ±√ . Also enx2 2n h (2nx2 − 3)i 00 2 g (x) = 2n x 2 < 0 x= √1 nx x= √1 2n e 2n

Thus g(x) attains its maximum value at x = √1 . Therefore, 2n

n nx o r n Mn = sup − 0 : x ∈ [0, ∞) = enx2 2e

Now , Mn 9 0 as n → ∞. Therefore hfn(x)i is not uniformly convergent on [0, ∞).

EXAMPLE 27. Thus for the sequence hfni, fn → f pointwise on a point set E ⊂ R, the convergence is uniform. √ 2 2 2 EXAMPLE 28. Verify that the sequence hfni, where fn(x) = n sin 4π n + x converges uniformly on [0, k], k > 0. Does hfni converge uniformly on R? √ Solution: The function sin 4π2n2 + x2 can be written as r p x2 sin 4π2n2 + x2 = sin 2nπ 1 + + 2nπ − 2nπ 4π2n2 r x2 = sin 2nπ 1 + − 1 4π2n2 p x2 = sin 4π2n2 + x2 − 2nπ = sin √ 4π2n2 + x2 + 2nπ Therefore, the limit function f is given by

x2 f(x) = lim fn(x) = lim n · sin √ n→∞ n→∞ 4π2n2 + x2 + 2nπ h x2 1 x2 3 i = lim n √ − √ + ··· n→∞ 4π2n2 + x2 + 2nπ 3! 4π2n2 + x2 + 2nπ h x2 1 x2 3 i x2 = lim q − q + ··· = n→∞ 2 x2 3!n 2 x2 4π 4π + n2 + 2π 4π + n2 + 2π

Dr. Prasun Kumar Nayak Home Study Materiel 14 Small Overview On Uniform Convergence

α3 Now, using the fact that sin α ≥ α − , we get for x ∈ [0, k], 3! x2 p x2 x2 − n sin 4π2n2 + x2 = − n sin √ 4π 4π 4π2n2 + x2 + 2nπ x2 x2 1 x2 3 ≤ − √ + √ 4π 4π2n2 + x2 + 2nπ 3! 4π2n2 + x2 + 2nπ k2 2 n k5 ≤ 1 − + 3 3 = Mn(say) 4π q k2 3! 8n π 1 + 1 + 4π2n2

Now, Mn → 0 as n → ∞, so the given sequence of functions is uniformly convergent on [0, k]. For x ∈ R, by the inequality | sin x| ≤ |x|, we obtain x2 p x2 x2 − n sin 4π2n2 + x2 = − n sin √ 4π 4π 4π2n2 + x2 + 2nπ x2 x2 x2 2

≥ − √ ≥ 1 − q , 4π 4π2n2 + x2 + 2nπ 4π x2 1 + 1 + 4π2n2 which shows that the convergence cannot be uniform on R. ∞ X THEOREM 3 (Weierstrass Mn test for the series of functions). The series un(x) deﬁned on I n=1 is uniformly convergent on I if ∃ a sequence of positive constants {Mn} such that

a) un(x) ≤ Mn, ∀n ΣMn is convergentb)

th P Proof: Let Sn is the n partial sum of the series un(x). Then

n X Sn(x) = u1(x) + u2(x) + ··· + un(x) = ur(x). r=1 P Let > 0 be chosen arbitrary,since Mn is convergent by Cauchy’s criteria, ∃N0() ∈ N such that

Pn − Pm < , for n > m > N0,

th P where, Pn = n partial sum of Mn. Therefore,

Mm+1 + Mm+2 + ··· + Mn < , for n > m > N0

Therefore, if n > m > N0 we obtain

n m n X X X Sn(x) − Sm(x) = ur(x) − ur(x) = ur(x) r=1 r=1 r=m+1 n n X X ≤ ur(x) ≤ Mr < , for n > m > N0, ∀x ∈ I r=m+1 r=m+1

Therefore ,Σun(x) converges uniformly on I.

Dr. Prasun Kumar Nayak Home Study Materiel 15 Small Overview On Uniform Convergence

∞ X sin nx EXAMPLE 29. Consider the series of functions on x ∈ (−∞, ∞). Here n2 n=1 sin nx 1 u (x) = ⇒ u (x) ≤ = M , ∀x ∈ (−∞, ∞) n n2 n n2 n X X 1 Now, M = is a hyperhermonic series with p = 2 > 1, so convergent. Therefore, by n n2 Weierstrass Mn test, the given series of functions is uniformly convergent on (−∞, ∞). ∞ X x EXAMPLE 30. Consider the series of functions ; x ∈ [a, b]. n(1 + nx2) n=1 x x 1 u (x) = ⇒ < n n(1 + nx2) n(1 + nx2) n2 P P 1 Now, Mn = n2 is a hyper hermonic series with p = 2 > 1,so convergent. Then by Mn test ∞ X x the given series of functions is uniformly convergent on [a, b]. n(1 + nx2) n=1 ∞ X sin nx EXAMPLE 31. Consider the series of functions , p > 1, x ∈ . Now np R n=1 sin nx 1 ≤ ; ∀x ∈ np np R ∞ X 1 The series is a Hyper-harmonic p-series with p > 1 and hence convergent. Therefore, by np n=1 Weierstrass M-test the given series of functions is uniformly convergent for all x ∈ R. ∞ X sin(nx2 + x2) EXAMPLE 32. Consider the series of functions , x ∈ . Now n(n + 1) R n=1 sin(nx2 + x2) 1 1 ≤ < ; ∀x ∈ n(n + 1) n(n + 1) n2 R ∞ X 1 The series is a Hyper-harmonic p-series with p = 2 > 1 and hence convergent. Therefore, n2 n=1 by Weierstrass M-test the given series of functions is uniformly convergent for all x ∈ R. x EXAMPLE 33. Test the covergence for Σ on any ﬁnite interval [a, b]. np + x2nq Solution: First let p > 1, q ≥ 0. Let α ≥ max{|a|, |b|}, then x α u (x) = ≤ = M (say) n np + x2nq np n X X α Now, M = is a hyper harmonic series with p > 2, so convergent. By M test n np n X x is uniformly convergent on [a, b]. np + x2nq Case -II: 0 < p ≤ 1 p + q ≥ 2 u (x) 1 Let and . Then n attains its maximum value 2n(p+q)/2 at the 2 q p P point where x n = n . Now , Mn is a hyper hermonic series converges as p + q ≥ 2. Hence by Mn test the given series uniformly convergent.

Dr. Prasun Kumar Nayak Home Study Materiel 16 Small Overview On Uniform Convergence

X −n n EXAMPLE 34. Prove or disprove: 2 cos(3 x) represents an everywhere continuous func- n tion. P −n n P −n n Solution: Let s(x) = n 2 cos(3 x) = n un(x), where un(x) = 2 cos(3 x); n ∈ N and x ∈ R. Thus 1 |u (x)| = |2−n cos(3nx)| ≤ = M , say, ∀n ∈ n 2n n N P P 1 P Now Mn = 2n = 1. So the series Mn is convergent. Hence, by Weierstrass M-test, the −n n given series is uniformly convergent on R. Again, un(x) = 2 cos(3 x) is continuous ∀x ∈ R and ∀n ∈ N. So the sum function f is everywhere continuous on R. P EXAMPLE 35. Consider the series un(x) for which the sum to ﬁrst n-terms is Sn(x) = ln(1 + n4x2) ; 0 ≤ x ≤ 1. Here 2n2

S(x) = lim Sn(x) = 0; 0 ≤ x ≤ 1 n→∞ 0 ∴ S (x) = 0; ∀x ∈ [0, 1] Again, P 0 Thus we see that un(x) does not converge uniformly on [0, 1] but the series may be differentiated term-by-term.  0; if x ≤ 0   1 nx2; if 0 < x ≤ EXAMPLE 36. Deﬁne a function f(x) = 2n The sequence of func-  1 1  x − ; if < x < ∞  4n 2n ( 0; if x ≤ 0 tions hfni converge uniformly on the entire real line R to the function f, where, f(x) = x; if x > 0 Notice that each of the functions fn is continuously differentiable on the entire real line, but f is not differentiable at 0.

THEOREM 4 (Cauchy’s criteria for Uniform Convergence). Let E ⊂ R and hfni ∈ F(E; R)N. Then hfnin converges uniformly on E0 ⊂ E if and only if, corresponding to an ε > 0, ∃N = N(ε) ∈ N, depends on ε only, such that

fm(x) − fn(x) < ε; whenever m, n ≥ N and ∀x ∈ E0

Proof: Necessary part : Let hfn(x)in converges uniformly on E0 to a limit function f(x). Then corresponding to an > 0, ∃ positive integers N1,N2 ∈ N independent of x such that f (x) − f(x) < , for n ≥ N and ∀x ∈ E n 2 1 0 f (x) − f(x) < , for m ≥ N and ∀x ∈ E m 2 2 0 Let N = max{N1,N2} ∈ N, then ∀x ∈ E0 we have

fm(x) − fn(x) = fm(x) − f(x) + f(x) − fn(x)

≤ fm(x) − f(x) + fn(x) − f(x) < + = , for n ≥ N. 2 2

Dr. Prasun Kumar Nayak Home Study Materiel 17 Small Overview On Uniform Convergence

Therefore, the condition holds. Sufﬁcient part : Conversely, if the condition of the theorem is satisﬁed, then, for each x ∈ E0, the numerical sequence hfn(x)in is a Cauchy sequence in R and hence converges. Let

f(x) = lim fn(x); ∀x ∈ E0. n→∞

For a given > 0, we can ﬁnd N ∈ N such that

fm(x) − fn(x) < ε; whenever m, n ≥ N and ∀x ∈ E0 For ﬁxed n, let m → ∞ in the above equation, we ﬁnd that

f(x) − fn(x) < ε; whenever n ≥ N and ∀x ∈ E0

Since ε > 0 was arbitrary, it follows that, sequence of functions hfn(x)in converges to f uniformly on E0, as desired.

Deduction 3.1. A sequence hfni ∈ F(E; R)N, where E ⊂ R; converges uniformly on E0 ⊂ E if n o and only if sup |fm(x) − fn(x)| : x ∈ E0 → 0 as m, n → ∞.

EXAMPLE 37. Determine whether the sequence hfni of functions converges uniformly on E: x2 √ a) f (x) = ; E = [0, 1] b) f (x) = n + 1 sinn x cos x; E = n x2 + (nx − 1)2 n R

pn n n c) fn(x) = 2 + |x| ; E = R Solution: a) The limit function f is given by

x2 f(x) = lim fn(x) = lim = 0. n→∞ n→∞ x2 + (nx − 1)2

0 2x(nx−1)3 0 1 Now, fn(x) = [x2+(nx−1)2]2 , so, fn(x) = 0 gives x = n . Thus

x2 1 M = sup − 0 = f = 1. n 2 2 n x∈[0,1] x + (nx − 1) n

As Mn 6→ 0 as n → ∞, hence by Weierstrass Mn-test the given sequence of functions is not uniformly convergent on E. b) The limit function f is given by √ n f(x) = lim fn(x) = cos x lim n + 1 sin x = 0. n→∞ n→∞

Here, we use Weierstrass Mn-test. For that √ √ n n Mn = sup n + 1 sin x cos x = n + 1 sup sin x cos x x∈R x∈R

Dr. Prasun Kumar Nayak Home Study Materiel 18 Small Overview On Uniform Convergence

Now we calculate supreme value of g(x) = sinn x cos x, Now, log g(x) = n log(sin x) + log(cos x), thus the supreme value of g is given by the equation tan2 x = n. Thus, √ √ n n/2 1 n Mn = n + 1 sup sin x cos x = n + 1 · · √ x∈R n + 1 n + 1 n n/2 1 = → √ 6→ 0 as n → ∞ n + 1 e

Hence by Weierstrass Mn-test the given sequence of functions is not uniformly convergent on E. c) The limit function f is given by ( r n pn n n n |x| 2; |x| ≤ 2 f(x) = lim fn(x) = lim 2 + |x| = lim 2 1 + = n→∞ n→∞ n→∞ 2 |x|; |x| > 2

We see that, all fn’s as well as the limit function is continuous, which implies that convergence is uniform.

EXAMPLE 38. Determine whether the sequence hfni of functions converges uniformly on E: 1 a) f (x) = ; E = [0, 1] b) f (x) = nxn(1 − x); E = [0, 1] n 1 + (nx − 1)2 n 2x c) f (x) = tan−1 ; E = n x2 + n3 R Solution: a) The limit function f is given by

( 1 1 2 ; x = 0 f(x) = lim fn(x) = lim = n→∞ n→∞ 1 + (nx − 1)2 0; x ∈ (0, 1]

We see that, all fn’s are continuous on E, while the limit function is not continuous, which implies that convergence is not uniform. b) Since xn(1 − x) → 0 as n → ∞, so, the limit function f is given by n f(x) = lim fn(x) = lim nx (1 − x) = 0. n→∞ n→∞ 0 n−1 0 n Now, fn(x) = nx [n − (n + 1)x], so, fn(x) = 0 gives x = n+1 . Thus n n n n Mn = sup |nx (1 − x) − 0| = fn = n+1 x∈[0,1] n + 1 (n + 1) 1 1 1 ∴ lim Mn = lim · n+1 = · 0 = 0 n→∞ n→∞ n 1 e 1 + n

Hence by Weierstrass Mn-test the given sequence of functions is uniformly convergent on E. c) The limit function f is given by

−1 2x f(x) = lim fn(x) = lim tan = 0 n→∞ n→∞ x2 + n3 0 2n3−2x2 0 √ Now, fn(x) = 4x2+(x2+n3)2 , so, fn(x) = 0 gives x = n n. Thus 2x √ 1 M = sup tan−1 − 0 = f n n = tan−1 √ n 2 3 n x∈R x + n n n

Therefore, Mn → 0 as n → ∞, so by Weierstrass Mn-test the given sequence of functions is uniformly convergent on E to 0.

Dr. Prasun Kumar Nayak Home Study Materiel 19 Small Overview On Uniform Convergence

∞ P EXAMPLE 39. Determine whether the series un(x) of functions converges uniformly on E: n=1 π 1 a) u (x) = −tan−1 n2(1+x2) ; E = b) u (x) = 2n sin ; E = (0, ∞) n 2 R n 3nx x2 c) un(x) = ln 1+ ; E = (−k, k), k > n · ln2 n 0

−1 −1 −1 −1 1 π Solution: a) We know, tan x + cot x = tan x + tan x = 2 . Using that identity, we get π 1 u (x) = − tan−1 n2(1 + x2) = tan−1 n 2 n2(1 + x2) 1 1 < ≤ = M ( say); ∀x ∈ . n2(1 + x2) n2 n R

∞ ∞ P P 1 Now, Mn = n2 is a hyperharmonic series with p = 2(> 1), so convergent. Therefore, by n=1 n=1 Weierstrass Mn-test the given series of functions is uniformly convergent on R. th b) For the problem, we use Cauchy criterion for uniform convergence. Let Sn(x) be the n partial sum of the series, then it is given by 1 1 1 S (x) = 2 sin + 22 sin + ··· + 2n sin n 3x 32x 3nx 1 π If 0 < 3nx ≤ 2 , then 1 1 |S (x) − S (x)| = 2n+1 sin + ··· + 2m+n sin n+m n 3n+1x 3m+nx 2 1 2 1 ≥ 2n+1 + ··· + 2m+n π 3n+1x π 3m+nx 2 1 ≥ 2n+1 π 3n+1x 1 Putting, x = 3n , we obtain 1 1 2n+2 23 S − S ≥ ≥ m+n 3n n 3n 3π 3π Thus by Cauchy criteria, the series of functions does not converge uniformly on E. c) Here, we use the Weierstrass Mn-test. Now x2 x2 x4 un(x) = ln 1 + = 1 + + + ··· n · ln2 n n · ln2 n n2 · ln4 n x2 k2 ≤ < = Mn n · ln2 n n · ln2 n

∞ ∞ 2 By Cauchy’s condensation test the series P M = P k is convergent. Therefore, the given n n·ln2 n n=1 n=1 series of functions is uniformly convergent on E. ∞ P EXAMPLE 40. Determine whether the series un(x) of functions converges uniformly on E: n=1

Dr. Prasun Kumar Nayak Home Study Materiel 20 Small Overview On Uniform Convergence

xn sin(nx) a) u (x) = ; E = b) u (x) = √ ; E = [0, 2π] n n! R n n cos2(nx) c) u (x) = ; E = n n2 R th Solution: a) Let Sn(x) be the n partial sum of the series. Then, for any n ≥ 1, we have nn sup |Sn(x) − Sn−1(x)| = sup |fn(x)| ≥ |fn(n)| = ≥ 1 x∈R x∈R n! Thus by Cauchy criteria, the series of functions does not converge uniformly on E. 2x b) First note that we do have pointwise convergence. Next notice that ≤ sin x for any π π √ π x ∈ [0, 4 ]. Let n ≥ 10 and h ∈ N such that 2n ≤ n + h < n n 4 . Thus for any k ∈ N with k π n ≤ k < n + h, we have √ < . Hence n n 4

n+h n+h k 2 k X 1 k X 1 2 k sin √ ≥ √ ⇒ √ sin √ ≥ √ √ n n π n n n n π n n k=n k k=n k n+h n+h √ n+h √ X 1 k X 2 k X 2 n 2 or, √ sin √ ≥ √ ≥ √ ≥ n n π n n π n n π k=n k k=n k=n

n+h X 1 k 2 This obviously show that sup √ sin √ ≥ for any n ≥ 0 and h ∈ such that n n π N x∈[0,2π] k=n k √ π 2n ≤ n + h < n n 4 . Therefore, the convergence will not be uniform on [0, 2π]. c) Here, we use the Weierstrass Mn-test. Now cos2(nx) 1 sup f (x) − 0 = sup ≤ = M ( say); ∀x ∈ . n 2 2 n R x∈R x∈R n n ∞ ∞ P P 1 Now, Mn = n2 is a hyperharmonic series with p = 2(> 1), so convergent. Therefore, by n=1 n=1 Weierstrass Mn-test the given series of functions is uniformly convergent on R.

THEOREM 5. [Dini’s Theorem of uniform convergence of a sequence ] Let I ⊂ R be a compact interval and suppose that hfni ∈ F(E; R)N is a sequence of continuous functions converging pointwise to a continuous function f : I → R. If hfni is increasing (i.e., fn(x) ≤ fn+1(x) for all x ∈ I and n ∈ N) or decreasing (i.e., fn(x) ≥ fn+1(x) for all x ∈ I and n ∈ N), then hfni converges to f uniformly on I.

Proof: The uniform convergence of hfni to f is equivalent to the uniform convergence of (f −fn) (or fn − f ) to 0. Let gn = f − fn (resp. gn = fn − f) if hfni is increasing (resp., decreasing). Then hgni is a decreasing sequence of continuous nonnegative functions converging pointwise to 0 on I. The theorem is proved if we show that this convergence is in fact uniform on I. Let ε > 0 ε be given. For each x ∈ I lim g (x) = 0 implies that we can pick N(x) ∈ with g (x) < . n N N(x) 2 Since gN(x)(x) is continuous at x, there is a δ(x) > 0 such that

gN(x)(t) < ε; ∀t ∈ (x − δ(x), x + δ(x)) (5)

Dr. Prasun Kumar Nayak Home Study Materiel 21 Small Overview On Uniform Convergence

Since I is compact, we can cover I by a ﬁnite number of intervals Ir = (xr − δ(xr), xr + δ(x )); 1 ≤ r ≤ k. Let N = max{N(x ),N(x ), ··· ,N(x )}. Now, for any t ∈ I, we have r 1 ε 2 k t ∈ I for some r and, by Eq. (5) g (t) < . But since N ≥ N(x ) and hg i is decreasing, r N(xr) 2 r n we have

0 ≤ gN ≤ gN(xr)(t) < ; ∀t ∈ I

Therefore, we indeed have

gn(x) ≤ gN (x) < ε; for n ≥ N and ∀x ∈ I and the proof is complete. 2

n−1 EXAMPLE 41. (i) Consider the sequence hfni, where, fn(x) = x (1 − x); x ∈ [0, 1]. Now

n−1 lim fn(x) = lim x (1 − x) = 0; for x ∈ [0, 1] n→∞ n→∞

The sequence hfni converges on [0, 1] to the function f(x) = 0 for x ∈ [0, 1]. Each fn is continuous on [0, 1], also f(x) is continuous on [0, 1]. Now, for each each x ∈ [0, 1]

n n+1 n−1 n fn+1(x) − fn(x) = (x − x ) − (x − x ) = −xn−1(x − 1)2 ≤ 0

i.e., hfni is monotone non-increasing for each x ∈ [0, 1]. By Dini’s Theorem 5, the convergence of the sequence is uniform on [0, 1]. √ p (ii) Consider the sequence hfni, where, f1(x) = x, fn(x) = xfn−1(x); for n ≥ 2, x ∈ √ 1 1 1 1 1 1/2 + 2 + 2 +···+ n [0, 1]. Therefore, f2(x) = x · x = x 2 2 , ··· , fn(x) = x 2 2 2 . 1− 1 At x = 0, lim fn(x) = lim x 2n = x, i.e., hfni converges to f on [0, 1], where n→∞ n→∞ f(x) = x, x ∈ [0, 1]. Each f(x) converges to f on [0, 1]. Each f(x) is continuous on [0, 1] and the limit function is also continuous on [0, 1]. Also 1 1 1 1 + 2 + ··· + n h n+1 i fn+1(x) − fn(x) = x2 2 2 x2 − 1

Therefore, hfni is monotone decreasing on [0, 1]. By Dini’s Theorem 5, the convergence of the sequence is uniform on [0, 1].

RESULT 2. The following examples shows that each of the conditions (compactness of E, continuity of the limit function, continuity of fn and monotonicity of the sequence hfni) in Dini’s theorem 5 is essential.

(i) To show that the compactness of E is essential, consider fn : (0, 1) → R deﬁned by 1 f (x) = ; for x ∈ (0, 1). The sequences has pointwise limit f = 0. Now n 1 + nx

Mn = sup |fn(x) − f(x)| = 1. x∈(0,1)

As Mn 6→ 0 as n → ∞, therefore, the convergence is not uniform.

Dr. Prasun Kumar Nayak Home Study Materiel 22 Small Overview On Uniform Convergence

The assumption of continuity of fn can not be omitted. Consider  1  0; if x = 0 or ≤ x ≤ 1 y 6 f (x) = n n 1 1  1; if 0 < x < n b b as depicted in the Fig. 14, are not continuous. They form a monotonic sequence 1 - pointwise convergent to zero on [0, 1], but n 1 x the convergence is not uniform. Figure 14: Graph of fn

(ii) The continuity of the limit function is also essential. Indeed, the sequence fn : [0, 1] → R, n deﬁned by fn(x) = x ; for x ∈ [0, 1] fails to converge uniformly on [0, 1] as in the Example 16.

(iii) Consider, fn : [0, 1] → R deﬁned by y 6 fn(x) = n  2 1  2n x; if 0 ≤ x ≤  2n  1 1 n − 2n2 x − 1 ; if < x ≤ 2n 2n n  -  1 O 1 1 x  0; if < x ≤ 1 2n n n as depicted in the Fig. 15 Figure 15: Graph of fn

The functions fn(x) are continuous and form a sequence which is pointwise convergent to R 1 zero function on [0, 1]. Therefore, 0 f(x)dx = 0. Now, for each n ∈ N,

Z 1 Z 1 Z 1/n Z 1 2n 2 h 2 1 i fn(x)dx = 2n x dx + n − 2n x − dx + 0 dx 1 2n 0 0 2n 1/n hx2 i 1 h x2 x i1/n 1 = 2n2 2n + nx − 2n2 − = 1 2 0 2 2n 2n 2

Z 1 1 Z 1 Therefore, lim fn(x)dx = 6= 0 f(x)dx. So the convergence is not uniform. n→∞ 0 2 0 x EXAMPLE 42. Let u : [1, 2] → be deﬁned by u (x) = . n R n (1 + x)n

∞ X (i) Show that un(x) converges for x ∈ [1, 2] n=1 (ii) Use Dini’s theorem to show that the convergence is uniform.

∞ ∞ Z 2 X X Z 2 (iii) Does the following hold: un(x) dx = un(x)dx? 1 n=1 n=1 1

Dr. Prasun Kumar Nayak Home Study Materiel 23 Small Overview On Uniform Convergence

1 Solution: (a) Let |1+x| < 1, that is, |1 + x| > 1. Then, we observe that ∞ ∞ X x X 1 x 1 = x = · = 1 (1 + x)n (1 + x)n 1 + x 1 − 1 n=1 n=1 1+x ∞ P So, in particular, un(x) is convergent for x ∈ [1, 2]. n=1 (b) Let E = [1, 2], is compact and un(x) → 0 pointwise. Clearly x x x2 u − u = − = − < 0 n+1 n (1 + x)n+1 (1 + x)n (1 + x)n+1 so that the sequence is monotonic. All the hypotheses of Dini’s Theorem 5 are satisﬁed and thus convergence is uniform. (c) Since the convergence is uniform we can interchange the integral and summation. Thus the equality holds.

Two Important Theorems regarding the Test of Uniform Convergence

THEOREM 6 (Abel’s test). If

(i) bn(x) is positive monotone decreasing function of n for each ﬁxed value x in [a, b]

(ii) bn(x) < k, ∀x ∈ [a, b]

(iii) The series Σun(x) is uniformly convergent on [a, b] then Σbn(x)un(x) converges uniformly on [a, b]

THEOREM 7 (Dirichlet’s test). The series Σun(x)vn(x) will be uniformly convergent on a set E ⊂ R if

(i) hvninis positive,a monotone decreasing sequence for every x ∈ E and converges uniformly to zero on E n X (ii) Sn(x) = un(x) < K for every x ∈ E and for ∀n ∈ N, where K is a constant. r=1 Consider the following examples: ∞ X sin nx (i) Consider the series of functions deﬁed on [a, b], where 0 < a ≤ x ≤ b < 2π. n n=1 1 Let un(x) = sin nx and vn(x) = n . Therefore n nx X sin n − 1 S (x) = u (x) = 2 sin x + x n r sin x 2 r=1 2 n X 1 1 x |S (x)| = | u (x)| ≤ = ; 0 < < π ∴ n r sin x sin x 2 r=1 2 2 1 x Now, x = cosec is bounded for all x ∈ [a, b], where 0 < a ≤ x ≤ b < 2π. sin 2 2 Also, hvninis positive,a monotone decreasing sequence for every x ∈ [a, b] and converges uniformly to zero for all x ∈ [a, b]. Hence by Dirichlet’s test the series converges uniformly on [a, b].

Dr. Prasun Kumar Nayak Home Study Materiel 24 Small Overview On Uniform Convergence

4 Uniform Convergence and Limit Theorems

As was pointed out in the previous section, even if all functions in a sequence have a nice property (such as continuity, differentiability, etc.), the (pointwise) limit function, if it exists, need not (in general) share this property. Our goal now is to show that, if the convergence is uniform, then many nice properties satisﬁed by all the functions in the sequence will also be satisﬁed by their (uniform) limit.

4.1 Uniform Convergence and Limit

In general, limits do not commute. Since the integral is deﬁned with a limit, and since we saw in the last section that integrals do not always respect limits of functions, we know some concrete instances of noncommutation of limits. The fact that continuity is deﬁned with a limit, and that the limit of continuous functions need not be continuous, gives even more examples of situations in which limits do not commute.

THEOREM 8. Let E0 ⊂ E ⊂ R and let hfni ∈ F(E; R)N. Suppose, hfni converges uniformly on E0 to a function f ∈ F(E0; R). Let x0 ∈ E0 and suppose that lim fn(x) = an ; (n = 1, 2, ··· ) x−→x0 then

(i) {an} of real constants converges.

(ii) lim f(x) = lim an i.e. lim { lim fn(x)} = lim { lim fn(x)} x−→x0 n−→∞ x−→x0 n−→∞ n−→∞ x−→x0

Proof: (i) Let > 0 be given. Since hfni converges uniformly on E0 to a function f ∈ F(E0; R), ∃ a positive number N(ε) ∈ N such that for all

fm(x) − fn(x) < ; whenever m, n ≥ N and ∀x ∈ E0

Keeping m, n ﬁxed and let x −→ x0 we get,

|am − an| < ; n ≥ N.

Hence, by Cauchy’s general principle of convergence of real sequence of constants hani converges, say to A, i.e., lim an = A. Therefore, hani of real constants converges. n−→∞ (ii) Let ε > 0 be chosen arbitrary. Since hani converges to A ∃ a positive number N = N(ε) ∈ N such that |a − A| < ; ∀n ≥ N n 3

Since, hfni converges uniformly on E0 to a function f ∈ F(E0; R), ∃ a positive number N = N(ε) ∈ N such that |f (x) − f(x)| < ; ∀n ≥ N and ∀x ∈ E n 3 0

Again since lim fn(x) = an for all n, so corresponding to > 0, ∃δ > 0 such that x−→x0 |f (x) − a | < ; whenever 0 < |x − x | < δ and ∀n ∈ n n 3 0 N

Dr. Prasun Kumar Nayak Home Study Materiel 25 Small Overview On Uniform Convergence

Hence for all n ≥ N we have,

|f(x) − A| ≤ |f(x) − fn(x)| + |fn(x) − an| + |an − A| < + + = ; whenever 0 < |x − x | < δ 3 3 3 0 ∴ lim f(x) = A = lim an. x−→x0 n−→∞ This proves the theorem. 2

∞ X THEOREM 9. Let E0 ⊂ E ⊂ R and let huni ∈ F(E; R)N. Suppose, the series un(x) n=1 converges uniformly on E0 to a sum function s ∈ F(E0; R). Let x0 ∈ E0 and suppose that lim un(x) = an ; (n = 1, 2, ··· ) then x−→x0

∞ X (i) an converges and n=1 ∞ ∞ ∞ X h X i X h i (ii) lim s(x) = an i.e. lim un(x) = lim un(x) x−→x0 x−→x0 x−→x0 n=1 n=1 n=1 ∞ X Proof: (i) Since the series un(x) converges uniformly on E0 corresponding to any > 0, ∃ a n=1 positive integer m such that ∀ x ∈ [a, b] and for any integer p ≥ 1,

n+p X ur(x) < ; ∀ n ≥ m , p ≥ 1 r=n+1

Keeping n, p ﬁxed, we let x −→ x0 and obtain

n+p X ar < ; ∀ n ≥ m , p ≥ 1 r=n+1

∞ X Hence it follows that the series an converges to a ﬁnite limit A (say). n=1 ∞ X (ii) Since un(x) converges uniformly to s(x) so corresponding to any > 0, ∃ a positive n=1 integer m such that ∀ x ∈ [a, b] and for any integer N1 ∈ N, such that ∀ x ∈ [a, b],

n X u (x) − s(x) < ; ∀ n ≥ N r 3 1 r=1

Similarly, ∃ a positive integer N2 ∈ N such that n X a − A < ; ∀ n ≥ N r 3 2 r=1

Dr. Prasun Kumar Nayak Home Study Materiel 26 Small Overview On Uniform Convergence

Again, since lim un(x) = an ; (n = 1, 2, 3, ··· ), ∃ a suitable δ such that x−→x0 u (x) − a < ; ∀ x ∈ |x − x | < δ n n 3n 0 n n n X X X ∴ ur(x) − ar ≤ ur(x) − ar r=1 r=1 r=1 < · n = ; ∀ x ∈ |x − x | < δ 3n 3 0

Let N = maxN1,N2 ∈ N we get for n ≥ N and for x ∈ |x − x0| < δ, we have n n n n X X X X s(x) − A = s(x) − ur(x) + ur(x) − ar + ar − A r=1 r=1 r=1 r=1 n n n n X X X X ≤ s(x) − ur(x) + ur(x) − ar + ar − A r=1 r=1 r=1 r=1 < + + = 3 3 3 Therefore, ∞ ∞ ∞ X X X n o lim f(x) = A = an i.e. lim un(x) = lim un(x) x−→x0 x−→x0 x−→x0 n=1 n=1 n=1 This proves the theorem. 2 4 cosn x π EXAMPLE 43. For n ≥ 1, deﬁne f : [0, π ] → by f (x) = ; x ∈ [0, ]. Then each n 2 R n 3 + cosn x 2 π n fn is continuous. For x ∈ (0, ], cos x → 0 as n → ∞, and so the sequence hfni converges (2 π 1; for x = 0 π pointwise on [0, 2 ] to f(x) = π , which is not continuous on [0, 2 ]. Since the 0; for x ∈ (0, 2 ] π limit function is not continuous, the sequence hfni cannot converge uniformly to f(x) on [0, 2 ]. ∞ X cos nx EXAMPLE 44. Evaluate lim x→0 n(n + 1) n=1 ∞ X cos nx Solution: The given series is of the form u (x), where u (x) = . Now, for all n n n(n + 1) n=1 n ∈ N cos nx 1 1 u (x) = ≤ < ; ∀x ∈ n n(n + 1) n(n + 1) n2 R X X 1 Now, M = is a hyperhermonic series with p = 2 > 1, so convergent. Therefore, by n n2 Weierstrass Mn test, the given series of functions is uniformly convergent on (−∞, ∞). Thus ∞ ∞ ∞ ∞ X cos nx X n cos nx o X 1 X h 1 1 i lim = lim = = − x→0 n(n + 1) x→0 n(n + 1) n(n + 1) n n + 1 n=1 n=1 n=1 n=1 1 1 1 1 1 = 1 − + − + ··· + − + ··· 2 2 3 n n + 1 1 = lim 1 − = 1. n→∞ n + 1

Dr. Prasun Kumar Nayak Home Study Materiel 27 Small Overview On Uniform Convergence

THEOREM 10. Let a function fn : E ⊆ R → R be bounded on E0 ⊆ E for all n ∈ N. If the sequence hfn(x)i of functions converges uniformly on E0, then the limit function f is bounded on E0 and the sequence hfn(x)i is uniformly bounded on E0.

Proof: Since hfn(x)i converges uniformly on E0 to the limit function f, then for a preassigned ε > 0, there exists a natural number N(ε) ∈ N such that

|fn(x) − f(x)| < ε, for all n ≥ N and ∀x ∈ E0

Therefore, |fM (x) − f(x)| < ε. Since fM (x) is bounded on E0, there exists a positive constant M such that |fN (x)| ≤ K. Thus, we have

|f(x)| = |fN (x) − {fN (x) − f(x)}|

≤ |fN (x)| + |fN (x) − f(x)| < M + ε; for all x ∈ E0

This proves that f is bounded on E0. Now, for every x ∈ E0 and each n ≥ n0

|fn(x)| = |fn(x) − f(x) + f(x)|

≤ |fn(x) − f(x)| + |f(x)| < M + 2ε

Again, fn being bounded on E0 for each n ∈ N, we have |fn(x)| ≤ Mn for every x ∈ E0 and n = 1, 2, ··· , (n0 − 1). Thus, if M0 = min{M1,M2, ··· ,Mn0−1,M + 2ε}, then |fn(x)| ≤ M0 for every x ∈ E0 and for all n ∈ N. Therefore, the sequence hfn(x)i is uniformly bounded on E0. 2

4.2 Uniform Convergence and Continuity

THEOREM 11. Let E0 ⊂ E ⊂ R and let hfni ∈ F(E; R)N. If each fn is continuous at some x0 ∈ E0 and hfni converges uniformly on E0 to a function f ∈ F(E0; R); then f is also continuous at x0. Thus, if each fn is continuous on E0, then so is the limit function f.

Proof: Let ε > 0 be chosen arbitrary. Since f is the uniform limit of hfni, we can ﬁnd N = N(ε) ∈ N such that f (x) − f(x) < ; ∀ n ≥ N, ∀ x ∈ E (i) n 3 0

With N as in (i), the continuity of fN at x0 implies that we can ﬁnd δ = δ() > 0 with |f (x) − f (x )| < ; ∀ x ∈ E ∩ (x − δ, x + δ)(ii) N N 0 3 0 0 0 Also, (i) implies that f (x) − f(x) < ; ∀ x ∈ E ∩ (x − δ, x + δ)(iii) N 3 0 0 0

Now (i), (ii) and (iii) imply that, for each x ∈ E0 ∩ (x0 − δ, x0 + δ), we have

|f(x) − f(x0)| = |f(x) − fN (x) + fN (x) − fN (x0) + fN (x0) − f(x0)|

≤ |f(x) − fN (x)| + |fN (x) − fN (x0)| + |fN (x0) − f(x0)| < + + = ε 3 3 3 and hence f is continuous at x0. Since x0 is taken arbitrary on E0, so the result holds. 2

Dr. Prasun Kumar Nayak Home Study Materiel 28 Small Overview On Uniform Convergence

THEOREM 12. Let E0 ⊂ E ⊂ R and let huni ∈ F(E; R)N. If each fn is continuous at some ∞ P x0 ∈ E0 and the series fn converges uniformly on E0 to a sum function s ∈ F(E0; R); then s n=1 is also continuous at x0. In particular, if each fn is continuous on E0, then so is the limit function s. P Proof: Let ε > 0 be chosen arbitrary. x0 be arbitrary point on E0. Since un(x) converges uniformly to s(x) on E0, therefore for > 0 we can chose N ∈ N such that ∀ x ∈ E0

n X u (x) − s(x) < ; ∀ n ≥ N r 3 r=1 and is particular, at x = x0 ∈ E0 and n = N,

N X u (x ) − s(x ) < r 0 0 3 r=1

n X Again, since each un(x) is continuous at x0, the sum of a ﬁnite number of functions ur(x) is r=1 also continuous at x = x0. Therefore, for > 0, ∃δ > 0 such that

N N X X u (x) − u (x ) < ; ∀ x ∈ E ∩ (x − δ, x + δ) r r 0 3 0 0 0 r=1 r=1

Hence for ∀ x ∈ E0 ∩ (x0 − δ, x0 + δ), we have

N N N N X X X X s(x) − s(x0) = s(x) − ur(x) + ur(x) − ur(x0) + ur(x0) − u(x0) r=1 r=1 r=1 r=1 N N N N X X X X ≤ s(x) − ur(x) + ur(x) − ur(x0) + ur(x0) − u(x0) r=1 r=1 r=1 r=1 < + + = 3 3 3

∴ s(x) −→ s(x0) as x −→ x0. Since x0 is arbitrary, so s(x) is continuous in E0. 2

RESULT 3. The converse of this theorem is not always true as may be seen in the following example

∞ X x (i) The series is uniformly convergent on any ﬁnite interval [a, b], (nx + 1){(n − 1)x + 1} n=1 when 0 < a < b. But the series is only point wise convergent but not uniformly convergent

Dr. Prasun Kumar Nayak Home Study Materiel 29 Small Overview On Uniform Convergence

x 1 1 on [a, b]. Here u (x) = = − . Therefore n (nx + 1){(n − 1)x + 1} (n − 1)x + 1 nx + 1

sn(x) = u1(x) + u2(x) + u3(x) + ··· + un(x) 1 1 1 1 1 = 1 − + − + − + x + 1 x + 1 2x + 1 2x + 1 3x + 1 h 1 1 i ··· + − (n − 1)x + 1 nx + 1 1 = 1 − nx + 1 ( 1 x 6= 0 s(x) = lim sn(x) = ∴ n−→∞ 0 x = 0

∴ The sum function s(x) is discontinuous on [a, b], and therefore the converges is uniform on [a, b], it is only point wise. When x 6= 0, let > 0 be given 1 |s (x) − s(x)| = 1 − − 1 n nx + 1 1 1 1 = < ; whenever n > − 1 nx + 1 x 1 1 h n 1 1 oi But − 1 decreases with x. Hence if we take N = max − 1 + 1 ∈ N, x x∈[a,b] x which is independent of x, then we obtain |sn(x) − s(x)| < , whenever n > N for all x ∈ [a, b] i.e. the series converges uniformly to s(x) = 1 on [a, b].

Below are some examples:

∞ X x4 x4 x4 x4 (i) Consider the series = x4 + + + + ··· on [0, 1]. (1 + x4)n 1 + x4 (1 + x4)2 (1 + x4)3 n=0 th Let {sn(x)}n be the sequence of n partial sums. Then n 1 1 o s (x) = x4 1 + + + ··· n 1 + x4 (1 + x4)2 1 1 − 4 n n 1 o = x4 (1+x ) = (1 + x4) 1 − , x 6= 0. 1 (1 + x4)n 1 − 1+x4

4 4 Therefore lim sn(x) = 1 + x i.e., s(x) = 1 + x , x 6= 0. Again, at x = 0, sn(0) = 0. n→∞ s(0) = lim sn(0) = 0. So n→∞ ( 0; x = 0 s(x) = 1 + x4; x 6= 0

lim s(x) = 1 6= s(0) ⇒ s is not continuous at x = 0 i.e., on [0, 1], and therefore n→∞ the converges is not uniform on [0, 1], it is only point wise. Given series is a series of continuous functions on [0, 1] but its sum function s is not so. Hence, the series does not converge uniformly on [0, 1].

Dr. Prasun Kumar Nayak Home Study Materiel 30 Small Overview On Uniform Convergence

∞ X x4 (ii) Consider the series on [0, 1]. Here, (1 + x4)n n=0 ( 1; x 6= 1 s(x) = 0; x = 0

Thus the sum function s(x) is discontinuous on [0, 1], and therefore the converges is not uniform on [0, 1], it is only point wise. ∞ X cos nx EXAMPLE 45. A function S deﬁned by S(x) = ; x ∈ . Show that S is continuous 5n R n=1 for any x ∈ R. P cos nx Solution: The given series is of the form un(x), where, un(x) = 5n . Now cos nx 1 |u (x)| = ≤ = M , say n 5n 5n n X 1 1 1 5 Mn = 1 + + 2 + ··· = 1 = 5 5 1 − 5 4 ∞ X cos nx So the series P M is convergent. Hence, by Weierstrass M-test the series is uni- n 5n n=1 cos nx formly convergent on . Again is continuous for all x ∈ and ∀n ∈ . So that the sum R 5n R N function S(x) is continuous on R.

4.3 Uniform Convergence and Integration Z b Z b n o Z b We shall investigate when do we have lim fn(x)dx = lim fn(x) dx = s(x)dx. n−→∞ a a n−→∞ a −nx2 Consider the sequence hfn(x)i given by, fn(x) = 2nxe .Each fn(x) is continuous on [0, 1] and hence integrable there. Now, 2nx f(x) = lim fn(x) = lim = 0 ; ∀ x ∈ [0, 1] n−→∞ n−→∞ enx2 But Z 1 Z 1 −nx2 fn(x)dx = − e (−2nx)dx 0 0 1 Z 2 h 2 i1 = − e−nx d(−nx2) = − e−nx = 1 − e−n 0 0 Z 1 −n ∴ lim fn(x)dx = lim (1 − e ) = 1 n−→∞ 0 n−→∞ Also, Z 1 n o Z 1 lim fn(x) dx = 0 · dx = 0 0 n−→∞ 0 Z 1 Z 1 n o ∴ lim fn(x)dx 6= lim fn(x) dx. n−→∞ 0 0 n−→∞ We show that Riemann integrability is preserved when we pass to uniform limits.

Dr. Prasun Kumar Nayak Home Study Materiel 31 Small Overview On Uniform Convergence

RESULT 4. [Lebesgue’s Integrability Criterion:] Let f :[a, b] → R be a bounded function. Then f is Riemann integrable if and only if it is continuous almost everywhere.

∞ 1 S Proof: For each N ∈ N, let Dn = {x ∈ [a, b]: ωf (x) ≥ N } and put D = Dn. Then D n=1 is the set of all discontinuity points of f in [a, b](?). Suppose that f ∈ R[a, b]. We want to prove that each Dn has measure zero. By Riemann’s Lemma, given any ε > 0 we can ﬁnd a partition n ε P = xk of [a, b] such that U(P ; f) − L(P ; f) < . Let us divide {1, 2, ··· , n} into two k=0 N parts, Gi = Gi(DN ); i = 1, 2

G1 = {j :(xj−1, xj) ∩ DN 6= φ}; G2 = {j :(xj−1, xj) ∩ DN = φ}

Now, with Mr = sup{f(x): x ∈ [xr−1, xr]}, mj = inf{f(x): x ∈ [xr−1, xr]} and δr = xr − xr−1, we have X X ε U(P ; f) − L(P ; f) = (M − m )δ + (M − m )δ < r r r r r r N r∈G1 r∈G2 ε Since (x , x ) ∩ D 6= φ, implies that M − m ≥ , we have r−1 r N r r N X X Nε δ ≤ N (M − m )δ < = ε r r r r N r∈G1 r∈G1

But the intervals (xr−1, xr) with r ∈ G1 cover DN . Therefore, DN has measure zero for each N; and hence, D has measure zero. Conversely, let us assume that D has measure zero and let ε > 0 be given. Each [a, b]/DN is (relatively) open. Therefore each DN is a closed (hence compact) subset of [a, b] and has n measure zero. Let N be such that (b − a)/ = N < ε/2 and pick a partition P = xk k=0 X ε of [a, b] such that δ < , where, M = sup{|f(x)| : x ∈ [a, b]}. Next, note that if r 4M r∈G1 K = S [x , x ], with G deﬁned as above, then K is a compact subset of [a, b] such that r∈G2 r−1 r 2 1 x ∈ K implies ωf (x) < N . Thus, we can pick a δ > 0, such that 1 |s − t| < δ ⇒ |f(s) − f(t)| < . N

n0 0 0 Let Q = xk0 be a refinement of P with mesh ν(P ) < δ. Then, with Mr0 , mr0 and δr0 , k0=0 0 0 0 defined as usual and the subsets G1,G2 ⊂ {1, 2, ··· n } defined as in the first part of the proof, we have

X 0 0 0 X 0 0 0 U(Q; f) − L(Q; f) = (Mr − mr)δr + (Mr − mr)δr 0 0 0 0 r ∈G1 r ∈G2 X b − a ε ε < 2M δ + < 2M · + = ε r N 4M 2 r∈G1 which shows indeed that f ∈ R ∈ [a, b] and completes the proof. 2

Dr. Prasun Kumar Nayak Home Study Materiel 32 Small Overview On Uniform Convergence

THEOREM 13 (Uniform Convergence and Integrability). Let hfni be a sequence of Riemann integrable functions on a compact interval [a, b] ⊂ R. If lim fn = f, uniformly on [a, b], then f is also Riemann integrable on [a, b] and we have Z x Z x f(t) dt = lim fn(t) dt; ∀x ∈ [a, b] a n→∞ a

Proof: Let ε > 0 be chosen arbitrary. The uniform convergence of hfni to f implies that, for some n ∈ N such that

|fn(x) − f(x)| < ε; ∀x ∈ [a, b] and n ≥ N

In particular, |fN (x) − f(x)| < ε, for all x ∈ [a, b]. Therefore,

|f(x)| = |f(x) − fN (x) + fN (x)| ≤ |f(x) − fN (x)| + |fN (x)|

< ε + |fN (x)|; ∀x ∈ [a, b]

Now, for each n ∈ N, fn is Riemann integrable and hence continuous on [a, b] except on a set Dn ∞ S of measure zero. Let D = Dn. Then D has measure zero. For each x ∈ [a, b]/D, all the fn n=1 are continuous at x. Since fn converges to f uniformly, Theorem implies that f is also continuous at x. Thus, f is indeed continuous on [a, b]/D and hence Riemann integrable. Next, given any ε > 0, by uniform convergence, we can ﬁnd N ∈ N such that |fN (t) − f(t)| < ε b−a , for all [a, b]. Z x Z x Z b

f(t) dt − fN (t) dt ≤ |f(t) − fN (t)| dt < ε a a a and the proof is complete. 2

THEOREM 14. Let hfn(x)i be a sequence of R-integrable functions on [a, b] where a, b are ﬁnite. If hfn(x)i converges uniformly to f(x) which is R-integrable in [a, b], then

Z b Z b n o Z b lim fn(x)dx = lim fn(x) dx = f(x)dx n−→∞ a a n−→∞ a

Proof: Let > 0 be any given positive number. Then by deﬁnition of U.C of the {sn(x)} on

[a, b] N() s (x) − s(x) < ∀ n > N We can ﬁnd a positive integer for which n b−a ; and for all x ∈ [a, b]. We chose n > N, we have Z b Z b Z b

sn(x)dx − s(x)dx = {sn(x) − s(x)} a a a Z b

≤ sn(x) − s(x) dx a Z b ≤ dx = a b − a

R b R b n o R b lim sn(x)dx = lim sn(x) dx = s(x)dx. Hence the theorem. ∴ n−→∞ a a n−→∞ a

Dr. Prasun Kumar Nayak Home Study Materiel 33 Small Overview On Uniform Convergence

THEOREM 15 (Term by Term Integration of an Uniform Convergent Series). Let hun(x)i be a ∞ X sequence of R-integrable functions on a compact interval [a, b] ⊂ R. If the infinite series un n=1 converges uniformly to sum s(x) on [a, b], then (i) s ∈ R[a, b], i.e., s is R-integrable on [a, b], and ∞ ∞ Z b Z b h X i X h Z b i (ii) s(x)dx = un(x) dx = un(x)dx a a n=1 n=1 a th Proof: Here sn(x) = u1(x) + u2(x) + ··· + un(x) = the n partial sum of the series. Let > 0 be any given positive number. Then by definition of uniformly convergent of the hsn(x)i on [a, b] We can find a positive integer m such that s (x) − s(x) < ; ∀ n ≥ m. n 3(b − a) In particular, s (x)−s(x) < i.e. − +s (x) < s(x) < +s (x). For this m 3(b − a) 3(b−a) m 3(b−a) m fixed m, since sm is R-integrable, we chose a partition of [a, b] such that U(P ; sm) − L(P ; sm) < 3 . s(x) < s (x) + ∵ m 3(b − a) U(P ; s) < U(P ; s ) + ∴ m 3 Again since, s(x) > s (x) − m 3(b − a) L(P ; s) > L(P ; s ) − ∴ m 3 Therefore 2 2 U(P ; s) − L(P ; s) < U(P ; s ) − L(P ; s ) + = + = m m 3 3 3 So, s(x) is R-integrable on [a, b]. Therefore, Z b Z b n o lim sn(x)dx = lim sn(x) dx n−→∞ a a n−→∞ ∞ ∞ X h Z b i Z b h X i un(x)dx = un(x) dx n=1 a a n=1 RESULT 5. It is observed here that term-by=term integration is not a sufficient condition for uniform convergence of a series of functions as may be seen in the following example 46.

∞ 1 X h EXAMPLE 46. Show that the series un(x), where u1(x) = x and un(x) = x2n − 1 − n=1 1 i x2n − 3 , n ≥ 2, is not uniformly convergent on [0, 1]. the series be integrated term-by-term on [0, 1] ?

Dr. Prasun Kumar Nayak Home Study Materiel 34 Small Overview On Uniform Convergence

Solution: The nth partial sum of the series is given by

1 Sn(x) = u1(x) + u2(x) + ··· + un(x) = x 2n−1 ; x ∈ [0, 1]

For all x ∈ (0, 1], lim Sn(x) = 1 and for x = 0, the sequence hSni converges to 0. Therefore, n→∞ ∞ X the series un(x) is convergent pointwise on [0, 1] to the limit function S, where S(x) = n=1 ( 0; if x = 0 . The limit function S(x) is discontinuous at x = 0. Since each u is 1; if x ∈ (0, 1] n P continuous on [0, 1] and the limit function S(x) is not continuous on [0, 1], so the series un is not uniformly convergent on [0, 1]. Now

∞ Z 1 X Z 1 Z 1 un(x) dx = S(x)dx = dx = 1 0 n=1 0 0 Z 1 Z 1 1 u1(x)dx = x dx = 0 0 2 1 1 Z 1 Z 1 h i 2n − 1 2n − 3 un(x)dx = dx = x2n − 1 − x2n − 3 = − ; n ≥ 2 (i) 0 0 2n 2n − 2 Z 1 Z 1 Z 1 Again, let In = u1(x)dx + u2(x)dx + ··· + un(x)dx, then using (i), we get 0 0 0 2n − 1 In = ⇒ lim In = 1 2n n→∞ ∞ ∞ X Z 1 Z 1 X ⇒ un(x)dx = 1 = un(x) dx n=1 0 0 n=1 i.e., the series can be integrated term-by-term on [0, 1].

∞ ∞ Z 1 X xn X 1 EXAMPLE 47. Prove that dx = n2 n2(n + 1) 0 n=1 n=1

xn Solution: Let un(x) = n2 ; x ∈ [0, 1], n ∈ N. For all x ∈ [0, 1], xn 1 |u (x)| = ≤ ; ∀n ∈ . n n2 n2 N 1 Let M = , then |u (x)| ≤ M for all x ∈ [0, 1] and for all n ∈ and P M is a convergent n n2 n n N n series of positive real numbers. P Therefore, by Weierstrass M-test, the series un is uniformly convergent on [0, 1]. Since each fn is integrable on [0, 1], the series can be integrated term by term on [0, 1]. Hence

∞ ∞ ∞ ∞ Z 1 X xn X Z 1 xn X h xn+1 i1 X 1 dx = dx = = n2 n2 (n + 1)n2 0 n2(n + 1) 0 n=1 n=1 0 n=1 n=1

Dr. Prasun Kumar Nayak Home Study Materiel 35 Small Overview On Uniform Convergence

2 x2 x ∞ − X h 1 − 1 2 i EXAMPLE 48. Show that the series u (x), where u (x) = 2x e n2 − e (n + 1) , n n n2 (n + 1)2 n=1 ∞ ∞ X Z 1 Z 1 X x ∈ [0, 1] is uniformly convergent on [0, 1] and further show that un(x)dx = un(x) dx. n=1 0 0 n=1 Solution: The nth partial sum of the series is given by

Sn(x) = u1(x) + u2(x) + ··· + un(x) x2 − h 2 1 2 i = 2x e−x − e (n + 1) (n + 1)2 −x2 For all x ∈ (0, 1], lim Sn(x) = 2xe . For x = 0, the sequence hSni converges to 0. Therefore, n→∞ ∞ X −x2 the series un(x) converges pointwise on [0, 1] to the function S, where S(x) = 2xe , n=1 x ∈ [0, 1]. Therefore, Z 1 ∞ Z 1 X 2 1 u (x) dx = 2xe−x dx = 1 − n e 0 n=1 0 For all x ∈ [0, 1], n 1 1 o |u (x)| ≤ 2 + = M (say ); ∀n ∈ n n2 (n + 1)2 n N ∞ X Then Mn is a convergent series of positive real numbers and for all x ∈ [0, 1], |un(x)| ≤ Mn n=1 ∞ X for all n ∈ N. Hence by Weierstrass M-test the series un(x) converges uniformly on [0, 1]. n=1 Now 2 x2 x 1 1 − Z Z h 1 − 1 2 i n2 (n + 1) un(x)dx = 2x 2 e − 2 e dx 0 0 n (n + 1) 2 x2 x 1 1 − − h − 2 i1 − 2 = − e n2 + e (n + 1) = −e n2 + e (n + 1) (i) 0 P Since each un is integrable on [0, 1] and un converges uniformly on [0, 1], then term-by-term integration for the series is possible and ∞ ∞ X Z 1 Z 1 X 1 u (x)dx = u (x) dx = 1 − n n e n=1 0 0 n=1 Z 1 Z 1 Z 1 Again, let In = u1(x)dx + u2(x)dx + ··· + un(x)dx, then using (i), we get 0 0 0 1 − 1 (n + 1)2 1 In = − + e ⇒ lim In = 1 − e n→∞ e ∞ X Z 1 1 ⇒ u (x)dx = 1 − n e n=1 0

Dr. Prasun Kumar Nayak Home Study Materiel 36 Small Overview On Uniform Convergence

∞ X h 2 −n2x2 2 −(n−1)2x2 i EXAMPLE 49. Let us consider the series un(x), where un(x) = x n e −(n−1) e , n=1 ∞ X x ∈ [0, 1]. Applying integration show that the series un(x) is not uniformly convergent on n=1 [0, 1].

Solution: The nth partial sum of the series is given by

2 −n2x2 Sn(x) = u1(x) + u2(x) + ··· + un(x) = n xe

For all x ∈ (0, 1],

4 4 2 2 n x 2 en x > > 0 ⇒ 0 < S (x) < ; ∀x ∈ (0, 1]. 2 n n3x3

Thus, by Sandwich theorem lim Sn(x) = 0 for all x ∈ (0, 1]. For x = 0, the sequence hSni n→∞ ∞ X converges to 0. Therefore, the series un(x) converges pointwise on [0, 1] to the function S, n=1 where S(x) = 0, x ∈ [0, 1]. Therefore,

∞ Z 1 X Z 1 un(x) dx = 0 dx = 0 0 n=1 0 Now Z 1 Z 1 h 2 −n2x2 2 −(n−1)2x2 i un(x)dx = x n e − (n − 1) e dx 0 0 1h 2 2 2 2 i 1h 2 2 i = e−(n−1) x − en x = e−(n−1) − en (i) 2 2 Z 1 Z 1 Z 1 Again, let In = u1(x)dx + u2(x)dx + ··· + un(x)dx, then using (i), we get 0 0 0

1 −n2 1 In = 1 − e ⇒ lim In = 2 n→∞ 2 ∞ ∞ X Z 1 1 Z 1 X ⇒ u (x)dx = 6= u (x)dx n 2 n n=1 0 0 n=1

∞ X Therefore, the series un(x) is not uniformly convergent on [0, 1]. n=1 ∞ X −nx EXAMPLE 50. If S(x) be the sum function of the series un(x), where un(x) = ne , x ∈ n=1 [a, b], 0 < a < b, then show that the series converges uniformly to S(x) on [a, b]. Evaluate Z log 3 S(x)dx. log 2

Dr. Prasun Kumar Nayak Home Study Materiel 37 Small Overview On Uniform Convergence

Solution: For all x ∈ [a, b] n 2n 2 |u (x)| = < < ; ∀n ∈ n enx n2x2 n2a2 N 2 2 P Let Mn = 2/a n , then un(x) converges uniformly on [a, b] where 0 < a < b to the sum function S(x), where

∞ X S(x) = un(x); x ∈ [a, b], 0 < a < b. n=1 P Now, each un is integrable on [a, b]. Also, the series un is uniformly convergent on [a, b] to the sum function S(x). Therefore

∞ Z b X Z b Z b Z b S(x)dx = un(x)dx = u1(x)dx + u2(x)dx + ··· a n=1 a a a Z log 3 Z log 3 Z log 3 Z log 3 S(x)dx = e−xdx + 2e−2xdx + 3e−2xdx + ··· log 2 log 2 log 2 log 2 1 1 1 1 1 1 = − + − + − + ··· 2 3 22 32 23 33 1 1 1 1 1 1 = 1 + + + + ··· − 1 + + + + ··· 2 22 23 3 32 33 3 1 = 2 − = 2 2

4.4 Uniform Convergence and Differentiation

Here, we look at the differentiability properties of the limit of a uniformly convergent sequence of differentiable functions. Here, the situation is more complicated. In fact, even the uniform limit of a sequence of differentiable functions need not be differentiable. −nx2 Consider the sequence hfn(x)i where fn(x) = xe , −1 ≤ x ≤ 1. Now

−nx2 f(x) = lim sn(x) = x lim e = 0. n−→∞ n−→∞

0 ∴ hfn(x)i converges uniformly to f(x) = 0 for all values of x in [−1, 1]. Hence f (x) = 0 ∀ x ∈ [−1, 1], so f 0(0) = 0. But

0 −nx2 −nx2 −nx2 2 fn(x) = e + x(−2nx)e = e (1 − 2nx ).

0 0 0 At x = 0 , sn(0) = 1 ; ∀ n. Therefore, fn(0) −→ 1 as n −→ ∞ and f (0) = 0. So uniform d n o n d o convergence of hfn(x)i is not enough to guarantee that lim fn(x) = lim fn(x) . dx n−→∞ n−→∞ dx

THEOREM 16. Let hfni be a real valued function deﬁned on [a, b] such that,

(i) each fn is continuously differentiable function on [a, b]

(ii) hfn(x)i converges at least at one point x0 ∈ [a, b]

0 (iii) hfni converges uniformly to a function σ(x) on [a, b].

Dr. Prasun Kumar Nayak Home Study Materiel 38 Small Overview On Uniform Convergence

Then

1. hfn(x)i must converges uniformly to a continuously differentiable function f(x) on [a, b], and

0 d n o n d o 2. f (x) = σ(x) i.e., lim fn(x) = lim fn(x) ; ∀x ∈ [a, b]. dx n−→∞ n−→∞ dx 0 Proof: Let > 0 be chosen number. Since hfni converges uniformly on [a, b], so, ∃ a positive integer N1() ∈ N such that f 0 (x) − f 0 (x) < ; ∀ m, n ≥ N , ∀ x ∈ [a, b] n m 2(b − a) 1

Also since hfn(x)i converges at x = x0 , corresponding to the same , ∃ a natural number N2(ε) ∈ N such that f (x ) − f (x ) < ; ∀ m, n ≥ N n 0 m 0 2 2

Let x and y be any points in [a, b]. Since fn(x) is differentiable and hence continuous on [a, b], by using Lagrange mean value theorem, we get

{fn(x) − fm(x)} − {fn(y) − fm(y)}

0 0 = (x − y){fn(ξ) − fm(ξ)} ; where, ξ ∈ (x, y) < (b − a) · < ; as |x − y| < b − a 2(b − a) 2

Let, N = max{N1,N2} ∈ N. Then, for ∀ m, n ≥ N and for all x ∈ [a, b], we have

|fn(x) − fm(x0)| = |{fn(x) − fm(x)} − {fn(x0) − fm(x0)} + {fn(x0) − fm(x0)}|

≤ {fn(x) − fm(x)} − {fn(x0) − fm(x0)} + |{fn(x0) − fm(x0)}| < + = ε 2 2

Therefore, by Cauchy criterion, hfni converges uniformly on [a, b] and f be the uniform limit of hfni on [a, b]. For ﬁxed x on [a, b], and for any y ∈ [a, b]; let us deﬁne f (y) − f (x) f(y) − f(x) φ (y) = n n ; n ∈ and φ(y) = ; y 6= x n y − x N y − x

Since each fn is differentiable on [a, b], so for each n ∈ N

fn(y) − fn(x) 0 lim φn(y) = lim = f (x). y→x y→x y − x n Now, for all m, n ≥ N, we have 1 |φ (y) − φ (y)| = · − {f (x) − f (x)} + {f (y) − f (y)} n m |y − x| n m n m 1 ε ε < · |x − y| · = . y − x 2(b − a) 2(b − a)

Dr. Prasun Kumar Nayak Home Study Materiel 39 Small Overview On Uniform Convergence

Therefore, {φn(y)} converges uniformly to φ(y) on [a, b] for y ∈ [a, b] but y 6= x. Since {fn(x)}n converges uniformly to f(x) on [a, b], we get

lim{ lim φn(y)} = lim {lim φn(y)} y→x n→∞ n→∞ y→x or, lim φ(y) = lim f 0 (x) y→x n→∞ n f(y) − f(x) or, lim = lim f 0 (x) y→x y − x n→∞ n 0 or, fn(x) = σ(x); ∀x ∈ [a, b]

0 d d n o 0 d But f (x) = {f(x)} = lim fn(x) and σ(x) = lim f (x) = lim fn(x). So, dx dx n→∞ n→∞ n n→∞ dx d n o d lim fn(x) = lim fn(x); ∀x ∈ [a, b] dx n→∞ n→∞ dx This completes the proof of the theorem. 2

2 −n2x2 EXAMPLE 51. Show that for the series whose partial sums are given by sn(x) = n xe

(i) the limit function is continuous.

(ii) term-by-term integration is valid

(iii) term-by-term differentiation is valid but

(iv) the series does not converge uniformly on any closed interval containing origin

Solution: Let [a, b] be a closed and bounded interval containing 0. Here s(x) = lim sn(x) = n→∞ 0∀x ∈ [a, b], a constant function. So, s is continuous on [a, b]. (ii)

Z b n o Z b lim sn(x) dx = 0dx = 0. a n→∞ a Z b Z b 2 2 2 1n −n2b2 −n2a2 o 2 2 sn(x)dx = n xe−n x dx = − e − e [ Put n x = u] a a 2 Z b ∴ lim sn(x)dx = 0 n→∞ a

Z b X X Z b So, { fn(x)}dx = fn(x)dx. Thus, term-by-term integration is valid. a a 0 0 2 −n2x2 2 2 (iii) Since s(x) = 0; ∀x ∈ [a, b]. ∴ s (x) = 0∀x ∈ [a, b]. Now, sn(x) = n e (1 − 2n x ). Therefore

lim s0 (x) = 0; x ∈ [a, b] n→∞ n 0 0 d n o d lim s (x) = s (x), i.e., lim sn(x) = lim (sn(x)) n→∞ n dx n→∞ n→∞ dx ∞ ∞ d n X o X or, f (x) = f 0 (x); ∀x ∈ [a, b] dx n n n=0 n=0

Dr. Prasun Kumar Nayak Home Study Materiel 40 Small Overview On Uniform Convergence

Thus term-by-term differentiation is valid for all x ∈ [a, b]. n2x (iv) M = sup |s (x) − s(x)| = sup = |g(x)|, say. Now n n n2 2 x∈[a,b] x∈[a,b] e x

2 2 1 g0(x) = x2e−n x (1 − 2n2x2) = 0 for x = ± √ n 2 2 2 1 g00(x) = n4x(4n2x2 − 6)en x < 0 at x = √ n 2 1 Therefore, M = ne−1/2 → ∞ as n → ∞. So the series is not uniformly convergent on [a, b], n 2 an interval containing 0.

THEOREM 17 (Term-by-Term Differentiation). Let huni be a sequence of differentiable functions ∞ ∞ P P 0 on [a, b], such that the series un(x0) converges for some x0 ∈ [a, b]. If the series un of n=1 n=1 ∞ P derivatives converges uniformly on [a, b], then the series un converges uniformly on [a, b] to a n=1 differentiable sum s and we have

∞ ∞ d X X s0(x) = u (x) = u0 (x); ∀x ∈ [a, b] dx n n n=1 n=1

∞ th P Proof: The n partial sum of the series un is sn(x) = u1(x) + u2(x) + u3(x) + ··· + un(x). n=1 Now,

(i) Each un is given to be differentiable on [a, b] and so sn is also differentiable on [a, b], and 0 0 0 0 0 sn(x) = u1 (x) + u2 (x) + u3 (x) + ··· + un (x). ∞ ∞ P 0 0 P 0 (ii) If un be the series of derived functions then this sn(x) = un(x) n=1 n=1 ∞ P 0 (iii) Given that the series un(x) converges uniformly to σ(x) on [a, b], so we may take n=1 0 {un(x)}n converges to σ(x) on [a, b]. ∞ P 0 (iv) Given that the series un(x) converges at least one point x0 ∈ [a, b], so we can take n=1 0 {sn(x)}n converges at least one point x0 ∈ [a, b].

Hence, by the previous theorem 16, the sequence {sn(x)} must converge uniformly to its limit function s(x) on [a, b] such that s0(x) = σ(x); ∀x ∈ [a, b].

EXAMPLE 52. We end this section by giving an example of a continuous function on R that is nowhere differentiable

(i) Consider the sawtooth function:  1  x − [x]; if x ≤ [x] +  2 f (x) = 0 1  [x] + 1 − x; if x > [x] + 2

Dr. Prasun Kumar Nayak Home Study Materiel 41 Small Overview On Uniform Convergence

Then f0(x) is the distance from x to the nearest integer, i.e., f0(x) = d(x, Z), and is a −n n continuous, periodic function on R with period 1. Now, deﬁne fn(x) = 4 f0(4 x) for all x ∈ R and n = 0, 1, 2, ··· . Then fn is also a continuous sawtooth function (with period −n 1 4 ), whose graph consists of line segments of slope ±1. Since 0 ≤ f0 ≤ 2 , we have 1 0 ≤ f (x) ≤ for all x ∈ and n ∈ . n 2 · 4n R N P RESULT 6. Only the uniform convergence of the series of functions n un(x) on [a, b] is not P sufﬁcient to ensure the validity of term-by-term differentiation of the series n un(x) on [a, b]. This situation is depicted in the following example 53. ∞ X th EXAMPLE 53. example We consider the series un(x), x ∈ [0, 1] whose n partial sum is n=1 x sn(x) = , x ∈ [0, 1]. Then lim sn(x) = 0 for all x ∈ [0, 1]. Hence, the sequence 1 + nx2 n→∞ hsn(x)i converges pointwise to the limit function s(x) where s(x) = 0, x ∈ [0, 1]. Let (as depicted in Example 20) x 1 M = sup s (x) − s(x) = sup = √ → 0 as n → ∞ n n 2 x∈[0,1] x∈[0,1] 1 + nx 2 n P Therefore hsn(x)i is uniformly convergent on [0, 1]. Thus, the series n un(x) converges uniformly to the limit function f on [0, 1]. Now

2 ( 0 d n o 1 − nx 0 0; 0 < x ≤ 1 s (x) = sn(x) = and lim s (x) = n dx (1 + nx2)2 n→∞ n 1; x = 0 ( 0; 0 < x ≤ 1 Hence the series P u0 converges to the function g, where g(x) = . Therefore n 1; x = 0

d d d h i u (x) + u (x) + ··· = 0 = u (x) + u (x) + ··· , for 0 < x ≤ 1 dx 1 dx 2 dx 1 2 d d d h i and u (x) + u (x) + ··· = 1 6= u (x) + u (x) + ··· , for x = 0 dx 1 dx 2 dx 1 2 P RESULT 7. If the series sn be convergent pointwise, then the uniform convergence of the series P 0 sn is only a sufﬁcient condition for the validity of term-by-term differentiation of the series P sn. This situation is depicted in the following example 54. ∞ X th EXAMPLE 54. We consider the series un(x), x ∈ [0, 1] whose n partial sum is sn(x) = n=1 log(1 + n4x2) , x ∈ [0, 1]. Then 2n2 log(1 + n4x2) lim sn(x) = lim = 0 for all x ∈ [0, 1] n→∞ n→∞ n2

Therefore, the sequence hsn(x)i converges pointwise to the limit function s(x) where s(x) = 0, x ∈ [0, 1]. Now n2x s0 (x) = u0 (x) + u0 (x) + ··· + u0 (x) = ; x ∈ [0, 1] n 1 2 n 1 + n4x2

Dr. Prasun Kumar Nayak Home Study Materiel 42 Small Overview On Uniform Convergence

Therefore, lim s0 (x) = 0 for all x ∈ [0, 1]. Thus the sequence hs0 (x)i converges point wise to n→∞ n n P 0 the limit function s(x) where s(x) = 0, x ∈ [0, 1] and hence the series sn(x) converges to the d d function g(x) on [0, 1]. Now [f(x)] = 0 for all x ∈ [0, 1] and also [f(x)] = g(x), x ∈ [0, 1]. dx dx Therefore, d d d h i [u (x)] + [u (x)] + ··· = u (x) + u (x) + ··· dx 1 dx 2 dx 1 2 P Thus term-by-term differentiation of the series un is valid. Let

n2x M = sup s0 (x) − s(x) = sup n n 4 2 x∈[0,1] x∈[0,1] 1 + n x

2 1 r n x + 2 1 1 For x > 0, we have n x ≥ n2x · , equality holds for x = . Therefore, 2 n2x n2 n2x 1 1 n2x ≤ for all x > 0, equality holds for x = . Again for x = 0, = 0. 1 + n4x2 2 n2 1 + n4x2 Hence n2x 1 Mn = sup 4 2 = x∈[0,1] 1 + n x 2

1 0 P 0 Since lim Mn = (6= 0), the sequence hs (x)i and hence the series u is not uniformly n→∞ 2 n n P 0 convergent on [0, 1]. Thus, although the series un is not uniformly convergent on [0, 1], term P by term differentiation of the series un is valid. ∞ X EXAMPLE 55. Show that term-by-term differentiation is not valid at x = 0 for the series un(x) n=1 nx (n − 1)x , where u (x) = − ; x ∈ [0, 1]. n 1 + n2x2 1 + (n − 1)2x2

th X Solution: Let sn(x) be the n partial sum of the series un(x), then nx s (x) = u (x) + u (x) + ··· + u (x) = ; x ∈ [0, 1] n 1 2 n 1 + n2x2

Therefore, lim sn(x) = 0 for all x ∈ [0, 1] and hence the sequence hsn(x)i converges pointwise n→∞ X to the limit function s(x), where s(x) = 0, x ∈ [0, 1]. Thus the series un(x) converges pointwise to the limit function s(x) for all x ∈ [0, 1]. Now

d X d u (x) = = 0; ∀x ∈ [0, 1] dx n dx d n − n3x2 (n − 1) − (n − 1)3x2 and u (x) = − dx n 2 h i2 1 + n2x2 1 + (n − 1)2x2

d X d At x = 0, u (x) = n − (n − 1) = 1 and hence u (x) = 1 + 1 + ··· , which is a dx n dx n d X X d divergent series. Therefore, at x = 0, u (x) 6= u (x) . dx n dx n

Dr. Prasun Kumar Nayak Home Study Materiel 43 Small Overview On Uniform Convergence

Problem Set

[Multiple Choice Questions] 1. Let lim xn = f(x), x ∈ [0, 1]. Then n→∞ ( ( 0; if 0 ≤ x < 1 1; if 0 ≤ x < 1 a) f(x) = b) f(x) = 1; if x = 1 0; if x = 0

c) f(x) = 1, x ∈ [0, 1] d) f(x) = 0; x ∈ [0, 1]

1 (a) ∞ X 2. The series xn, converges pointwise to the sum function s(x), x ∈ [0, 1]. Then n=1 1 1 a) f(x) = , x ∈ [0, 1] b) f(x) = , x ∈ [0, 1] 1 − x 1 + x c) f(x) = x, x ∈ [0, 1] f(x) = 0 does not existd)

2 (d) 2m 3. For each n ∈ , let fn(x) = lim cos n!πx ; x ∈ . Then the sequence of functions N m→∞ R hfni converges on R to the function f deﬁned by ( ( 0; if x ∈ 1; if x ∈ a) f(x) = Q b) f(x) = Q 1; if x ∈ R − Q 0; if x ∈ R − Q ( ( π; if x ∈ −π; if x ∈ c) f(x) = Q d) f(x) = Q 1; if x ∈ R − Q 1; if x ∈ R − Q 3 (b)

4. Which of the following sequences of functions is uniformly convergent on (0, 1)? n x 1 a) xn b) c) d) nx + 1 nx + 1 nx + 1 4 (c)

1/n 5. Let fn(x) = x for x ∈ [0, 1]. Then

a) lim fn(x) exists for all x ∈ [0, 1] b) lim fn(x) deﬁnes a continuous func- n→∞ n→∞ tion on [0, 1]

c) {fn(x)} converges uniformly on [0, 1] d) lim fn(x) = 0 for all n ∈ [0, 1] n→∞ 5 (a)

6. Let lim xe−nx = f(x), x ≥ 0. Then n→∞ a) f(x) = 0, x ≥ b) f(x) = 1, x ≥ c) f(x) = e−1, x ≥ none of the aboved) 0 0 0

Dr. Prasun Kumar Nayak Home Study Materiel 44 Small Overview On Uniform Convergence

6 (a)

−1 7. Let fn(x) = tan nx; x ∈ . Then lim fn(x) is R n→∞ π π π a) 0 b) c) − d) sgnx 2 2 2 7 (d) 1 2x 4x3 2nx2n−1 8. For x ∈ (−1, 1), the sum of the series + + + ··· + + ··· is 1 + x 1 + x2 1 + x4 1 + x2n 1 1 1 − x a) b) c) 0 d) 1 + x 1 − x 1 + x 8 (b) e−2x e−4x e−6x 9. The series 1 − + − + ··· is 22 − 1 42 − 1 62 − 1

a) converges uniformly for all x ≥ 0 b) converges uniformly for all x ∈ R c) converges uniformly for all x ∈ (−1, 1) d) converges uniformly on [−1, 1]

9 (a)

10. Which of the following sequence hfni of functions does not converge uniformly on [0, 1]? e−x a) f (x) = b) f (x) = (1 − x)n n n n x2 + nx sin(nx + n) c) f (x) = d) f (x) = n n n n 10 (b)

11. Which one of the following series of functions is uniformly convergent for all real x ?

X (−1)nx2n X (−1)nx2n X (−1)nx2n a) √ b) c) None of thesed) n(1 + x2n) n3/2(1 + x2n) n2/3(1 + x2n) 11 (b) ( 1 − nx; for 0 ≤ x ≤ 1 12. Let f (x) = n n 1 0; for n < x ≤ 1

a) lim fn(x) deﬁnes a continuous func- b) {fn} converges uniformly on [0, 1] n→∞ tion on [0, 1]

c) lim fn(x) = 0 for all x ∈ [0, 1] d) lim fn(x) exists for all x ∈ [0, 1] n→∞ n→∞ 12 (d)

( 1 1 − nx; for 0 ≤ x ≤ n 13. Let fn(x) = and lim fn = f. Then 1 n→∞ 0; for n < x ≤ 1

Dr. Prasun Kumar Nayak Home Study Materiel 45 Small Overview On Uniform Convergence

a) f is continuous on [0, 1] b) f is bounded on [0, 1]

convergence of hfni is uniformc) none of thesed)

13 (b) ( n(1 − nx); for 0 ≤ x ≤ 1 14. Let f (x) = n . Then hf i n 1 n 0; for n < x ≤ 1

a) converges pointwise to f(x) = 0, x ∈ b) converges uniformly to f(x) = 0, x ∈ [0, 1] [0, 1] Z 1 Z 1 c) lim fn(x)dx = f(x)dx none of the aboved) n→∞ 0 0 14 (a) xn 15. Let f (x) = ; x ∈ [0, 3]. Then n 1 + xn

a) Convergence of hfni is uniform on [0, 3] b) The limit function is continuous on [0, 3] c) the limit function is bounded on [0, 3] d) The convergence is not pointwise on [0, 3]

15 (c) ∞ X xn 16. The series of functions converges 1 + xn n=1

a) uniformly on [0, 1] b) pointwise on [0, 1] c) sum function on [0, 1] all of thesed)

16 (a)

n 17. Let fn : [1, 2] → [0, 1] be given by fn(x) = (2 − x) for all non-negative integers n. Let f(x) = lim fn(x) for 1 ≤ x ≤ 2. Then which of the following is true? n→∞

a) f is a continuous function on [1, 2] b) fn converges uniformly to f on [1, 2] as n → ∞ Z 2 Z 2 0 c) lim fn(x)dx = f(x)dx d) For any a ∈ (1, 2) we have lim fn(a) 6= n→∞ 1 1 n→∞ f 0(a)

17 (c)

18. Let {bn} and {cn} be sequences of real numbers. Then a necessary and sufﬁcient condition 2 for the sequence of polynomials fn(x) = bnx + cnx to converges uniformly to 0 on the real line is

Dr. Prasun Kumar Nayak Home Study Materiel 46 Small Overview On Uniform Convergence

∞ ∞ X X a) lim bn = 0 and lim cn = 0 b) |bn| < ∞ and |cn| < ∞ n→∞ n→∞ n=1 n=1

c) There exists a positive integer N such that bn = 0 and cn = 0 for all n > N

d) lim cn = 0 n→∞ 18 (c) 1 19. Which one of the statement is true for the sequence of functions: f (x) = , n = n n2 + x2 1, 2, . . . , x ∈ [1/2, 1]?

a) The sequence is monotonic and has 0 as the limit for all x ∈ [1/2, 1] as n → ∞ 1 b) The sequence is not monotonic but has f(x) = as the limit as n → ∞ x2 1 c) The sequence is monotonic and has f(x) = as the limit as n → ∞ x2 The sequence is not monotonic but has 0 as the limitd)

19 (a)

−nx2 20. For n ≥ 1, let fn(x) = xe , x ∈ R. Then the sequence {fn} is

a) Uniformly convergent on R b) Uniformly convergent only on compact subset of R c) Bounded and not uniformly convergent A sequence of unbounded functionsd) on R 20 (a) Z π/2 Z π/2 2n+1 21. Let fn(x) = n sin x cos x. Then the value of lim fn(x)dx− lim fn(x)dx n→∞ 0 0 n→∞ is

1 1 a) 2 b) 0 c) − 2 d) −∞ 21 (a)

22. Let {fn} be a sequence of continuous real-valued functions deﬁned on [0, ∞). Suppose fn(x) → f(x) for all x ∈ [0, ∞) and that f is integrable. Then Z ∞ Z ∞ a) fn(x)dx → f(x)dx as n → ∞ 0 0 Z 1 Z 1 b) If fn → f uniformly on [0, ∞),then fn(x)dx → f(x)dx 0 0 Z ∞ Z ∞ c) If fb → f uniformly on [0, ∞), then fn(x)dx → f(x)dx 0 0 Z 1 d) If |fn(x) − f(x)| → 0, then fn → f uniformly on [0, 1] 0

Dr. Prasun Kumar Nayak Home Study Materiel 47 Small Overview On Uniform Convergence

22 (b)

23. Let f : R → R be strictly increasing continuous function. If hani is a sequence in [0, 1], then the sequence hf(an)i is

increasinga) boundedb) convergentc) not necessarily boundedd)

23 (b) ∞ X 1 24. The series converges uniformly for n2 + n3x2 n=1

a) all x ∈ R b) x ≥ 0 c) x ∈ [0, 1] d) [a, ∞), a > 0 24 (a)

25. Which of the following conditions below imply that a function f : [0, 1] → R is necessarily of bounded variation?

a) f is a monotone function on [0, 1] b) f is a continuous and monotone function on [0, 1]

c) f has a derivative at each x ∈ (0, 1) d) f has a bounded derivative on the interval (0, 1)

25 (a), (b), (c)

26. Let f : R → [0, ∞) be a non-negative real valued continuous function. Let φn(x) =  ( k h k k + 1 n if f(x) ≥ n  n if f(x) ∈ n , n , φ (x) = 2 2 2 and g (x) = φ (x) + n,k h k k + 1 n n 0 if f(x) < n 0 if f(x) 6∈ , 2n 2n n2n−1 P φn,k(x). As n ↑ ∞, which of the following are true? k=0

a) gn(x) ↑ f(x) for every x ∈ R

b) Given any C > 0, gn(x) ↑ f(x) uniformly on the set {x : f(x) < C}

c) gn(x) ↑ f(x) uniformly for x ∈ R

d) Given any C > 0, gn(x) ↑ f(x) uniformly on the set {x : f(x) ≥ C}

26 (a), (b) ( 0 if x 6∈ An 27. Let An ⊆ R for n ≥ 1, and χn : R → {0, 1} be the function χn(x) = . 1 if x ∈ An Let g(x) = lim supχn(x) and h(x) = lim χn(x) n→∞ n→∞

Dr. Prasun Kumar Nayak Home Study Materiel 48 Small Overview On Uniform Convergence

a) If g(x) = h(x) = 1,then there exists m such that for all n ≥ m we have x ∈ An

b) If g(x) = 1 and h(x) = 0. then there exist m such that for all n ≥ m we have x ∈ An

c) If g(x) = 1 and h(x) = 0 then there exists a sequence n1, n2,... of distinct integers

such that x ∈ Ank for all k ≥ 1

d) If g(x) = h(x) = 0 then there exists m such that for all n ≥ m we have x 6∈ An

27 (a), (c), (d)

28. Let {fn} be a sequence of continuous functions on R Z ∞ Z ∞ a) If {fn} converges to f pointwise on R then lim fn(x)dx = f(x)dx n→∞ −∞ −∞ Z ∞ Z ∞ b) If {fn} converges to f uniformly on R then lim fn(x)dx = f(x)dx n→∞ −∞ −∞

If {fn} converges to f uniformly on R then f is continuous on R.c)

d) There exists a sequence of continuous functions {fn} on R, such that {fn} converges Z ∞ Z ∞ to f uniformly on R, but lim fn(x)dx 6= f(x)dx n→∞ −∞ −∞ 28 (c), (d) x 2 1 R 29. For n ≥ 1, let gn(x) = sin (x = n , x ∈ [0, ∞) and fn(x) = gn(t)dt. Then 0

a) {fn} converges pointwise to a function f on [0, ∞), but does not converge uniformly on [0, ∞)

b) {fn} does not converge pointwise to nay function on [0, ∞)

c) {fn} converges uniformly on [0, 1]

d) {fn} converges uniformly on [0, ∞)

29 (c), (d)

n 2 2 30. Let t and a be positive real numbers. Deﬁne Ba = {x = (x1, x2, . . . , xn) ∈ R |x1 + x2 + 2 2 n ··· + xn ≤ a }. Then for any compactly supported continuous function f on R which of the following are correct? Z Z Z Z a) f(tx)dx = f(x)t−ndx b) f(tx)dx = f(x)tdx Ba Bta Ba Btna Z Z Z Z c) f(x+y)dx = f(x)dx for some d) f(tx)dx = f(x)tndx Rn Rn Rn Rn n y ∈ R 30 (a), (c)

31. Consider all sequences {fn} of real valued continuous functions on [0, ∞). Identify which of the following statements are correct.

Dr. Prasun Kumar Nayak Home Study Materiel 49 Small Overview On Uniform Convergence

Z ∞ Z ∞ a) If {fn} converges to f pointwise on [0, ∞), then lim fn(x)dx = f(x)dx n→∞ 0 0 Z ∞ Z ∞ b) If {fn} converges to f uniformly on [0, ∞),then lim fn(x)dx = f(x)dx n→∞ 0 0

c) If {fn} converges to f uniformly on [0, ∞), then f is continuous on [0, ∞)

d) There exists a sequence of continuous functions {fn} on [0, ∞),such that {fn} con- Z ∞ Z ∞ verges to f uniformly on [0, ∞) but lim fn(x)dx 6= f(x)dx n→∞ 0 0 31 (c), (d)

32. Find out which of the following series converge uniformly for x ∈ (−π, π)

∞ ∞ ∞ ∞ X e−n|x| X sin(xn) X xn X 1 a) b) c) d) n3 n5 n ((x + π)n)2 n=1 n=1 n=1 n=1 32 (a), (b), (c)

33. Which one of the following is not uniformly convergent for all x ∈ R ? ∞ ∞ ∞ ∞ X cos nx X cos nx X sin nx X sin nx a) b) c) d) √ n2 n3 n2 n n=1 n=1 n=1 n=1 33 (d)

n 34. Let fn(x) = (−x) , x ∈ [0, 1]. Then decide which of the following are true.

a) There exist a pointwise convergent sub sequence of fn

fn has no pointwise convergent sub sequenceb)

fn converges pointwise everywhere.c)

fn has exactly one pointwise convergent sub sequenced)

34 (a) ∞ X 35. The series 2−n sin(2nx) n=0

converges pointwise on R but not uniformlyb)a) R converges uniformly on π c) converges uniformly on [0, ] but not d) does not converge pointwise on 2 R on R 35 (b) √ 2 −2 36. Let fn(x) = x + n ; ∀x ∈ R and n ∈ N. Then

Dr. Prasun Kumar Nayak Home Study Materiel 50 Small Overview On Uniform Convergence

a) hfn(x)i converges to x uniformly only b) hfn(x)i converges pointwise to |x| on on a ﬁnite interval of R R but not uniformly there

c) hfn(x)i converges to |x| uniformly on d) hfn(x)i converges pointwise to x on R R 36 (b) ∞ P n 37. If a function f(x) be such that f(x) = φn(x), where φn(x) = (1 − x) , 0 ≤ x ≤ 1. n=0 Then

a) the series does not converge uniformly on [0, 1] b) f(x) is continuous on [0, 1] c) the series may or may not converge uniformly on [0, 1]

d) the series converges uniformly on 0, 1] to x on R 37 (a) 2x2 38. Let f [0, 1] → be given by f (x) = ; n ∈ . Then the sequence {f } n R n x2 + (1 − 2nx)2 N n

converges uniformly on [0, 1].a) b) does not converge uniformly on [0, 1] but has a subsequence that converges uniformly on [0, 1]

c) does not converge pointwise on [0, 1] d) converges pointwise [0, 1] but does not has a sequence that converges uniformly on [0, 1]

38 (d) n x X 39. Let f (x) = and s (x) = f (x) for x ∈ [0, 1]. Then the n {(n − 1)x + 1}(nx + 1) n j j=1 sequence {sn}

converges uniformly on [0, 1].a) [0, 1] but notconverges uniformly pointwiseb) on c) converges pointwise for x = 0 but not does not converge for x ∈ [0, 1].d) for x ∈ [0, 1] 39 (b)

40. Let f be uniformly continuous on R and hani converges to a in R. Let fn(x) = f(x + an) for all x ∈ R. Then

hfn(x)i is only pointwise convergenta) b) hfn(x)i is uniformly convergent in R

hfn(x)i is divergent sequencec) d) hfn(x)i is only bounded but not pointwise convergent

Dr. Prasun Kumar Nayak Home Study Materiel 51 Small Overview On Uniform Convergence

40 (b) 1 41. Let f : → be a non-zero function such that |f(x)| ≤ for all x ∈ . Deﬁune R R 1 + 2x2 R ∞ P real valued functions fn on R for all n ∈ N by fn(x) = f(x+n). Then the series fn(x) n=1 converges uniformly

a) on [0, 1] but not on [−1, 0] b) on [−1, 0] but not on [0, 1] c) on both [0, 1] and [−1, 0] d) neither on [0, 1] nor on [−1, 0]

41 (c) sin nx 42. Let fn(x) = √ ; n ∈ N and x ∈ R. Then as lim fn(x) =, n n→∞

a) 0 a continuous functionb) a bounded functionc) does not existd)

42 (d) sin nx 43. Let f (x) = √ ; n ∈ and x ∈ [−1, 1]. Then as n → ∞, n n N Z n a) hfni does not converge uniformly in [−1, 1]b) lim fn(x)dx 6= 0 n→∞ −1 0 c) hfn(x)i does not converge uniformly in d) fn(x), n ∈ N is not uniformly continu- [−1, 1] ous in [−1, 1] 43 (c) ∞ X sin nx 44. The series √ ; n ∈ converges uniformly on n N n=1 hπ 3π i a) [5, 2π − 5] b) [10, 2π − 10] c) , d) does not converge 2 2 uniformly 44 (c) ∞ X cos nx 45. The series ; n ∈ converges uniformly on n N n=1 hπ 7π i a) [10, 2π − 10] b) [4, 2π − 4] c) , none of thesed) 4 4 45 (c) 1 46. Let f (x) = for n ∈ , x ∈ . Which of the following are true ? n 1 + n2x2 N R

a) fn converges uniformly on [0, 1] b) fn converges pointwise on [0, 1] to a continuous function Z 1 Z 1 1 c) fn converges uniformly on [ 2 , 1] d) lim fn(x)dx = ( lim fn(x))dx n→∞ 0 0 n→∞

Dr. Prasun Kumar Nayak Home Study Materiel 52 Small Overview On Uniform Convergence

46 (c), (d)

47. Let Cc(R) = {f : R → R| f is continuous and there exists a compact set K such that c −x2 f(x) = 0 for all x ∈ K }. Let g(x) = e for all x ∈ R. For which of the following statements are true?

There exists a sequence {fn} in Cc(R) such that fn → g uniformly.a)

There exists a sequence {fn} in Cc(R) such that fn → g pointwiseb)

c) If a sequence in Cc(R) converges pointwise to g then it must converge uniformly to g

There does not exist any sequence in Cc(R) converging pointwise to g.d) 47 (a), (b)

48. Which of the following sequence hfn(x)i of functions does not converge uniformly on [0, 1]? e−x x2 + nx sin(nx + n) a) b) (1 − x)n c) d) n n n 48 (b)

n 49. Let fn : [1, 2] → [0, 1] be given by fn(x) = (2 − x) ; n ∈ . Let f(x) = lim fn(x), N n→∞ 1 ≤ x ≤ 2. Then which of the following is true?

a) f is continuous function on [1, 2] b) fn converges uniformly to f on [1, 2] Z 2 Z 2 0 0 c) lim fn(x) dx = f(x) dx d) ∀a ∈ (1, 2), lim fn(a) 6= f (a) n→∞ 1 1 n→∞ 49 (d) x2n 50. Let fn :[−1, 1] → R be given by fn(x) = ; n ∈ N and f(x) = lim fn(x). Let 1 + x2n n→∞ f(x) = lim fn(x), 1 ≤ x ≤ 2. Then n→∞

a) f is continuous function on [−1, 1] b) fn → f uniformly to on [−1, 1] c) f is integrable on [−1, 1] none of thesed)

50 (c)

2 2 −nx 51. Let fn(x) = n x e ; n ∈ N and x ≥ 0. Then hfni converges uniformly on a) [0, ∞) b) [a, ∞) for a > 0

convergence is not uniform anywherec) fn is discontinuousd) each

51 (b)

2 2 52. Let fn(x) = log(n + x ); n ∈ N and x ∈ R. Then

Dr. Prasun Kumar Nayak Home Study Materiel 53 Small Overview On Uniform Convergence

0 a) hfni converges uniformly on R b) hfni converges uniformly on R 0 c) hfni converges pointwise on R d) hfni is not convergent on R 52 (b) ∞ X (−1)nx2n 53. The series is uniformly convergent for all x ∈ if p is np(1 + x2n) R n=1 1 3 a) 0 b) c) d) 2 2 4 53 (d) ∞ X cos nx 54. If the series converges uniformly on . Then a value of p is np R n=1 2 3 a) 1 b) c) d) −1 3 2 54 (c) ∞ X sin nx 55. If the series converges uniformly on . Then a value of p is np R n=1 7 5 1 a) 1 b) c) d) 5 7 2 55 (b)

[Short Answer Type Questions]

1. Find the limit function f, for the following sequence hfni of functions x (a) f (x) = ; 0 ≤ x < ∞. Ans: f(x) = 0, ∀x ∈ [0, ∞) n 1 + nx 2 2 n (b) fn(x) = n x(1 − x ) ; 0 ≤ x ≤ 1. Ans: f(x) = 0 nx (c) f (x) = ; x ∈ . Ans: f(x) = 0, ∀x ∈ n 1 + n2x2 R R cos nx (d) f (x) = ; x ∈ +. Ans: f(x) = 0, ∀x ∈ + n n R R x2n (e) f (x) = ; x ∈ . Ans: f(x) = 0, |x| < 1, 1 , |x| = 1, 1, |x| > 1 n 1 + x2n R 2 x4n (f) f (x) = ; x ∈ [−1, 1]. Ans: f(x) = 0, |x| < 1, 1 , x = ±1 n 1 + x4n 2

2. Use the deﬁnition to examine uniform convergence of the sequence hfn(x)i on [0, ∞), where fn(x) = x a) b) xe−nx c) n2x2e−nx x + n Ans : a) NUC b) UC c) NUC

Dr. Prasun Kumar Nayak Home Study Materiel 54 Small Overview On Uniform Convergence

3. Discuss the uniform convergence of the following sequence of functions hfn(x)i deﬁned by setting fn(x) = x nx a) ; x ∈ [0, 1] b) ; x ∈ [0, 1] c) xe−nx; x ≥ 0 1 + nx2 1 + n3x2 ( 1 nx; 0 ≤ x ≤ n d) 1 1; n < x ≤ 1 Ans : a) UC b) UC c) UC d) NUC

4. Study the uniform convergence on [0, 1] of the sequence of functions hfn(x)i, deﬁned by setting fn(x) = 1 x2 a) b) c) xn(1 − x) 1 + (nx − 1)2 x2 + (nx − 1)2 nx2 d) nxn(1 − x) e) n3xn(1 − x)4 f) 1 + nx 1 g) 1 + xn

5. Study the uniform convergence of hfn(x)i on A and B, deﬁned by setting fn(x) =

n n h π i h π π i a) cos x(1 − cos x); A = 0, 2 , B = 4 , 2

n 2n h π i b) cos x sin x; A = R, B = 0, 4

∞ X cos nx 6. A function f deﬁned by f(x) = , x ∈ . Show that f is continuous for any 10n R n=1 x ∈ R. ∞ X 7. Prove or disprove: 2−n cos(3nx) represents an everywhere continuous function. n=1

∞ ∞ n X anx 8. If P a is absolutely convergent, prove that the series converges uniformly n 1 + x2n n=1 n=1 for all x ∈ R. ∞ X 1 9. Show that √ is uniformly convergent throughout the positive x-axis. 2n−1 1 + nx n=1 ∞ P 10. Discuss the convergence and uniform convergence of the series un(x), where un(x) is n=1 given by 1 1 x a) ; x ∈ b) ; x ∈ /{0} c) sin ; x ∈ x2 + n2 R n2x2 R n2 R 1 xn (−1)n d) ; x ∈ /{0} e) ; x ≥ 0 f) ; x ≥ 0 xn + 1 R 1 + xn n + x 1 g) rn sin nx; 0 < r < 1, x ∈ h) ; x ∈ n3 + n4x2 R R

Dr. Prasun Kumar Nayak Home Study Materiel 55 Small Overview On Uniform Convergence

Ans : a) UC b) NUC c) UC on |x| ≤ a, a > 0 d) UC on [a, ∞], a > 1 e) NUC f) NUC g) UC h) UC [nf(x)] 11. For a function f deﬁned on [a, b] set f (x) = ; n ∈ and x ∈ [a, b]. Show that n n N hfn(x)i converges uniformly to f(x) on [a, b]. ∞ X h cos(n2x)i 12. Examine whether 4−n sin(3nπx) + is uniformly convergent on , where pn R n=p ∞ P p is a positive integer ≥ 2. Hints : If the given series is of the form un(x), then n=p 1 1 |u (x)| ≤ + = M . n 4n pn n 13. Find where the following series converge pointwise: ∞ ∞ X 1 X xn a) ; x 6= −1 b) ; x 6= −1 1 + xn 1 + xn n=1 n=1 ∞ ∞ X 2n + xn 1 X xn−1 c) ; x 6= − d) ; x 6= −1, 1 1 + 3nxn 3 (1 − xn)(1 − xn+1) n=1 n=1 ∞ n−1 ∞ X x2 X ln xx e) ; x 6= −1, 1 f) 1 − x2n n n=1 n=2 ∞ ∞ X X p g) xln n; x > 0 h) sin2(2π n2 + x2) n=1 n=0 14. Study the uniform convergence of the following series on the given set A:

∞ X hπ i 1 1 a) − tan−1 n2(1 + x2) ; A = Hints: tan−1 < < 2 R n2(1 + x2) n2(1 + x2) n=2 1 = M n2 n ∞ X ln(1 + nx) 1 b) ; A = [2, ∞) Hints: M = nxn n 2n−1 n=1 ∞ X 2 4 c) n2x2e−n |x|; A = Hints: M = R n e2n2 n=1 ∞ ( X n−1 1; x ∈ [−1, 1]/{0} d) x2 1 − x2 ; A = [−1, 1] Hints: f(x) = is not 0; x = 0 n=1 continuous ∞ X n2 1 n2 e) √ xn + x−n ; A = {x ∈ : ≤ |x| ≤ 2} Hints: M = √ 2n+1 R 2 n n=1 n! n! ∞ X 1 f) 2n sin ; A = (0, ∞) 3nx n=1 ∞ X x2 a2 g) ln 1 + ; A = (−a, a), a > 0 Hints: Mn = n ln2 n n ln2 n n=2

Dr. Prasun Kumar Nayak Home Study Materiel 56 Small Overview On Uniform Convergence

Ans : a) UC b) UC c) UC d) NUC e) UC f) NUC g) UC

∞ X x 15. Study the continuity on [0, ∞) of the function f deﬁned by f(x) = . {(n − 1)x + 1}(nx + 1) n=1 ( 1; x > 1 Hints: f(x) = is not continuous 0; x = 0

16. Study the continuity of the sum of the following series on the domain of its pointwise convergence:

∞ n ∞ ∞ ∞ X x X 2 X X a) sin(nx) b) xn c) n2nxn d) lnn(x + 1) n! n=0 n=0 n=1 n=1 Ans : a) Converges absolutely on R and the sum function is also continuous on R b) Converges absolutely on (−1, 1) and the sum function is also continuous on (−1, 1) c) Converges absolutely on (−1/2, 1/2) and the sum function is also continuous on (−1/2, 1/2) 1 d) Converges absolutely on ( e − 1, e − 1) and the sum function is also continuous on 1 ( e − 1, e − 1) ∞ X x sin(n2x) 17. Show that the series converges pointwise to a continuous function on . n2 R n=1 ∞ √ n X 18. Determine whether the series x converges pointwise, and study the continuity of n=1 the sum. Ans : Converges in (0, 1) and the sum function is also continuous there.

n 19. Let φ be continuous on [0, 1]. Then the sequence hfni deﬁned by fn(x) = x φ(x) converges uniformly on [0, 1] if and only if φ(1) = 0. √ 2 2 2 20. Verify that the sequence hfn(x)i, where fn(x) = n sin 4π n + x converges uniformly on [0, a], a > 0. Does hfn(x)i converge uniformly on R ? ∞ X 21. (a) Suppose that the series un(x); x ∈ A, converges uniformly on A and that s : A → n=1 ∞ P R is bounded. Prove that the series s(x)un(x) converges uniformly on A. n=1 (b) Show by example that boundedness of s is essential. Under what assumption concern- ∞ X ing s does the uniform convergence of the series s(x)un(x) imply the uniform n=1 ∞ P convergence of un(x) on A? n=1

22. Assume that hfn(x)i is a sequence of functions deﬁned on A and such that

a) fn(x) ≥ 0 for x ∈ A and n ∈ N b) fn(x) ≥ fn+1(x) for x ∈ A and n ∈ N

supx∈A fn(x) −→ 0 as n → ∞.c)

Dr. Prasun Kumar Nayak Home Study Materiel 57 Small Overview On Uniform Convergence

∞ X n+1 Prove that (−1) fn(x) converges uniformly on A. Hints : For x ∈ A, the series n=1 ∞ X n (−1) fn(x) converges by the Leibnitz Theorem. Moreover, n=1 ∞ X k+1 sup |Rn(x)| = sup (−1) fk(x) ≤ sup fn+1(x) x∈A x∈A k=n+1 x∈A

23. Prove that the following series converge uniformly on R: ∞ ∞ ∞ X (−1)n+1 X (−1)n+1 X (−1)n+1 a) b) √ c) √ n + x2 3 2 2 n + cos x n=1 n=1 n + x + x n=2 Hints : Use the Exercise 22.

24. Determine the domain A of pointwise convergence and the domain B of absolute convergence of the series given below. Moreover, study the uniform convergence on the indicated set C. ∞ ∞ X 1 h i X 1 x + 1n a) 2n(3x − 1)n; C = 1 , 1 b) , C = [−2, −1] n 6 3 n x n=1 n=1

h 1 1 1 1 1 Ans: a) A = 6 , 2 and B = 6 , 2 UC on C b) A = (−∞, − 2 ] and B = 1 (−∞, − 2 ), UC on C

25. Assume that the functions fn, gn : A → R, n ∈ N, satisfy the following conditions: ∞ X (a) the series fn+1(x) − fn(x) is uniformly convergent on A n=1 (b) sup |fn(x)| → 0 as n → ∞ x∈A n P (c) the sequence hGn(x)i, where Gn(x) = gk(x), is uniformly bonded on A. k=1 ∞ X Prove that the series fn(x)gn(x) is uniformly convergent on A. n=1 26. Show that the following series converge uniformly on the indicated set A: ∞ ∞ X xn X sin(nx) a) (−1)n+1 ; A = [0, 1] b) ; A = [α, 2π−α], 0 < α < n n n=1 n=1 π ∞ ∞ X sin(n2x) sin(nx) X sin(nx) tan−1(nx) c) ; A = d) ; A = [α, 2π − n + x2 R n n=1 n=1 α], 0 < α < π ∞ ∞ X 1 X e−nx e) (−1)n+1 ; A = [a, ∞), a > 0 f) (−1)n+1 √ ; A = [0, ∞) nx 2 n=1 n=1 n + x

Dr. Prasun Kumar Nayak Home Study Materiel 58 Small Overview On Uniform Convergence

27. Let f, fn : [0, 1] → R be continuous functions. Compute the following sentence such that both statements (a) and (b) below are true: Let fn → f: Z 1 Z 1 a) lim fn(x)dx = f(x) dx b) lim lim fn(x) = lim lim fn(x) n→→∞ 0 0 n→→∞ x→0 x→0 n→→∞ Ans : 27 Uniformly on [0, 1]

n 28. Let fn(x) = x for n ∈ N. Which of the following statements are ture? 1 1 a) The sequence hfn(x)i converges uniformly on [ 4 , 2 ]

b) The sequence hfn(x)i converges uniformly on [0, 1]

c) The sequence hfn(x)i converges uniformly on (0, 1)

Ans : 28 (a)

29. Give examples to illustrate that all the hypotheses in Dini’s Theorem (Theorem 5) are essential.

30. Let hrni be a sequence consisting of all the rational numbers and for n = 1, 2, ··· , we ( 1; x = rn deﬁne the functions fn on R by fn(x) = . Prove that hfn(x)i converges 0; otherwise pointwise but not uniformly on every interval of R.

[Long Answer Type Questions]

1. Determine whether the sequence hfn(x)i converges uniformly on A, deﬁned by setting fn(x) = 2x x2 a) tan−1 ; A = b) n ln 1 + ; A = x2 + n3 R n R

1 + nx 2pn c) n ln ; A = (0, ∞) d) 1 + x2n; A = nx R √ pn n e) 2n + |x|n; A = R f) n + 1 sin x cos x; A = R √ g) n( n x − 1); A = [1, a], a > 1

2. Prove that for the sequence hfni, fn → f pointwise on a point set E ⊂ R, the convergence is uniform.

3. (a) Suppose that hfni ∈ F(E; R)N converges to f uniformly on E and that each fn is bounded on E. Show that hfni is uniformly bounded; i.e., there is an M > 0 such that |fn(x)| ≤ M for all n ∈ N and all x ∈ E. 1 x (b) Let f (x) = + for all x ∈ (0, 1]. Show that hf i converges uniformly to f(x) = n x n n −1 x on (0, 1] and yet the fn and f are all unbounded on (0, 1].

4. Let E ⊂ R and hfni ∈ F(E; R)N. Prove that, a sequence of functions hfn(x)in deﬁned on E0 converges uniformly on E0 ⊂ E if and only if corresponding to an > 0, ∃N0() ∈ N

such that fn+p(x) − fn(x) < , for n ≥ N and ∀x ∈ E0.

Dr. Prasun Kumar Nayak Home Study Materiel 59 Small Overview On Uniform Convergence

5. Prove that, a sequence hfni ∈ F(E; R)N, where E ⊂ R; converges uniformly on E0 ⊂ E n o if and only if lim sup |fn(x) − f(x)| : x ∈ E0 = 0. n−→∞

6. Prove that, a sequence hfni ∈ F(E; R)N, where E ⊂ R; converges uniformly on E0 ⊂ E if and only if, corresponding to an ε > 0, ∃N = N(ε) ∈ N, depends on ε only, such that n o sup |fm(x) − fn(x)| : x ∈ E0 < ε; whenever m, n ≥ N

7. Find the convergence set of each of the following sequences hfn(x)i of functions on [0, 1] n 2 n x n a) fn(x) = n x (1 − x) b) fn(x) = 1 + n fn(x) = nx(1 − x) .c) 6 y 6 y 6 y f1

1 f2 f1

x - -x -x O (a)O (b)1 O (c) 1

Figure 16: Figures of fn(x) as depicted in the Figures 16 (a), (b) and (c) respectively. Also ﬁnd sets on which these converge uniformly.

8. If a sequence of functions hfni converges uniformly to f on [a, b] and g is a bounded function on [a, b], show that the sequence of functions hgfni converges uniformly to gf on [a, b].

9. If a sequence of functions hfni converges uniformly on [a, b] to a function f and if c ∈ [a, b] such that lim fn(x) = αn, n ∈ , show that the sequence hαni is convergent. n→c N

10. If hfni be a sequence of continuous real valued functions converging uniformly to f on a set E(⊂ ). Show that lim fn(xn) = f(x) for any sequence hαni in E converging to a R n→∞ point x in E.

11. Suppose K is a compact set and let hfn(x)i be a sequence of continuous functions converging pointwise to a continuous functions f and fn ≥ fn+1, then fn → f uniformly. Hints : Theorem 5

12. (Dini’s Theorem of uniform convergence of a series of functions) Let I ⊂ R be a compact interval and suppose that huni ∈ F(E; R)N is a sequence of continuous functions such ∞ P that the series un converging pointwise to a continuous function s : I → R. If the n=1 fn is increasing (i.e., un(x) ≤ un+1(x) for all x ∈ I and n ∈ N) or decreasing (i.e., ∞ P un(x) ≥ un+1(x) for all x ∈ I and n ∈ N), then prove that un converges to s uniformly n=1 on I.

Dr. Prasun Kumar Nayak Home Study Materiel 60 Small Overview On Uniform Convergence

13. For each n ≥ 1, let fn(x) be a monotonic increasing real valued function on [0, 1] such that the sequence hfn(x)i converges pointwise to a function f ≡ 0. Pick out the true statements from the following:

Sequence hfn(x)i converges to f uniformlya)

b) If the functions fn are non-negative, then fn must be continuous for sufﬁciently large ( 0; x ∈ [0, 1) n Hints : fn(x) = 1 n ; x = 1 Ans: 13 (a)

14. Let fn and f be continuous functions on an interval [a, b] and assume that fn → f uniformly on [a, b]. Pick out the true statements:

If fn is Riemann-integrable, then f is Riemann-integrable.a)

b) If f is continuously differentiable, then f is continuously differentiable.Hints : fn(x) = r 1 x2 + in [−1, 1] n2

If xn → x in [a, b], then fn(x) → f(x).c)

Ans: 14 (a), (c)

15. Pick out the sequence hfni which are uniformly convergent NBHM’09 sin nx a) f (x) = nxe−nx on (0, ∞)b) f (x) = xn on [0, 1] c) f (x) = √ on n n n n R Ans: 15 (c)

16. Test the following for uniformly convergent ∞ D xn E X sin nx a) The sequence of functions i b) The series f(x) = over [0, 2] 1 + xn n2 + 1 n=1 over [0, 2] D E c) The sequence of functions n2x2e−nx i over (0, 1)

Ans: 16 (b)

17. In each of the following cases, examine whether the given sequence (or series) of functions converges uniformly over the given domain NBHM’11 ∞ nx X n sin nx xn a) f (x) = ; x ∈ b) ; x ∈ [0, π] c) f (x) = ; x ∈ n 1 + nx en n 1 + xn n=1 (0, 1) [0, 2]

Ans: 17 (b)

18. Which of the following sequence (or series) of functions are uniformly convergent on [0, 1] ? NBHM’13

Dr. Prasun Kumar Nayak Home Study Materiel 61 Small Overview On Uniform Convergence

∞ n X cos(n6x) a) f (x) = cos(πn!x) b) f(x) = c) f (x) = n2x(1 − x2)n n n3 n n=1 Ans: 18 (b) 1 h 1 i 19. Let g (x) = f x + − f(x) , where f : → a continuous function. Which of the n n n R R following statements are ture? NBHM’14

3 0 a) If f(x) = x , then gn → f uniformly on R as n → ∞ 2 0 b) If f(x) = x , then gn → f uniformly on R as n → ∞ 0 0 c) If f is differentiable and f is uniformly continuous on R, then gn → f uniformly on R as n → ∞ Ans: 19 (b)

20. Which of the following statements are true ? NBHM’14

∞ X x2 a) The series does not converge uniformly on 1 + n2x2 R n=1 ∞ X x2 b) The series converges uniformly on 1 + n2x2 R n=1 ∞ X sin nx2 c) The sum of the series deﬁnes a continuously differentiable function on 1 + n2 R n=1 Ans: 20 (b), (c)

21. Which of the following statements are true ? NBHM’17

∞ X x2 Z π sin nx a) The series is UC on [−1, 1] b) lim = π (1 + x2)n n→∞ nx5 n=1 π/2 ∞ X sin nx2 c) Deﬁne x ∈ , f(x) = , then f is a continuously differentiable function. R 1 + n3 n=0 Ans: 21 (c)

22. Let hfn(x)i be a sequence of continuous functions deﬁned on [0, 1]. Assume that fn(x) → f(x) for each x ∈ [0, 1]. Which of the following conditions imply that this convergence is uniform ? NBHM’18

The function f is continuousa) b) fn(x) ↓ f(x) for every x ∈ [0, 1] Hints 1/n : fn(x) = 1 − x

c) The function f is continuous and fn(x) ↓ f(x) for every x ∈ [0, 1] Hints : Dini’s Theorem 5

Ans: 22 (c)

Dr. Prasun Kumar Nayak Home Study Materiel 62 Small Overview On Uniform Convergence

23. Which of the following statements are true ? NBHM’19

n a) The sequence of functions hfn(x)i, deﬁned by fn(x) = x (1 − x), is uniformly convergent on [0, 1] x2 b) The sequence of functions hf (x)i, deﬁned by f (x) = n log 1 + , is uniformly n n n convergent on R ∞ X 1 c) The series 2n sin is uniformly convergent on [1, ∞) 3nx n=1 Ans: 23 (a), (c)

24. Test the uniform convergence of the sequence of functions hfni, deﬁned by fn :[−k, k] → log(1 + n2x2) , where, f (x) = ; x ∈ [−k, k], k > 0. R n n n 25. Show that the sequence fn : x → x converges for each x ∈ I = {x : 0 ≤ x ≤ 1} but that the convergence is not uniform.

26. In each of the following problems, show that the sequence hfni converges to f for each x ∈ I and determine whether or not the convergence is uniform: 2x (a) f : x −→ ; f(x) ≡ 0; I = {x : 0 ≤ x ≤ 1} n 1 + nx cos nx (b) f : x −→ √ ; f(x) ≡ 0; I = {x : 0 ≤ x ≤ 1} n n n3x (c) f : x −→ ; f(x) ≡ 0; I = {x : 0 ≤ x ≤ 1} n 1 + n4x n3x (d) f : x −→ ; f(x) ≡ 0; I = {x : a ≤ x < ∞, a > 0} n 1 + n4x2 nx2 (e) f : x −→ ; f(x) ≡ x; I = {x : 0 ≤ x ≤ 1} n 1 + nx 1 1 x 1 (f) f : x −→ √ + cos ; f(x) ≡ √ ; I = {x : 0 < x ≤ 2} n x n n x sin nx (g) f : x −→ ; f(x) ≡ 0; I = {x : 0 < x < ∞} n 2nx n √ (h) fn : x −→ x (1 − x) n; f(x) ≡ 0; I = {x : 0 ≤ x ≤ 1} 1 − xn 1 1 1 (i) f : x −→ ; f(x) ≡ ; I = {x : − ≤ x ≤ } n 1 − x 1 − x 2 2 −nx2 (j) fn : x −→ nxe ; f(x) ≡ 0; I = {x : 0 ≤ x ≤ 1} ∞ X 27. (a) (xe−x)n; x ∈ [0, 2] n=1 X sin(x2 + n2x) (b) n(n + 1) ∞ X x2n (c) (−1)n np(1 + x2n) n=1

Dr. Prasun Kumar Nayak Home Study Materiel 63 Small Overview On Uniform Convergence

x4 x4 x4 28. Show that the series x4 + + + +··· is not uniformly convergent 1 + x4 (1 + x4)2 (1 + x4)3 on [0, 1].

29. Prove that if a series of continuous functions converges uniformly then the sum function is also continuous.

30. Formulate and prove a result about the derivative of the sum of a convergent series of differentiable functions.

31. Let huni be a sequence of R-integrable functions on a compact interval [a, b] ⊂ R. If the ∞ X inﬁnite series un converges uniformly to sum s on [a, b], then n=1 (a) s ∈ R[a, b], i.e., s is R-integrable on [a, b], and ∞ ∞ Z x Z x h X i X h Z x i (b) s(t)dt = un(t) dt = un(t)dt a a n=1 n=1 a 1 32. Let hfn(x)i be a sequence of functions in C [0, 1] such that fn(0) = 0 for all n ∈ N. Which of the following statements are true ? NBHM’19

a) If the sequence hfn(x)i converges uniformly on [0, 1], then the limit function is in C1[0, 1]

0 b) If the sequence hfn(x)i is uniformly convergent over [0, 1], then the sequence hfn(x)i is also uniformly convergent over the same interval.

c) If the sequence hfn(x)i converges uniformly on [0, 1], then the limit function is in C1[0, 1] ∞ X 0 d) If the series fn(x) converges uniformly on [0, 1] to a function g, then g is Riemann n=1 ∞ Z 1 X integrable and g(t)dt = fn(x) 0 n=1 Ans: 32 (b), (c)

33. Let {r1, r2, ··· , rn, ···} be an enumeration of the rationals in the interval [0, 1]. Deﬁne, for ( 1; if x ∈ {r1, r2, ···} n ∈ N, and for each x ∈ [0, 1] fn(x) = Which of the following 0; otherwise statements are true? NBHM’19

a) The function fn is Riemann integrable over [0, 1] for each n ∈ N

b) The sequence hfn(x)i is pointwise convergent and the limit function is Riemann in- ( 1; if x ∈ tegrable over the interval [0, 1] Hints : f(x) = Q 0; otherwise

c) The sequence hfn(x)i is pointwise convergent but the limit function is not Riemann integrable over [0, 1]

Dr. Prasun Kumar Nayak Home Study Materiel 64 Small Overview On Uniform Convergence

Ans: 33 (c)

34. Let hfni be a sequence of continuous real valued functions deﬁned on R converging uniformly on R to a function f. Which of the following statements are true? NBHM’16

If each of the functions fn is bounded, then f is also bounded.a)

b) If each of the functions fn is uniformly continuous, then f is also uniformly continuous. Z ∞ Z ∞ c) If each of the functions fn is integrable, then lim fn(x) dx = f(x)dx. n→∞ −∞ −∞ Ans: 34 (a),(b)

35. Let hfn(x)i be a sequence of non-negative continuous functions deﬁned on [0, 1]. Assume that fn(x) → f(x) for each x ∈ [0, 1]. Which of the following conditions imply that Z 1 Z 1 lim fn(x)dx = f(x)dx ? NBHM’18 n→∞ ) 0

a) fn(x) ↑ f(x) for every x ∈ [0, 1]

b) fn(x) ≤ f(x) for every x ∈ [0, 1]. Hints : Actually the limit of non-negative continuous functions may not be R integrable function but the additional condition fn(x) ≤ f(x) makes not only f R integrable but the limit of integration is also convergent.

c) The function f is continuous Hints : Consider the functions fn : [0, 1] → R, deﬁned ( n − n2x; if 0 < x < 1 by f (x) = n n 0; otherwise

Ans: 35 (a), (b)

36. Let hfni be a sequence of bounded real valued functions on [0, 1] converging to f at all points of this interval. Which of the following statements are ture? NBHM’14 Z 1 Z 1 a) If fn and f are all continuous, then lim fn(x)dx = f(x)dx Hints : Example n→∞ 0 0 10 Z 1 Z 1 b) If fn → f uniformly, on [0, 1], then lim fn(x)dx = f(x)dx. Hints : Theo- n→∞ 0 0 rem 14 Z 1 Z 1 Z 1

c) If fn(x) − f(x) dx → 0 and n →→ ∞, then lim fn(x)dx = f(x)dx 0 n→∞ 0 0 Z 1 Z 1 Z 1

Hints : Using the inequality lim fn(x)dx − f(x)dx ≤ lim |fn(x) − n→∞ 0 0 n→∞ 0 f(x)| dx

Ans: 36 (b),(c)

Dr. Prasun Kumar Nayak Home Study Materiel 65 Small Overview On Uniform Convergence

Z 1 37. Let hfn(x)i and f be integrable functions on [0, 1] such that lim |fn(x)−f(x)|dx = 0. n→∞ 0 Which of the following statements are true? NBHM’19

a) fn(x) → f(x), as n → ∞, for almost every x ∈ [0, 1] Z 1 Z 1 b) lim fn(x)dx = f(x)dx n→∞ 0 0

c) If hgn(x)i is a uniformly bounded sequence of continuous functions converging point- Z 1 wise to a function g, then |fn(x)gn(x) − f(x)g(x)|dx = 0 as n → ∞ 0 Ans: 37 (b), (c)

38. Let hfni be a sequences of continuous real valued functions defined on [a, b] and also let hani and hbni be two sequences on [a, b] such that lim an = a and lim bn = b. If hfni n→∞ n→∞ Z bn Z b converges uniformly to f on [a, b], then show that lim fn(x)dx = f(x)dx. n→∞ an a ( a ; if x = n ∈ [0, 2014] ∩ 39. Let f(x) = n+1 Z , where ha i is a real sequence. Is f 0; otherwise n Z 2014 integrable on [0, 2014]? If so, find f(x)dx. Ans: 0 0  n2x; 0 ≤ x ≤ 1  n  2 1 2 40. Let hfni be a sequence defined by fn(x) = 2n − n x; n < x ≤ n for ∀n ≥ 2 . Then   2 0; n < x ≤ 1 show that

(a) The sequence hfni is not uniformly convergent on [0, 1].

(b) each fn is Riemann integrable on [0, 1]

(c) hfni has a pointwise limit f which is also Riemann integrable on [0, 1] and Z 1 Z 1 (d) lim fn(x)dx 6= f(x)dx. n→∞ 0 0 nx 41. For what values of p, the sequence hf i deﬁned on [0, 1] by f (x) = (p > 0) n n 1 + n2xp for x ∈ [0, 1] converges uniformly on [0, 1]. Examine further for p = 2 and p = 4 if Z 1 Z 1 lim fn(x)dx = lim fn(x)dx. n→∞ 0 0 n→∞ 2 2 n 42. Show that the sequence of functions hfni deﬁned on [0, 1] by fn(x) = n x(1−x ) , n ∈ N Z 1 for x ∈ [0, 1] converges pointwise to a function f on [0, 1]. By establishing lim fn(x)dx 6= n→∞ 0 Z 1 f(x)dx, show that the sequence hfni is not uniformly convergent on [0, 1]. 0

Dr. Prasun Kumar Nayak Home Study Materiel 66 Small Overview On Uniform Convergence

( 1 1 − nx; x ∈ [0, n ] 43. Show that for the sequence of continuous functions hfni, where fn(x) = 1 0; x ∈ ( n , 1] the pointwise limit function is not continuous in [0, 1] and hence deduce that lim{ lim fn(x)}= 6 x→0 n→∞ lim {lim fn(x)}. n→∞ x→0 Z x t dt 44. Show that the sequence hfni of functions deﬁned by fn(x) = 2 , x ≥ 0 converges 0 1 + n t uniformly to 0 on [0, a), a > 0; but not on [0, ∞).

5 The Weierstrass Approximation Theorem

The name Weierstrass has occurred frequently in this chapter. In fact Karl Weierstrass (1815-1897) revolutionized analysis with his examples and theorems. This section is devoted to one of his most striking results. We introduce it with a motivating discussion.

Deﬁnition 5. Bernstein polynomial : For f : [0, 1] −→ R, let Bn(f, x) be the Bernstein polynomial of order n associated with the function f, deﬁned by

n X n k B (f, x) = f xk(1 − x)n−k. (6) n k n k=0

The Bernstein polynomial of the continuous function f0(x) = 1 is given by

n X n k B (f ) = f xk(1 − x)n−k n 0 k 0 n k=0 n X n n = xk(1 − x)n−k = x + (1 − x) = 1 (7) k k=0

The Bernstein polynomial of the continuous function f1(x) = x is given by

n n X n k X n k B (f ) = f xk(1 − x)n−k = · xk(1 − x)n−k n 1 k 1 n k n k=0 k=0 n X n − 1 n−1 = x xk−1(1 − x)n−k = x x + (1 − x) = x = f (x) (8) k − 1 1 k=1

Dr. Prasun Kumar Nayak Home Study Materiel 67 Small Overview On Uniform Convergence

2 The Bernstein polynomial of the continuous function f2(x) = x is given by n n X n k 2 X k n − 1 B (f ) = · xk(1 − x)n−k = · · xk(1 − x)n−k n 2 k n n k − 1 k=0 k=0 n X n k − 1 n − 1 1 o n − 1 = · + · · xk(1 − x)n−k n − 1 n n k − 1 k=0 n X hn 1 o n − 2 1 n − 1i = 1 − · + · xk(1 − x)n−k n k − 2 n k − 1 k=0 n n 1 X n − 2 1 X 1 n − 1 = 1 − · xk(1 − x)n−k + xk(1 − x)n−k n k − 2 n n k − 1 k=2 k=1 1 1 1 1 = 1 − x2 + x = 1 − f (x) + f (x) (9) n n n 2 n 1

This shows that the Bernstein polynomial Bn(f2) converges to f2(x) on any bounded subset of R. n 2 x(1 − x) X k n−k k EXAMPLE 56. Prove that = x (1 − x) x − . n n k=0 Solution: Differentiate both sides of Eq. (7), we obtain

n X nh i 0 = kxk−1(1 − x)n−k + xk(n − k)(1 − x)n−k−1 · (−1) k k=0 n X n = (k − nx)xk−1(1 − x)n−k−1 (10) k k=0 Multiply both sides of Eq. (10) by x(1 − x), we get

n X n 0 = (k − nx)xk(1 − x)n−k (11) k k=0 Differentiate both sides of Eq. (11), we obtain

n n X n X n n o 0 = −n xk(1 − x)n−k + (k − nx) (k − nx)xk−1(1 − x)n−k−1 k k k=0 k=0 n X n = −n + (k − nx)2xk−1(1 − x)n−k−1 (12) k k=0 Multiply both sides of Eq. (12) by x(1 − x), we get

n X n 0 = −nx(1 − x) + (k − nx)2xk(1 − x)n−k k k=0 n X n or, nx(1 − x) = (k − nx)2xk(1 − x)n−k k k=0 n x(1 − x) X k 2 or, = xk(1 − x)n−k x − . n n k=0

Dr. Prasun Kumar Nayak Home Study Materiel 68 Small Overview On Uniform Convergence

THEOREM 18. For f : [0, 1] −→ R, let Bn(f, x) be the Bernstein polynomial of order n of the function f as in Eq. (6). If f is continuous on [0, 1], then hBn(f, x)i converges uniformly on [0, 1] to f.

Proof: Using the equality (7), we get

n X n f(x) = f(x) xk(1 − x)n−k. k k=0 Consequently,

n X k n B (f, x) − f(x) ≤ f − f(x) xk(1 − x)n−k (13) n n k k=0

By the uniform continuity of f on [0, 1], given ε > 0, there is δ > 0 such that

|f(x) − f(x0)| < ε; whenever |x − x0| < δ; x, x0 ∈ [0, 1]

Clearly, there is M > 0 such that |f(x)| ≤ M for x ∈ [0, 1]. Then the set {0, 1, 2, ··· , n} can be n k o n k o decomposed into the two sets : A = k : − x < δ and B = k : − x ≥ δ . If k ∈ A, n n k then f − f(x) < ε and so n

X k n X n f − f(x) xk(1 − x)n−k < ε xk(1 − x)n−k ≤ ε (14) n k k x∈A x∈A

(k − nx)2 If k ∈ B, then ≥ 1, then n2δ2

X k n f − f(x) xk(1 − x)n−k n k k∈B 2M X n M ≤ (k − nx)2 xk(1 − x)n−k ≤ (15) n2δ2 k 2nδ2 k∈B This Equation (15) combined with Eqs. (13) and (14) yields M B (f, x) − f(x) ≤ ε + ; x ∈ [0, 1] n 2nδ2 This proves the theorem. 2

THEOREM 19 (Approximation theorem of Weierstrass). If f :[a, b] → R be a continuous function on an interval [a, b], then for ε > 0 there is a polynomial p(x) such that

|f(x) − p(x)| < ; x ∈ [a, b]

In particular, there exists a sequence hpn(x)i of polynomial functions such that pn(x) −→ f(x) on [a, b].

Dr. Prasun Kumar Nayak Home Study Materiel 69 Small Overview On Uniform Convergence

Proof: We know that for each n ∈ N, ∃ a polynomial pn(x) such that 1 |p (x) − f(x)| < ; ∀x ∈ [a, b] n n

Taking n = 1, 2, 3, ··· , we get the sequence of polynomials hpn(x)i. Let ε > 0 be chosen 1 arbitrary. Choose N ∈ N such that N > ε . Then 1 1 |p (x) − f(x)| < ≤ < ε; ∀n ≥ N and ∀x ∈ [a, b] n n N

N depends only on ε. Therefore, hpn(x)i converges uniformly to f on [a, b]. The geometrical signiﬁcance of this this theorem lies in the fact that, the graph (Fig. 17) of the y 6 y = f(x) + ε

y = f(x)

y = pn(x) y = f(x) − ε

- x Figure 17: The Weierstrass Approximation Theorem polynomial pn(x) is conﬁned within the region bounded by y = f(x) − ε and y = f(x) + ε for all x ∈ (a, b). This theorem does not guarantee the existence of an polynomial, even if how to construct the polynomial.

EXAMPLE 57. Let f ∈ C[0, 1]. Determine the cases where the given condition implies that f ≡ 0: NBHM’07,’10 Z π Z π a) xnf(x)dx = 0 for all n ≥ 0 b) xnf(x) cos nxdx = 0 for all n ≥ 0 0 0 Z π c) xnf(x) sin nxdx = 0 for all n ≥ 1 0 Solution: (a) For any given f ∈ C[0, 1], by Weierstrass polynomial approximation theorem, there m Z π P k n exists a function φ(x) = akx such that |f(x) − φ(x)| < ε. Since x f(x)dx = 0 for all k=0 0 n ≥ 0, we have

Z π m Z π X k φ(x)f(x)dx = ak x f(x)dx = 0 0 k=0 0 Thus, as f is continuous on a compact set, it must be bounded (say bounded by M). Consider Z π Z π Z π f 2(x) dx = f(x)[f(x) − φ(x)]dx + φ(x)f(x)dx 0 0 0 < εMπ + 0

Dr. Prasun Kumar Nayak Home Study Materiel 70 Small Overview On Uniform Convergence

Z π Since ε > 0 is arbitrary, we have f 2(x) dx = 0, implies f ≡ 0 as it is continuous. So this 0 option is correct. (b) Extend f to f e on [−π, π] so that f e(−x) = f(x) for x ∈ [0, π]. Then, we have Z π Z π f e(x) cos nx dx = 2 f(x) cos nxdx = 0; n ≥ 0 −π 0 Z π as f e and cos both are even functions. Similarly, we have f e(x) sin nx dx = 0, as sin is odd, −π Z π 2 e e n ≥ 1. Thus all the Fourier coefﬁcients of f are zero. By Parseval’s theorem, f (x) dx = −π 0. The continuity of f e then imply f e = 0. Because of f e = f on [0, π], we have f ≡ 0. Therefore option (b) is also correct. (c) This option is again correct, similarly as option (b).

Problem Set

Short answer type questions 1. Let P denote the set of all polynomials in the real variable x which varies over the interval [0, 1]. What is the closure of P in C[0, 1] ? Ans : C[0, 1] Hints : Weierstrass approximation theorem.

Long answer type questions

n x(1 − x) X k 2 1. Establish the identity: = xk(1 − x)n−k x − . Hints : Differentiate Eq. n n k=0 (7) Z 1 n 2. Let f : [0, 1] −→ R be continuous. Assume that x f(x)dx = 0 for n ∈ N, then prove ) that f = 0.

3. Show that there exists no sequence of polynomials hpni such that pn → f on R, where a) f(x) = sin x b) f(x) = ex

4. Let f : R → R be continuous. Show that we can ﬁnd a sequence of polynomials hpni such that pn → f on any bounded subset of R. 1 5. Let f : (0, 1) → be deﬁned by f(x) = . Show that there does not exist a sequence of R x polynomials hpni such that pn → f on (0, 1).

6. Keep the hypothesis of the Weierstrass approximation theorem, can we ﬁnd a sequence of P polynomials hpni such that pn = f on [0, 1] ?

Dr. Prasun Kumar Nayak Home Study Materiel 71 Small Overview On Uniform Convergence

6 Sequence of Functions in a Compact Set

The following elementary lemma is needed to prove certain important theorems for the sequence of complex functions in a compact set.

LEMMA 1. Suppose that the sequence

(i) hfn(x)i converges uniformly to f(x) on a compact set D and g(x) is a continuous function on D. Then hg(x)fn(x)i converges to g(x)f(x) uniformly on D.

(ii) hun(x)i converges uniformly to S(x) on a compact set D and g(x) is a continuous function ∞ P on D. Then g(x)un(x) converges to g(x)S(x) uniformly on D. n=1

THEOREM 20. Let hfn(x)i be a sequence of differentiable functions on a domain D. If fn(x) → (k) (k) f(x) uniformly on every compact subset of D, then, for any k ≥ 1, fn (x) → f (x) for all x ∈ D; i.e., the limit of the kth derivative is the kth derivative of the limit. Moreover, for each D (k) E (k) k ≥ 1, the differentiated sequence fn (x) converges to f (x) uniformly on every compact subset of D.

Proof:

RESULT 8. The above Theorem 20 does not hold if D is assumed to be an arbitrary set instead nsin nxo of a domain. The sequence converges uniformly to zero on the real axis; however, the n nsin nz o sequence of its derivative {cos nx} converges only at x = 0. Thus, the sequence cannot n converge uniformly on any domain containing points of the real axis.

Deﬁnition 6. A sequence hfn(x)i of differentiable functions of a domain D ⊆ C converges normally to the differentiable function f(z) on D if it converges uniformly to f(z) on each compact subsets in D.

6.1 Convergence in the Space of Differentiable Functions

In this section, we shall prove two important theorems namely, Ascoli-Arzela theorem and the Motel’s theorem which guarantees that, given a family F of functions in R that any sequence in F have a uniformly convergent subsequence.

Deﬁnition 7. [ Equicontinuous]: Let F be a family of collection of functions, deﬁned and continuous and real-valued on a compact subset E ⊂ R. A function f ∈ F is continuous at x0 ∈ E if given ε > 0, ∃δ = δ(f, x0, ε) > 0 such that

|x − x0| < δ and x ∈ E ⇒ f(x) − f(x0) < ε.

If δ, independent of f and depending on x0 and ε can be found ∀ε > 0, we say that the family F is equicontinuous at x0.

Dr. Prasun Kumar Nayak Home Study Materiel 72 Small Overview On Uniform Convergence

Deﬁnition 8. [ Uniformly bounded]: Let hfni be a sequence of differentiable functions on a domain D ⊂ R and let U ⊂ D. Then, we say that hfni is uniformly bounded on U, if ∃ a M > 0 such that

|f(x)| ≤ M; ∀x ∈ D and ∀ n ∈ N.

Let F be a family of collection of functions, deﬁned and continuous and real-valued on a compact subset E ⊂ R. We say that F is uniformly bounded on E if

|f(x)| ≤ M; ∀x ∈ E and ∀f ∈ F

Deﬁnition 9. [Locally bounded]: A set F ⊂ H(D) is locally bounded if for each point α ∈ D, there are constants M and ρ > 0 such that

|f(x)| ≤ M; |x − α| < ρ for all f ∈ F n o i.e., sup |f(x)| : |x − α| < ρ, f ∈ F < ∞

That is, F is locally bounded if about a point α ∈ D, there is a interval on which F is uniformly bounded, which immediately extends to the requirement that F is uniformly bounded on a compact sets in D.

Deﬁnition 10. [ Normal convergence]: If a sequence of real functions hfni converges on a compact subsets, it is called normal convergence. If a sequence of functions is uniformly bounded on compact subsets of the domain, it is said to be normally bounded.

THEOREM 21. Ascoli-Arzela Theorem : Let F be a family of collection of continuous, real- valued function on a compact subset E ⊂ R. Suppose F is uniformly bounded on E. Then the followings are equivalent:

(i) F is equicontinuous at each point of E D E D E (ii) Every sequence fn ⊂ F has a uniformly convergent subsequence fnk

Proof: Let EQ be the set of points of E with rational coordinates as n o EQ = rn : rn ∈ Q = E ∩ Q

Since Q is countable and dense subset in E, then E is countable and E = E. As E is Q D QE Q countable, we may enumerate the points of EQ by r1, r2, ··· , i.e., EQ = rn . n≥1 D E Let fn be a sequence of F. Since F is uniformly bounded on each compact subset of E and D E hence it is pointwise bounded. Thus, if we consider fn(r1) , then it is a bounded sequence n≥1 of real numbers, and so |fn(r1)| ≤ M. By Bolzano-Weierstrass theorem, we have a convergent D E D E subsequence fn1 (r1) of fn(r1) . n ≥1 D 1 E

Now, consider fn1 (r2) . Again fn1 (r2) ≤ M and so by Bolzano-Weierstrass theo- n ≥1 D 1 E D E D E rem, ∃ a subsequence fn2 (r2) of fn1 (r2) which converges. Note that fn2 (r1) also n ≥1 converges. 1

Dr. Prasun Kumar Nayak Home Study Materiel 73 Small Overview On Uniform Convergence

D E By induction, we get for ∀n ≥ 1, a subsequence hfnk i of hfnk−1 i such that fnk (rj) converges for j ≤ k, and

hfn1 i ⊇ hfn2 i ⊇ · · · ⊇ hfnk−1 i ⊇ hfnk i ⊇ · · ·

We get a list of lists:

Converge at

fn1 : fi1 (r1) & fi2 (r1) fi3 (r1) ··· r1

fn2 : fj1 (r2) fj2 (r2) & fj3 (r3) ··· r1, r2

fn3 : fk1 (r3) fk2 (r3) fk3 (r3) &··· r1, r2, r3 . . . . . ··· .

∞ D E T D E Consider the subsequence hFli, where Fl = lth member of fnl , then hFli ⊂ fnk , where k=1 D E D E fnl converge at the points r1, r2, r3, ··· . If x ∈ EQ, then at x = rl and since fnl (rl) D E converges we have, the sequence Ft(rl) converges. t≥l Since F is equicontinuous on E, so is hFlil≥1. So given x0 ∈ E, and an ε > 0, ∃δ > 0 with δ = δ(x0, ε) such that ε f(x) − f(x ) < , whenever |x − x | < δ, ∀f ∈ F (16) 0 3 0 n δ(x , ε) o The collection |x − x | < 0 : x ∈ E is an open cover of E, which is compact, so 0 2 0 admits a ﬁnite subcover: n δ δ δ o |x − ζ | < (ζ , ε); |x − ζ | < (ζ , ε); ··· , |x − ζ | < (ζ , ε) 1 2 1 2 2 2 k 2 k

n δ o n δ o n Choose xi1 ∈ |x − ζ1| < 2 (ζ1, ε) ; xi2 ∈ |x − ζ2| < 2 (ζ2, ε) ··· ; xik ∈ |x − ζk| < δ o D E (ζk, ε) where, xij ∈ EQ = rn . Note that 2 n≥1

If x ∈ E, then |x − ζl| < δ(ζl, ε)/2 for some l with 1 ≤ l ≤ k and xil is also in |x − ζk| < δ(ζk, ε)/2. Let m, n be two positive integers, greater than a large positive integer. Now

|F (x) − F (x)| = F (x) − F (x ) + F (x ) − F (x ) + F (x ) − F (x) m n m m il m il n il n il n

≤ F (x) − F (x ) + F (x ) − F (x ) + F (x ) − F (x) m m il m il n il n il n

Since hFmi converge on EQ and xil ∈ EQ, for the given ε, ∃N (ε, xil ) such that ε F (x ) − F (x ) < for m, n ≥ N (ε, x ) m il n il 3 il

Dr. Prasun Kumar Nayak Home Study Materiel 74 Small Overview On Uniform Convergence

n o Thus for m, n ≥ max N (ε, xil ) : 1 ≤ l ≤ k we have

|Fm(x) − Fn(x)| < ε independent of x ∈ E

⇒ hFmi converges uniformly on E.

This proves the theorem. 2

THEOREM 22. Let fn :[a, b] → R be continuous such that hfni are uniformly bounded on 0 [a, b] and the derivatives fn exist and are uniformly bounded on (a, b0. Then fn has a uniformly convergent subsequence.

0 0 Proof: Since fn are uniformly bounded on (a, b), there exists M > 0 such that |fn(x)| ≤ M, ∀n ∈ N and for any x ∈ (0, 1). Using the mean value theorem, we get

|fn(x) − fn(y)| ≤ M|x − y|; for any x, y ∈ [a, b] and n ∈ N

So, if ε > 0 is given, set δ = ε/(M + 1). Then

|fn(x) − fn(y)| ≤ M|x − y|; for any n ∈ N with |x − y| < δ

This shows that hfni is equicontinuous. Let us prove that hfni has a subsequence which converges uniformly. The Arzela-Ascoli theorem then supplies the uniformly convergent subsequence.

EXAMPLE 58. Which of the following sequences hfni in C[0, 1] must contain a uniformly convergent subsequence?

a) When {fn(x)} ≤ 3 for all x ∈ [0, 1] and for all n ∈ N

1 0 b) When fn ∈ C [0, 1], |fn(x)| ≤ 3 and |fn(x)| ≤ 5 for all x ∈ [0, 1] and for all n ∈ N

n Solution: (a) Let fn(x) = x , for C, implies ||fn|| = 1 for all n ∈ N. To prove that there is no convergent subsequence for this sequence, it is sufﬁcient to show that any subsequence of hfni is not Cauchy. (Since every convergent sequence is a Cauchy sequence). Observe that:

n 2n n n 2 2 1 ||f2n − fn|| = sup (x − x ) = sup (x − (x ) ) = sup (t − t ) = x∈[0,1] x∈[0,1] t∈[0,1] 4

Since hfni is monotonic, we see that 1 k ≥ 2n ⇒ ||f − f || ≥ . k n 4

Now, we have any subsequence of hfni then the above estimate shows that this subsequence is not 0 1 Cauchy. For any given k , we can ﬁnd k > k such that n 0 > 2n and ||f − f || ≥ . So 0 0 k k0 nk0 nk0 4 this option is incorrect. (b) Let hfni be a uniformly bounded sequence of real-valued differentiable functions on [a, b] such 0 that the derivatives hfni are uniformly bounded. Then by Arzela-Ascoli theorem, we conclude that there exists a subsequence of hfni that converges uniformly on [a, b]. Thus this option is correct.

Dr. Prasun Kumar Nayak Home Study Materiel 75 Small Overview On Uniform Convergence

Problem Set

Short answer type questions 1. If f : R → R is continuous and the sequence fn(x) = f(nx) is equicontinuous, what can be said about f?

Long answer type questions

1. Is the sequence of functions fn : R → R deﬁned by

1 2 n fn(x) = cos(n + x) + log 1 + √ sin (n x) n + 2 equicontinuous? Prove or disprove. Z ∞ 2. Let f : R → R is continuous and |f(x)|dx < ∞. Show that there is a sequence hxni −∞ in R such that xn → ∞, xnf(xn) → 0, and xnf(−xn) → 0 as n → ∞.

3. Let hfni be a sequence of functions are continuous over [0, 1] and continuously differen- 0 tiable in (0, 1). Assume that |fn(x)| ≤ 1 and that |fn(x)| ≤ 1 for all x ∈ (0, 1) and for each n ∈ N. Pick the true statements: NBHM’09

a) fn is uniformly continuous for each n Hints : Since fn is continuous on a compact set [0, 1]

b) hfni contain a subsequence which converges in C[0, 1] Hints : Arzela-Ascoli theorem n c) hfni is a convergent sequence in C[0, 1]. Hints : fn(x) = (−1)

Ans: 3 (a), (b)

4. Which of the following sequences hfni in C[0, 1] must contain a uniformly convergent subsequence? NBHM’15

a) When {fn(x)} ≤ 3 for all x ∈ [0, 1] and for all n ∈ N 1 0 b) When fn ∈ C [0, 1], |fn(x)| ≤ 3 and |fn(x)| ≤ 5 for all x ∈ [0, 1] and for all n ∈ N Z 1 1 c) When fn ∈ C [0, 1] and |fn(x)|dx ≤ 1 for all n ∈ N Hints Take the same 0 example 58 as taken in option (a) Ans: 4 (b)

References

[1] Brian S. Thomson, Judith B. Bruckner, Andrew M. Bruckner. Elementary Real Analysis. Prentice Hall (Pearson), 2001.

[2] Houshang H. Sohrab, Basic Real Analysis. Springer New York Heidelberg Dordrecht Lon- don, 2010

Dr. Prasun Kumar Nayak Home Study Materiel 76 Small Overview On Uniform Convergence

[3] Kenneth R. Davidson Allan P. Donsig, Real Analysis and Applications Theory in Practice, Springer New York Dordrecht Heidelberg London, 2000.

[4] Steven G. Krantz, Real Analysis and Foundations, CHAPMAN & HALUCRC, 2006.

[5] M.H.Protter, C.B.Morrey, A First Course in Real Analysis, Springer, 2004.

[6] Kumaresan, S., Topology of Metric Spaces, Narosa Publishing House, 2005.

[7] Charles Chapman Pugh, Real Mathematical Analysis, Springer, 2004.

Dr. Prasun Kumar Nayak Home Study Materiel