46 1. THE THEORY OF CONVERGENCE

7. Properties of uniformly convergent Here a relation between continuity, differentiability, and Riemann integrability of the sum of a functional or the limit of a functional and uniform convergence is studied.

7.1. Uniform convergence and continuity. Theorem 7.1. (Continuity of the sum of a series) N The sum of the series un(x) of terms continuous on D R is continuous if the series converges uniformly on D. ⊂ P Let Sn(x)= u1(x)+u2(x)+ +un(x) be a sequence of partial sum. It converges to some S···(x) because every uniformly converges pointwise. Continuity of S at a point x means (by definition) lim S(y)= S(x) y→x Fix a number ε> 0. Then one can find a number δ such that S(x) S(y) <ε whenever 0 < x y <δ | − | | − | In other words, the values S(y) can get arbitrary close to S(x) and stay arbitrary close to it for all points y = x that are sufficiently close to x. Let us show that this condition follows6 from the hypotheses. Owing to the uniform convergence of the series, given ε > 0, one can find an integer m such that ε S(x) Sn(x) sup S Sn , x D, n m | − |≤ D | − |≤ 3 ∀ ∈ ∀ ≥

Note that m is independent of x. By continuity of Sn (as a finite sum of continuous functions), for n m, one can also find a number δ > 0 such that ≥ ε Sn(x) Sn(y) < whenever 0 < x y <δ | − | 3 | − | So, given ε> 0, the integer m is found. Then for an integer n m, ε, and a point x D, the number δ is found such that ≥ ∈ S(x) S(y) = S(x) Sn(x)+ Sn(x) Sn(y)+ Sn(y) S(y) | − | | − − − | S(x) Sn(x) + Sn(x) Sn(y) + Sn(y) S(y) ≤ | − | | − | | − | ε whenever 0 < x y <δ ≤ | − | This means that S is continuous at x, but the above argument holds for any x in D as n does not depend on x. Therefore the sum S(x) is continuous on D. 7. PROPERTIES OF UNIFORMLY CONVERGENT SEQUENCES 47

The same argument can be used to show that the limit function of a uniformly convergent sequence of continuous functions is continuous: lim sup u u = 0 lim lim u (y)= lim lim u (y) →∞ n → →∞ n →∞ → n n D | − | ⇒ y x n n y x for any x D. In other words, the order of limits with respect to the index and∈ parameters of the sequence can be changed, provided the sequence converges uniformly.

Continuity of . The sum of an absolutely convergent trigonometric series

inx f(x)= ane , an < , n n | | ∞ X X is continuous for all real x. Indeed, by Weierstrass test, an absolutely convergent trigonometric series also converges uniformly on R. There- fore its sum is continuous because the terms einx are continuous on R. By Dirichlet’s test for uniform convergence, the series ∞ einx n n=1 X converges uniformly on any closed interval that does not contain points 2πm where m = 0, 1, 2,.... Therefore its sum is continuous at any point where it exists.± Note± that the series converges everywhere except x = 2πm.

7.2. Continuity of a . By the stated theorem, the sum of a power series ∞ n f(z)= anz n=0 X is a in any disk of radius b < R where R is the ra- dius of convergence, because it converges uniformly on it (see Section 6.3) and terms of the series are continuous (power functions) every- where. Suppose the series converges at a point on the circle z = R (the boundary of the convergence disk). Is f continuous at this| point?| The answer is affirmative for a power series of a real variable x. Theorem 7.2. (Abel) Suppose a series of complex terms an converges. Then the power P 48 1. THE THEORY OF CONVERGENCE

n series anx converges uniformly on the interval [0, 1] and n P lim anx = an x→1− X X First note that if the series an converges absolutely,

Pan < , | | ∞ then the conclusion of Abel’sX theorem follows from the Weierstrass test n for uniform convergence because anx an if 0 x 1. Abel’s | |≤| | ≤ ≤ theorem is less obvious if an converges conditionally. By the Cauchy criterion, the convergence of an means that for any preassigned (arbitraryP small) number ε> 0, one can find an integer N such that P

am + am + + am p ε, p 0 , m N | +1 ··· + |≤ ∀ ≥ ∀ ≥ The uniform convergence of the power series and, hence, its continuity on [0, 1], would follow from the Cauchy criterion for uniform conver- gence if m m+1 m+p amx + am x + + am px ε, | +1 ··· + |≤ p 0 , m N, x [0, 1] ∀ ≥ ∀ ≥ ∀ ∈ Let us show that the second inequality follows from the first one. Put

Ap = am + am + + am p , p = 0, 1, 2,... +1 ··· + Then the first inequality is restated as

Ap ε, p 0 | |≤ ∀ ≥ By factoring out xm and using that x 1, ≤ m m+p p amx + + am px am + + am px | ··· + |≤| ··· + | Using the identity am p = Ap Ap− , one infers that + − 1 p p am + + am+px = A0 +(A1 A0)x + +(Ap Ap−1)x ··· 2 − ···p−1 p − p = A (1 x)+ A (x x )+ + Ap− (x x )+ Apx 0 − 1 − ··· 1 − On the interval [0, 1], the terms xp−1 xp 0 are non-negative. There- fore by the − ≥ p am + + am+px | ··· | 2 p−1 p p A 1 x + A x x + + Ap− x x + Ap x ≤| 0|| − | 1|| − | ··· | 1|| − | | || | ε(1 x)+ ε(x x2)+ + ε(xp−1 xp)+ εxp ≤ − − ··· − = ε, p 0 , x [0, 1] ∀ ≥ ∀ ∈ and the desired inequality follows. 7. PROPERTIES OF UNIFORMLY CONVERGENT SEQUENCES 49

Abel’s theorem allows us to calculate sums of some conditionally convergent numerical series. For example, ∞ 1 = ( 1)nxn , x < 1 1+ x − | | n=0 X In the next section it will be proved that a power series can be inte- grated term-by-term and the series of integrals converges to the integral of the sum and has the same . By applying this assertion to the above geometric series, ∞ x dt x = ( 1)n tn dt , x < 1 1+ t − | | Z0 n=0 Z0 X∞ ( 1)n+1 ln(1 + x) = − xn , x < 1 n | | n=1 X The alternating harmonic series ∞ ( 1)n+1 − n n=1 X converges by the test. Therefore by Abel’s theorem and continuity of the logarithm function, the sum of the alternating harmonic series is obtained: ∞ ( 1)n+1 − = lim ln(1 + x) = ln(2) n x→1− n=1 X Suppose that f(z), z C, is defined by a power series that converges for all z < R where R is∈ the radius of convergence. Suppose that the power| series| converges for some z , z = R. Then by Abel’s theorem 0 | 0| ∞ ∞ n n lim f(z0x)= lim an(z0x) = anz0 x→1− x→1− n=0 n=0 X X In other words, f(z) is continuous along the straight line segment con- necting z = 0 and z = z0. The latter does not imply that limz→z0 f(z) even exists! A neighborhood of z0 in the disk of convergence contains infinitely many paths leading to z0. Generally, there can exist paths along which the limit would fail to exist. It can be proved that if z z0 within the so called Stolz sector, defined by the condition →

z z M z z | 0 − |≤ | 0|−| |   50 1. THE THEORY OF CONVERGENCE for some M, then ∞ n lim f(z)= anz0 . z→z0 n=0 X Any z in the interior of the convergence disk can be written in the form iθ z = z0xe where 0 x< 1 and θ is the difference of phases of z and z0. The limit z z means≤ taking the two-variable limit (x,θ) (1−, 0). → 0 → A path leading to z0 can be defined by a continuous function θ = θ(x) with the property that θ(x) 0 as x 1−. If θ(x) = 0, then the path → → is the line segment from 0 to z0. The Stolz sector is then defined by 1 xeiθ M(1 x) | − |≤ − This condition basically determines how fast θ(x) should approach 0 as x 1−. Indeed, using the identity → 1 xeiθ = 1 2x cos(θ)+ x2 = (1 x)2 + x sin2(θ/2) | − | − − q the Stolz conditionp can be restated as x sin2(θ/2) 1+ M 2 (1 x)2 ≤ − If θ(x) = O((1 x)p), p > 0, then the Stolz condition is fulfilled only if p 1. For− example, any path for which θ(x) √1 x, does ≥ ∼ − not approach z0 from within the Stolz sector (because the left side of the inequality is not bounded). The power series is guaranteed to be continuous along any path leading to z0 that lies within the Stolz sector. So, it is a sufficient, not necessary condition for continuity of a power series.

7.3. Uniform convergence and Riemann integrability. Here only one-di- mensional integration is discussed. A general case will be studied later in the framework of the theory. Suppose that a series of continuous terms ∞

S(x)= un(x) , a x b ≤ ≤ n=1 X converges uniformly on [a,b]. Then its sum S(x) is continuous on [a,b] and, hence, integrable on [a,b]. Moreover,

b b ∞ ∞ b (7.1) S(x) dx = un(x) dx = un(x) dx a a n=1 ! n=1 a Z Z X X Z 7. PROPERTIES OF UNIFORMLY CONVERGENT SEQUENCES 51 that is, a uniformly convergent series can be integrated term-by-term. Indeed, let Sn(x)= u1(x)+ + un(x) be a sequence of partial sums. Then ···

b b S(x) Sn(x) dx S(x) Sn(x) dx − ≤ | − | Za Za   b

sup S Sn =(b a) sup S Sn ≤ | − | − | − | [a,b] Za [a,b]

This shows that the left side converges to zero because sup S Sn 0 as n by uniform convergence of the series, which means| − that|→ Eq. (7.1)→∞ holds. One can also established a more general result.

Theorem 7.3. Let un be a sequence of functions that are Rie- { } mann integrable on [a,b] such that the series un converges uniformly on [a,b]. Then the sum of the series is Riemann integrable on [a,b] and Eq. (7.1) holds. P

A major part of the proof is to show that the sum of a uniformly con- vergent series with integrable terms is a Riemann integrable function5. In the previous argument, the integrability followed from continuity, but there are integrable functions that are not continuous, and, hence, the integrability of the sum is not obvious if the terms are not contin- uous. Once the integrability is established, Equation (7.1) is proved in the same way as above.

Integration of a power series. A power series of a real variable x that converges for x < R can be integrated term-by-term over any interval R

5see, e.g., W. Rudin, Principles of , Chapter 7 52 1. THE THEORY OF CONVERGENCE

However, the power series for sin(x) has infinite radius of convergence and so does the power series for the integrand: ∞ ∞ sin(x) 1 ( 1)nx2n+1 ( 1)nx2n = − = − , x x (2n + 1)! (2n + 1)! n=0 n=0 ∞X X π sin(x) ( 1)nπ2n+1 dx = − x (2n + 1)!(2n + 1) 0 n=0 Z X 7.4. Weierstrass theorem. An approximation of a continuous function by a polynomial is a standard problem in the approximation theory. Naturally, an important question here is an assessment of the error of the approximation. The answer was first given by Weierstrass. Theorem 7.4. (Weierstrass) If f is a real or complex continuous function on [a,b], there exists a sequence of polynomials Pn that converges to f uniformly on [a,b]: lim sup f(x) P (x) = 0 →∞ n n [a,b] | − |

If f is real, the polynomials Pn may be taken real.

6 A proof is based on an explicit construction of Pn for a given f.

7.5. Exercises.

1. Let ∞ 1 f(x)= , x> 0 1+ n2x n=1 X (i) Does the series converge uniformly on (0, )? (ii) On what intervals does the series converge uniformly? ∞ (iii) Is f continuous wherever the series converges? (iv) If f bounded?

2. Let ∞ sin(nx)+ n f(x)= ( 1)n − n2 n=1 X (i) Show that f(x) is continuous everywhere; (ii) Does the series converge absolutely?

6see, e.g., W. Rudin, Principles of mathematical analysis, Chapter 7. 7. PROPERTIES OF UNIFORMLY CONVERGENT SEQUENCES 53

3. Show the sums of the following series are continuous in (0, 2π) if 0 < p 1: ≤ ∞ sin(nx) (i) np n=1 X∞ cos(nx) (ii) np n=1 X Find the largest interval of continuity of the sums of the above series if p> 1.

4. Let n u (x)= , n = 1, 2,... n 1+ x2n2 (i) Find the pointwise limit u(x) of the sequence un . (ii) Find { } b lim un(x) dx n→∞ Za for any interval [a,b]. Is the equation b b lim un(x) dx = lim un(x) dx n→∞ n→∞ Za Za true or false?

5. Suppose that a sequence of continuous non-negative functions un(x) on [0, ) has the following properties ∞ un u uniformly on [0,b] • → un(x) g(x) , x [0, ) • ∞ ≤ ∀ b ∈ ∞ g(x)= lim g(x) dx < • b→∞ ∞ Z0 Z0 Show that the improper Riemann integral of u over [0, ) exists and ∞ ∞ b ∞ u(x)= lim u(x) dx = lim un(x) dx →∞ →∞ 0 b 0 n 0 Z Z x Z Hint: Consider the sequence vn(x) = 0 un(t) dt. Investigate its con- vergence and find its limit function v in terms of u. Investigate the limit of v(x) as x . R →∞