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7. Properties of Uniformly Convergent Sequences

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7. Properties of Uniformly Convergent Sequences 46 1. THE THEORY OF CONVERGENCE 7. Properties of uniformly convergent sequences Here a relation between continuity, differentiability, and Riemann integrability of the sum of a functional series or the limit of a functional sequence 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 function S···(x) because every uniformly convergent series 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 trigonometric series. 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 power series. By the stated theorem, the sum of a power series ∞ n f(z)= anz n=0 X is a continuous function 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 triangle inequality − ≥ 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 radius of convergence. 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 alternating series 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 Lebesgue integration 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].
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