6. Uniform Convergence the Next Important Question Is: What Are Conditions Under Which the Limit Function Or the Sum of a Series

Home , Absolute convergence, Uniform convergence

6. Uniform convergence The next important question is: What are conditions under which the limit function or the sum of a series inherits some properties of the terms of a sequence or series? In particular, under what conditions can b the order of operations lim (or ) and d/dx (and/or a ) be changed? This question is answered usingP the concept of uniformR convergence Definition 6.1. (Uniform convergence) A sequence of functions un is said to converge uniformly to a function u on a set D RN if { } ⊂ lim sup un(x) u(x) = 0 n→∞ D | − | A series ∞

un(x)= u(x) Xn=1 is said to converge uniformly to u(x) on a set D if the sequence of partial sums of the series converges uniformly to u(x) on D.

For example, the sequence un(x) = sin(nx)/√n converges uniformly to u(x) = 0 on R because sin(nx) 1 un(x) u(x) = | | 0 | − | √n ≤ √n → as n . →∞ Suppose un(x) converges to u(x) pointwise for each x D. If D is ∈ closed and bounded and un and the limit function are continuous on D, then by the extreme value theorem the diﬀerence un u take its | − | maximal value at some xn D. The uniform convergence means that ∈ these maximal values un(xn) u(xn) tend to zero as n . In other | − | →∞ words, the maximal error of the approximation u(x) un(x) tends to zero with increasing n. This error does not depend on≈ x, that is, it is uniform for all x D because ∈ un(x) u(x) max un u = un(xn) u(xn) , x D. | − |≤ D | − | | − | ∀ ∈ Therefore the approximation error tends to zero uniformly on D. Of course, if terms un are not continuous, or the limit function is not continuous, or D is not bounded or closed, then the error un(x) u(x) may not attain its maximal value on D, but the supremum| − always| exists and deﬁnes the least upper bound for the approximation error uniformly for all x D. This error is required to converge to zero. Not every sequence that∈ converges pointwise on D has the latter property. 40 1. THE THEORY OF CONVERGENCE

For example, the series ∞ x2 1+ x2 , x = 0 = u(x)= 6 (1 + x2)n 0 , x = 0 Xn=0 (see Example 1 in the previous section) does not converge uniformly on any interval whose closure contains x = 0. Indeed, if Sn(x) are partial sums, then as shown in the previous section 1 Sn(x) u(x) = 2 n+1 sup Sn(x) u(x) = 1 | − | (1 + x ) ⇒ I | − |

lim sup Sn(x) u(x) = 1 = 0 ⇒ n→∞ I | − | 6 for any interval (closed or open) I for which x = 0 is an endpoint. However, the series converges uniformly for 0 pointwise convergence depends only on a particular value of the argument; A sequence or series that converges uniformly on D converges • pointwise on D, and the converse is false.

6.1. The Cauchy criterion for uniform convergence. Let Sn(x) be a sequence of partial sums of a series un(x). Here x D denotes a set N ∈ of parameters (real, or complex, orP from R ). Then

Sk(x) Sn(x) sup Sk Sn , x D | − |≤ D | − | ∀ ∈ Fix ε> 0. If one can ﬁnd an integer m such that

sup Sk Sn ε, k, n m D | − |≤ ∀ ≥ then the sequence Sn(x) converges for each x D by the Cauchy criterion { } ∈ Sk(x) Sn(x) ε, k, n m | − |≤ ∀ ≥ and there exists a limit function lim Sn(x) = S(x). Since above inequality holds for all n m and all x D, one can take the limit n in the right side≥ to obtain ∈ →∞ Sk(x) S(x) ε, k m, x D | − |≤ ∀ ≥ ∀ ∈ 6. UNIFORM CONVERGENCE 41

This inequality holds for all x and, hence, one can take the supremum in the right side to infer that

sup Sk(x) S(x) ε, k m, D | − |≤ ∀ ≥ which means that

lim sup Sk(x) S(x) = 0 k→∞ D | − | and, hence, the sequence Sn converges uniformly to S(x). This proves the Cauchy criterion for{ uniform} convergence:

A series n u(x) converges uniformly if and only if for any ε > 0 there existsP an integer m such that

k sup uj(x) ε, k n m D ≤ ∀ ≥ ≥ Xj=n

Note that, just like in the case of pointwise convergence, the Cauchy criterion does not require any knowledge about the sum of the series. It is a necessary and suﬃcient condition for the uniform convergence. Unfortunately, it is not always easy to verify it in practice. It turns out that in many instances the question about uniform convergence of a series can be reduced to the equation about convergence of a numerical series.

6.2. Suﬃcient condition for uniform convergence.

Theorem 6.1. (Weierstrass) Suppose that un(x) Mn for all x D and n = 1, 2,.... If the series | |≤ ∈ of the upper bounds Mn < converges, then the series un(x) ∞ n converges uniformly onP D. P So if the terms of the series are uniformly bounded for all values of parameters x, and the numerical series of the bounds converges, then the series converges uniformly. If the series satisﬁes the stated hypotheses, then it satisﬁes the Cauchy criterion for uniform convergence, which follows from the inequality

k k uj(x) Mj , x D ≤ ∀ ∈ Xj=n Xj=n

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Indeed, since the inequality holds for all x, it implies that k k sup uj(x) Mj . D ≤ Xj=n Xj=n but the right side of the latter inequality can be made arbitrary small for all suﬃciently large n and k because Mn < (the series of ∞ bounds converges). Thus, the series convergesP uniformly by the Cauchy criterion.

∞ n 6.3. Uniform convergence of a power series. a power series n=0 anz converges uniformly on any disk z b < R, where R is theP radius of | |≤ n convergence of the power series. Indeed, put un(z)= anz . Then n n un(z) = an z an b = Mn , z b. | | | || | ≤| | ∀| |≤ The series of upper bounds converges Mn < by the root test: n ∞ P n n b lim sup Mn = b lim sup an = < 1 n→∞ n→∞ R p p| | where the deﬁnition of the radius of convergence has been used. There- fore the power series converges uniformly in any disk of radius less than the radius of convergence.

6.4. Uniform convergence of trigonometric series. Since cos(nx) 1 , sin(nx) 1 | |≤ | |≤ The trigonometric series ∞ 1 a + a cos(nx)+ b sin(nx) 2 0 n n Xn=1 if the series of coeﬃcients converge absolutely:

an < , bn < X | | ∞ X | | ∞ For example, the series ∞ einx n2p + 1 nX=−∞ 1 converges uniformly for all real x and p> 2 because einx 1 1 1 1 and < 1+2 < , p> n2p + 1 ≤ n2p + 1 n2p + 1 n2p ∞ 2 X Xn≥1

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6.5. Dirichlet’s tests for uniform convergence. The Weierstrass test for uniform convergence resembles tests for absolute convergence. So, one might guess that it would fail to detect uniform convergence for con- ditionally convergent series. This turns out be the case. There are a few extensions of Dirichlet’s and Abel’s tests for investigating uniform convergence. Theorem 2. n 6. Let the sequence An(x)= k=1 ak(x), be bounded, P An(x) M, x D | |≤ ∀ ∈ where M is independent of x and n, and the real sequence bn(x) is decreasing monotonically and converges to zero uniformly on{D: }

bn(x) bn+1(x) , x D and lim sup bn(x) = 0 ≥ ∀ ∈ n→∞ D then the series

an(x)bn(x) X converges uniformly on D. This is an extension of Dirichlet’s test to the case of uniform convergence (it is often called Dirichlet’s test for uniform convergence). A proof is based on the same identity as the proof of Dirichlet’s test: m m−1

an(x)bn(x) = An(x)(bn(x) bn (x)) − +1 Xn=k Xn=k +Ak(x)bk(x) Am− (x)bm(x) − 1 Therefore it follows from the above identity and non-negativity and monotonicity of bn(x) that (cf. the proof of Dirichlet’s test) m an(x)bn(x) 2Mbk(x) 2M sup bk , x D ≤ ≤ D ∀ ∈ Xn=k

The inequality holds for all x in D and therefore, by taking the supremum of the left side, m sup an(x)bn(x) 2M sup bk D ≤ D Xn=k

By the hypothesis, the right side of this inequality tends to zero and, hence, can be made arbitrary small for all suﬃciently large k. This implies that the series in question satisﬁes the Cauchy criterion for uniform convergence. This completes the proof. 44 1. THE THEORY OF CONVERGENCE

Example. The trigonometric series

∞ einx n Xn=1 converges uniformly on any interval [δ, 2π δ], π>δ> 0. Note that the Weierstrass test would fail to detect uniform− convergence because

einx 1 1 = , = n n n ∞ X

inx Put an(x)= e . Then (see Section 4)

inx ix 2ix inx ix 1 e An(x) = e + e + + e = e − ··· 1 eix − 1 einx 1 cos(nx) 1 An(x) = − = − | | 1 eix r 1 cos x ≤ sin(x/2) − − | | 1 An(x) M = , x D =[δ, 2π δ] | | ≤ sin(δ/2) ∀ ∈ − | where M is just the maximal value of 1/ sin(x/2) on the interval D. 1 | | Put bn(x)= . So, bn 0 monotonically and uniformly on any interval n → (because bn are independent of x). Thus, the hypotheses of Dirichlet’s test for uniform convergence are fulﬁlled and the trigonometric series in question converges uniformly on the said interval.

6.6. Exercises.

n 1. Let D = [0, ) and un(x)= √x. ∞ (i). Show that limn→∞ un(x) = u(x) where u(x) = 1if x > 0 and u(0) = 0. (ii). Show that the convergence in not uniform on D. To do so, show that the approximation error un(x) u(x) cannot be made arbitrary small simultaneously for all x| 0 and− all| suﬃciently large n. Hint: put x = e−n > 0. ≥

2. Show that each of the following trigonometric series converges uniformly on any closed interval [a,b] that lies in the open interval (0, 2π) if 0 < p 1 and, if p> 1, then they converge uniformly on the whole ≤ 6. UNIFORM CONVERGENCE 45 real axis: ∞ sin(nx) (i) np Xn=1 ∞ cos(nx) (ii) np Xn=1 3. Find an interval in which the following trigonometric series converge uniformly: ∞ ( 1)n sin(nx) (i) − np Xn=1 ∞ ( 1)n cos(nx) (ii) − np Xn=1 where p> 0.

4 . Put 1 2n−1 an(x)= x , n = 1, 2,..., x R ∈ (i). Show that the series ∞

an(x) an (x) − +1 Xn=1 converges pointwise on R. (ii) Find its sum and show that it is not continuous at x = 0. (iii) Show that the series does not converge uniformly in any interval containing x = 0. Hint: Find partial sums Sn(x). Find S(x) = limn→∞ Sn(x). Show that the approximation error S(x) Sn(x) cannot be made arbitrary small for all x in the said interval| for− all large| enough n. For example, eval- uate the error at x = e−(2n−1).

5. Repeat the analysis (i) and (ii) of Problem 3 if 1 (1 + z)n an(z)= − , z C 1+(1+ z)n ∈ Investigate the uniform convergence of the series in any disk that contains z = 0.