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Uniform Convergence

Module by: Lawrence Baggett. E-mail the author

Summary: We introduce now two different notions of the limit of a sequence of functions. Some definitions cover uniform convergence and pointwise convergence. The Weierstrass M-Test is stated and proven, as well the theorem of Abel.

We introduce now two different notions of the limit of a sequence of functions. Let SS be a set of complex numbers, and let {fn}{fn} be a sequence of complex-valued functions each having domain S.S.

Definition 1:

We say that the sequence {fn}{fn}converges or converges pointwise to a function f:S→Cf:S→C if for every x∈Sx∈S and every ϵ>0ϵ>0 there exists a natural number N,N, depending on xx and ϵ,ϵ, such that for every n≥N,n≥N,|fn(x)-f(x)|<ϵ.|fn(x)-f(x)|<ϵ. That is, equivalently, {fn}{fn} converges pointwise to ff if for every x∈Sx∈S the sequence {fn(x)}{fn(x)} of numbers converges to the number f(x).f(x).

We say that the sequence {fn}{fn}converges uniformly to a function ff if for every ϵ>0,ϵ>0, there exists an N,N, depending only on ϵ,ϵ, such that for every n≥Nn≥N and every x∈S,x∈S,|fn(x)-f(x)|<ϵ.|fn(x)-f(x)|<ϵ.

If {un}{un} is a sequence of functions defined on S,S, we say that the infinite series ∑un∑unconverges uniformly if the sequence {SN=∑n=0Nun}{SN=∑n=0Nun} of partial sums converges uniformly.

These two definitions of convergence of a sequence of functions differ in subtle ways. Study the word order in the definitions.

Exercise 1

  1. Prove that if a sequence {fn}{fn} of functions converges uniformly on a set SS to a function ff then it converges pointwise to f.f.
  2. Let S=(0,1),S=(0,1), and for each nn define fn(x)=xn.fn(x)=xn. Prove that {fn}{fn} converges pointwise to the zero function, but that {fn}{fn} does not converge uniformly to the zero function. Conclude that pointwise convergence does not imply uniform convergence. HINT: Suppose the sequence does converge uniformly. Take ϵ=1/2,ϵ=1/2, let NN be a corresponding integer, and consider xx's of the form x=1-hx=1-h for tiny hh's.
  3. Suppose the sequence {fn}{fn} converges uniformly to ff on S,S, and the sequence {gn}{gn} converges uniformly to gg on S.S. Prove that the sequence {fn+gn}{fn+gn} converges uniformly to f+gf+g on S.S.
  4. Suppose {fn}{fn} converges uniformly to ff on S,S, and let cc be a constant. Show that {cfn}{cfn} converges uniformly to cfcf on S.S.
  5. Let S=R,S=R, and set fn(x)=x+(1/n).fn(x)=x+(1/n). Does {fn}{fn} converge uniformly on S?S? Does {fn2}{fn2} converge uniformly on S?S? What does this say about the limit of a product of uniformly convergent sequences versus the product of the limits?
  6. Suppose aa and bb are nonnegative real numbers and that |a-b|<ϵ2.|a-b|<ϵ2. Prove that |a-b|<2ϵ.|a-b|<2ϵ. HINT: Break this into cases, the first one being when both aa and bb are less than ϵ.ϵ.
  7. Suppose {fn}{fn} is a sequence of nonnegative real-valued functions that converges uniformly to ff on S.S. Use part (f) to prove that the sequence {fn}{fn} converges uniformly to f.f.
  8. For each positive integer n,n, define fnfn on (-1,1)(-1,1) by fn(x)=|x|1+1/n.fn(x)=|x|1+1/n. Prove that the sequence {fn}{fn} converges uniformly on (-1,1)(-1,1) to the function f(x)=|x|.f(x)=|x|. HINT: Let ϵ>0ϵ>0 be given. Consider |x||x|'s that are <ϵ<ϵ and |x||x|'s that are ≥ϵ.≥ϵ. For |x|<ϵ,|x|<ϵ, show that |fn(x)-f(x)|<ϵ|fn(x)-f(x)|<ϵ for all n.n. For |x|≥ϵ,|x|≥ϵ, choose NN so that |ϵ1/n-1|<ϵ.|ϵ1/n-1|<ϵ. How?

Exercise 2

Let {fn}{fn} be a sequence of functions on a set S,S, let ff be a function on S,S, and suppose that for each nn we have |f(x)-fn(x)|<1/n|f(x)-fn(x)|<1/n for all x∈S.x∈S. Prove that the sequence {fn}{fn} converges uniformly to f.f.

We give next four important theorems concerning uniform convergence. The first of these theorems is frequently used to prove that a given function is continuous. The theorem asserts that if ff is the uniform limit of a sequence of continuous functions, then ff is itself continuous.

Theorem 1: The uniform limit of continuous functions is continuous.

Suppose {fn}{fn} is a sequence of continuous functions on a set S⊆C,S⊆C, and assume that the sequence {fn}{fn} converges uniformly to a function f.f. Then ff is continuous on S.S.

Proof

This proof is an example of what is called by mathematicians a “3ϵ3ϵ argument.”

Fix an x∈Sx∈S and an ϵ>0.ϵ>0. We wish to find a δ>0δ>0 such that if y∈Sy∈S and |y-x|<δ|y-x|<δ then |f(y)-f(x)|<ϵ.|f(y)-f(x)|<ϵ.

We use first the hypothesis that the sequence converges uniformly. Thus, given this ϵ>0,ϵ>0, there exists a natural number NN such that if n≥Nn≥N then |f(z)-fn(z)|<ϵ/3|f(z)-fn(z)|<ϵ/3 for all z∈S.z∈S. Now, because fNfN is continuous at x,x, there exists a δ>0δ>0 such that if y∈Sy∈S and |y-x|<δ|y-x|<δ then |fN(y)-fN(x)|<ϵ/3.|fN(y)-fN(x)|<ϵ/3. So, if y∈Sy∈S and |y-x|<δ,|y-x|<δ, then

| f ( y ) - f ( x ) | = | f ( y ) - f N ( y ) + f N ( y ) - f N ( x ) + f N ( x ) - f ( x ) | ≤ | f ( y ) - f N ( y ) | + | f N ( y ) - f N ( x ) | + | f N ( x ) - f ( x ) | < ϵ 3 + ϵ 3 + ϵ 3 = ϵ . | f ( y ) - f ( x ) | = | f ( y ) - f N ( y ) + f N ( y ) - f N ( x ) + f N ( x ) - f ( x ) | ≤ | f ( y ) - f N ( y ) | + | f N ( y ) - f N ( x ) | + | f N ( x ) - f ( x ) | < ϵ 3 + ϵ 3 + ϵ 3 = ϵ .
(1)

This completes the proof.

REMARK Many properties of functions are preserved under the taking of uniform limits, e.g., continuity, as we have just seen. However, not all properties are preserved under this limit process. Differentiability is not, integrability is sometimes, being a power series function is, and so on. We must be alert to be aware of when it works and when it does not.

Theorem 2: Weierstrass M-Test

Let {un}{un} be a sequence of complex-valued functions defined on a set S⊆C.S⊆C. Write SNSN for the partial sum SN(x)=∑n=0Nun(x).SN(x)=∑n=0Nun(x). Suppose that, for each n,n, there exists an Mn>0Mn>0 for which |un(x)|≤Mn|un(x)|≤Mn for all x∈S.x∈S. Then

  1. If ∑Mn∑Mn converges, then the sequence {SN}{SN} converges uniformly to a function S.S. That is, the infinite series ∑un∑un converges uniformly.
  2. If each function unun is continuous, and ∑Mn∑Mn converges, then the function SS of part (1) is continuous.

Proof

Because ∑Mn∑Mn is convergent, it follows from the Comparison Test that for each x∈Sx∈S the infinite series ∑n=0∞un(x)∑n=0∞un(x) is absolutely convergent, hence convergent. Define a function SS by S(x)=∑n=0∞un(x)=limSN(x).S(x)=∑n=0∞un(x)=limSN(x).

To show that {SN}{SN} converges uniformly to S,S, let ϵ>0ϵ>0 be given, and choose a natural number NN such that ∑n=N+1∞Mn<ϵ.∑n=N+1∞Mn<ϵ. This can be done because ∑Mn∑Mn converges. Now, for any x∈Sx∈S and any m≥N,m≥N, we have

| S ( x ) - S m ( x ) | = | lim k → ∞ S k ( x ) - S m ( x ) | = | lim k → ∞ ( S k ( x ) - S m ( x ) ) | = lim k → ∞ | S k ( x ) - S m ( x ) | = lim k → ∞ | ∑ n = m + 1 k u n ( x ) | ≤ lim k → ∞ ∑ n = m + 1 k | u n ( x ) | ≤ lim k → ∞ ∑ n = m + 1 k M n = ∑ n = m + 1 ∞ M n ≤ ∑ n = N ∞ M n < ϵ . | S ( x ) - S m ( x ) | = | lim k → ∞ S k ( x ) - S m ( x ) | = | lim k → ∞ ( S k ( x ) - S m ( x ) ) | = lim k → ∞ | S k ( x ) - S m ( x ) | = lim k → ∞ | ∑ n = m + 1 k u n ( x ) | ≤ lim k → ∞ ∑ n = m + 1 k | u n ( x ) | ≤ lim k → ∞ ∑ n = m + 1 k M n = ∑ n = m + 1 ∞ M n ≤ ∑ n = N ∞ M n < ϵ .
(2)

This proves part (1).

Part (2) now follows from part (1) and Theorem 1, since the SNSN's are continuous.

Theorem 3

Let f(z)=∑n=0∞anznf(z)=∑n=0∞anzn be a power series function with radius of convergence r>0,r>0, and let {SN(z)}{SN(z)} denote the sequence of partial sums of this series:

S N ( z ) = ∑ n = 0 N a n z n . S N ( z ) = ∑ n = 0 N a n z n .
(3)

If 0<r'<r,0<r'<r, then the sequence {SN}{SN} converges uniformly to ff on the diskBr'(0).Br'(0).

Proof

Define a power series function gg by g(z)=∑n=0∞|an|zn,g(z)=∑n=0∞|an|zn, and note that the radius of convergence for gg is the same as that for f,f, i.e., r.r. Choose tt so that r'<t<r.r'<t<r. Then, since tt belongs to the disk of convergence of the power series function g,g, we know that ∑n=0∞|an|tn∑n=0∞|an|tn converges. Set mn=|an|tn,mn=|an|tn, and note that ∑mn∑mn converges. Now, for each z∈Br'(0),z∈Br'(0), we have that

| a n z n | ≤ | a n | r ' n ≤ | a n | t n = m n , | a n z n | ≤ | a n | r ' n ≤ | a n | t n = m n ,
(4)

so that the infinite series ∑anzn∑anzn converges uniformly on Br'(0)Br'(0) by the Weierstrass M-Test.

Exercise 3

Let f(z)=∑n=0∞zn.f(z)=∑n=0∞zn. Recall that the radius of convergence for ff is 1. Verify that the sequence {SN}{SN} of partial sums of this power series function fails to converge uniformly on the full open disk of convergence B1(0),B1(0), so that the requirement that r'<rr'<r is necessary in the preceding theorem.

The next theorem shows that continuous, real-valued functions on closed bounded intervals are uniform limits of step functions. Step functions have not been mentioned lately, since they aren't continuous functions, but this next theorem will be crucial for us when we study integration in (Reference).

Theorem 4

Let ff be a continuous real-valued function on the closed and bounded interval [a,b].[a,b]. Then there exists a sequence {hn}{hn} of step functions on [a,b][a,b] that converges uniformly to f.f.

Proof

We use the fact that a continuous function on a compact set is uniformly continuous ((Reference)).

For each positive integer n,n, let δnδn be a positive number satisfying |f(x)-f(y)|<1/n|f(x)-f(y)|<1/n if |x-y|<δn.|x-y|<δn. Such a δnδn exists by the uniform continuity of ff on [a,b].[a,b]. Let Pn={x0<x1<...<xmn}Pn={x0<x1<...<xmn} be a partition of [a,b][a,b] for which xi-xi-1<δnxi-xi-1<δn for all 1≤i≤mn.1≤i≤mn. Define a step function hnhn on [a,b][a,b] as follows:

If xi-1≤x<xi,xi-1≤x<xi, then hn(x)=f(xi-1).hn(x)=f(xi-1). This defines hn(x)hn(x) for every x∈[a,b),x∈[a,b), and we complete the definition of hnhn by setting hn(b)=f(b).hn(b)=f(b). It follows immediately that hnhn is a step function.

Now, we claim that |f(x)-hn(x)|<1/n|f(x)-hn(x)|<1/n for all x∈[a,b].x∈[a,b]. This is clearly the case for x=b,x=b, since f(b)=hn(b)f(b)=hn(b) for all n.n. For any other x,x, let ii be the unique index such that xi-1≤x<xi.xi-1≤x<xi. Then

| f ( x ) - h n ( x ) | = | f ( x ) - f ( x i - 1 ) | < 1 / n | f ( x ) - h n ( x ) | = | f ( x ) - f ( x i - 1 ) | < 1 / n
(5)

because |x-xi-1|<δn.|x-xi-1|<δn.

So, we have defined a sequence {hn}{hn} of step functions, and the sequence {hn}{hn} converges uniformly to ff by Exercise 2.

We close this chapter with a famous theorem of Abel concerning the behavior of a power series function on the boundary of its disk of convergence. See the comments following (Reference).

Theorem 5: Abel

Suppose f(z)=∑n=0∞anznf(z)=∑n=0∞anzn is a power series function having finite radius of convergence r>0,r>0, and suppose there exists a point z0z0 on the boundary of Br(0)Br(0) that is in the domain of f;f; i.e., ∑anz0n∑anz0n converges to f(z0).f(z0). Suppose gg is a continuous function whose domain contains the open disk Br(0)Br(0) as well as the point z0,z0, and assume that f(z)=g(z)f(z)=g(z) for all zz in the open disk Br(0).Br(0). Then f(z0)f(z0) must equal g(z0).g(z0).

Proof

For simplicity, assume that r=1r=1 and that z0=1.z0=1. See the exercise that follows this proof. Write SnSn for the partial sum of the anan's: Sn=∑n=0nan.Sn=∑n=0nan. In the following computation, we will use the Abel Summation Formula in the form

∑ n = 0 N a n z n = S N z N + ∑ n = 0 N - 1 S n ( z n - z n + 1 ) . ∑ n = 0 N a n z n = S N z N + ∑ n = 0 N - 1 S n ( z n - z n + 1 ) .
(6)

See (Reference). Let ϵϵ be a positive number. Then, for any 0<t<10<t<1 and any positive integer N,N, we have

| g ( 1 ) - f ( 1 ) | = | g ( 1 ) - f ( t ) + f ( t ) - ∑ n = 0 N a n t n + ∑ n = 0 N a n t n - f ( 1 ) | ≤ | g ( 1 ) - g ( t ) | + | f ( t ) - ∑ n = 0 N a n t n | + | ∑ n = 0 N a n t n - f ( 1 ) | ≤ | g ( 1 ) - g ( t ) | + | f ( t ) - ∑ n = 0 N a n t n | + | S N t N + ∑ n = 0 N - 1 S n ( t n - t n + 1 ) - f ( 1 ) | = | g ( 1 ) - g ( t ) | + | f ( t ) - ∑ n = 0 N a n t n + | S N t N + ∑ n = 0 N - 1 ( S n - S N ) ( t n - t n + 1 ) + S N ∑ n = 0 N - 1 ( t n - t n + 1 ) - f ( 1 ) | = | g ( 1 ) - g ( t ) | + | f ( t ) - ∑ n = 0 N a n t n + | ∑ n = 0 N - 1 ( S n - S N ) ( t n - t n + 1 ) + S N ( t N + ∑ n = 0 N - 1 ( t n - t n + 1 ) ) - f ( 1 ) | ≤ | g ( 1 ) - g ( t ) | + | f ( t ) - ∑ n = 0 N a n t n | + | ∑ n = 0 N - 1 ( S n - S N ) ( t n - t n + 1 ) | + | S N - f ( 1 ) | ≤ | g ( 1 ) - g ( t ) | + | f ( t ) - ∑ n = 0 N a n t n | + | ∑ n = 0 P ( S n - S N ) ( t n - t n + 1 ) | + | ∑ n = P + 1 N - 1 ( S n - S N ) ( t n - t n + 1 ) | + | S N - f ( 1 ) | ≤ | g ( 1 ) - g ( t ) | + | f ( t ) - ∑ n = 0 N a n t n | + | ∑ n = 0 P ( S n - S N ) ( t n - t n + 1 ) | + ∑ n = P + 1 N - 1 | S n - S N | ( t n - t n + 1 ) + | S N - f ( 1 ) | = t 1 + t 2 + t 3 + t 4 + t 5 . | g ( 1 ) - f ( 1 ) | = | g ( 1 ) - f ( t ) + f ( t ) - ∑ n = 0 N a n t n + ∑ n = 0 N a n t n - f ( 1 ) | ≤ | g ( 1 ) - g ( t ) | + | f ( t ) - ∑ n = 0 N a n t n | + | ∑ n = 0 N a n t n - f ( 1 ) | ≤ | g ( 1 ) - g ( t ) | + | f ( t ) - ∑ n = 0 N a n t n | + | S N t N + ∑ n = 0 N - 1 S n ( t n - t n + 1 ) - f ( 1 ) | = | g ( 1 ) - g ( t ) | + | f ( t ) - ∑ n = 0 N a n t n + | S N t N + ∑ n = 0 N - 1 ( S n - S N ) ( t n - t n + 1 ) + S N ∑ n = 0 N - 1 ( t n - t n + 1 ) - f ( 1 ) | = | g ( 1 ) - g ( t ) | + | f ( t ) - ∑ n = 0 N a n t n + | ∑ n = 0 N - 1 ( S n - S N ) ( t n - t n + 1 ) + S N ( t N + ∑ n = 0 N - 1 ( t n - t n + 1 ) ) - f ( 1 ) | ≤ | g ( 1 ) - g ( t ) | + | f ( t ) - ∑ n = 0 N a n t n | + | ∑ n = 0 N - 1 ( S n - S N ) ( t n - t n + 1 ) | + | S N - f ( 1 ) | ≤ | g ( 1 ) - g ( t ) | + | f ( t ) - ∑ n = 0 N a n t n | + | ∑ n = 0 P ( S n - S N ) ( t n - t n + 1 ) | + | ∑ n = P + 1 N - 1 ( S n - S N ) ( t n - t n + 1 ) | + | S N - f ( 1 ) | ≤ | g ( 1 ) - g ( t ) | + | f ( t ) - ∑ n = 0 N a n t n | + | ∑ n = 0 P ( S n - S N ) ( t n - t n + 1 ) | + ∑ n = P + 1 N - 1 | S n - S N | ( t n - t n + 1 ) + | S N - f ( 1 ) | = t 1 + t 2 + t 3 + t 4 + t 5 .
(7)

First, choose an integer M1M1 so that if PP and NN are both larger than M1,M1, then t4<ϵ.t4<ϵ. (The sequence {Sk}{Sk} is a Cauchy sequence, and ∑(tk-tk+1∑(tk-tk+1 is telescoping.)

Fix such a P>M1.P>M1. Then choose a δ>0δ>0 so that if 1>t>1-δ,1>t>1-δ, then both t1t1 and t3<ϵ.t3<ϵ. How?

Fix such a t.t. Finally, choose a N,N, greater than M1,M1, and also large enough so that both t2t2 and t5t5 are less than ϵ.ϵ. (How?)

Now, |g(1)-f(1)|<5ϵ.|g(1)-f(1)|<5ϵ. Since this is true for every ϵ>0,ϵ>0, it follows that f(1)=g(1),f(1)=g(1), and the theorem is proved.

Exercise 4

Let f,g,r,f,g,r, and z0z0 be as in the statement of the preceding theorem. Define f^(z)=f(z0z)f^(z)=f(z0z) and g^(z)=g(z0z).g^(z)=g(z0z).

  1. Prove that f^f^ is a power series function f^(z)=∑n=0∞bnzn,f^(z)=∑n=0∞bnzn, with radius of convergence equal to 1, and such that ∑n=0∞bn∑n=0∞bn converges to f^(1);f^(1); i.e., 1 is in the domain of f^.f^.
  2. Show that g^g^ is a continuous function whose domain contains the open disk B1(0)B1(0) and the point z=1.z=1.
  3. Show that, if f^(1)=g^(1),f^(1)=g^(1), then f(z0)=g(z0).f(z0)=g(z0). Deduce that the simplification in the preceding proof is justified.
  4. State and prove the generalization of Abel's Theorem to a function ff that is expandable in a Taylor series around a point c.c.

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