uniform convergence

UNIFORM CONVERGENCE

MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE

1. Pointwise Convergence of a Sequence

Let E be a set and Y be a metric space. Consider functions fn : E → Yfor n = 1, 2, . . . . We say that the sequence (fn) converges pointwise onE if there is a function f : E → Y such that fn(p) → f(p) for every p ∈ E.Clearly, such a function f is unique and it is called the pointwise limit of(fn) on E. We then write fn → f on E. For simplicity, we shall assumeY = R with the usual metric.

Let fn → f on E. We ask the following questions:

(i) If each fn is bounded on E, must f be bounded on E? If so, mustsupp∈E |fn(p)| → supp∈E |f(p)|?

(ii) If E is a metric space and each fn is continuous on E, must f becontinuous on E?

(iii) If E is an interval in R and each fn is differentiable on E, must fbe differentiable on E? If so, must f ′

n → f ′ on E?(iv) If E = [a, b] and each fn is Riemann integrable on E, must f be

Riemann integrable on E? If so, must∫ b

afn(x)dx →

∫ b

af(x)dx?

These questions involve interchange of two processes (one of which is‘taking the limit as n → ∞’) as shown below.

(i) limn→∞

supp∈E

|fn(p)| = supp∈E

∣

∣

∣lim

n→∞

fn(p)∣

∣

∣.

(ii) For p ∈ E and pk → p in E, limk→∞

limn→∞

fn(pk) = limn→∞

limk→∞

fn(pk).

(iii) limn→∞

d

dx(fn) =

d

dx

(

limn→∞

fn

)

.

(iv) limn→∞

∫ b

a

fn(x)dx =

∫ b

a

(

limn→∞

fn(x))

dx.

Answers to these questions are all negative.

Examples 1.1. (i) Let E := (0, 1] and define fn : E → R by

fn(x) :=

{

0 if 0 < x ≤ 1/n,1/x if 1/n ≤ x ≤ 1.

Then |fn(x)| ≤ n for all x ∈ E, fn → f on E, where f(x) := 1/x.Thus each fn is bounded on E, but f is not bounded on E.

(ii) Let E := [0, 1] and define fn : E → R by fn(x) := 1/(nx+1). Theneach fn is continuous on E, fn → f on E, where f(0) := 1 andf(x) := 0 if 0 < x ≤ 1. Clearly, f is not continuous on E.

(iii) (a) Let E := (−1, 1) and define fn : E → R by fn(x) := 1/(nx2+ 1).Then each fn is differentiable on E and fn → f on (−1, 1),

1

2 MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE

where f(0) := 1 and f(x) := 0 if 0 < |x| < 1. Clearly, f is notdifferentiable on E.

(b) Let E := R and define fn : R → R by fn(x) := (sin nx)/n.Then each fn is differentiable, fn → f on R, where f ≡ 0. Butf ′

n(x) = cos nx for x ∈ R, and (f ′

n) does not converge pointwiseon R. For example, (f ′

n(π)) is not a convergent sequence.(c) Let E := (−1, 1) and define

fn(x) :=

{

[2 − (1 + x)n]/n if −1 < x < 0,(1 − x)n/n if 0 ≤ x < 1.

Then each fn is differentiable on E. (In particular, we havef ′

n(0) = −1 by L’Hopital’s Rule.) Also, fn → f on (−1, 1),where f ≡ 0. Also,

f ′

n(x) =

{

−(1 + x)n−1 if −1 < x < 0,−(1 − x)n−1 if 0 ≤ x < 1.

Further, f ′

n → g on (−1, 1), where g(0) := −1 and g(x) := 0for 0 < |x| < 1. Clearly, f ′ 6= g.

(iv) (a) Let E := [0, 1] and define fn : [0, 1] → R by

fn(x) :=

{

1 if x = 0, 1/n!, 2/n!, . . . , n!/n! = 1,0 otherwise.

Then each fn is Riemann integrable on [0, 1] since it is discon-tinuous only at a finite number of points.

Also, fn → f on [0, 1], where f(x) :=

{

1 if x is rational,0 if x is irrational.

For if x = p/q with p ∈ {0, 1, 2, . . . , q} ⊂ N, then for all n ≥ q,we have n!x ∈ {0, 1, 2, . . .} and so fn(x) = 1, while if x is anirrational number, then fn(x) = 0 for all n ∈ N. We have seenthat the Dirichlet function f is not Riemann integrable.

(b) Let E := [0, 1], and define fn : [0, 1] → R by fn(x) := n3xe−nx.Then each fn is Riemann integrable and fn → f on [0, 1],where f ≡ 0. (Use L’Hopital’s Rule repeatedly to show thatlimt→∞ t3/est = 0 for any s ∈ R with s > 0.) However, usingIntegration by Parts, we have∫ 1

0xe−nxdx =

1

n2− 1

n2en− 1

nenfor each n ∈ N,

so that∫ 10 fn(x)dx = n − (n/en) − (n2/en) → ∞.

(c) Let E := [0, 1] and define fn : [0, 1] → R by fn(x) := n2xe−nx.As above, each fn is Riemann integrable, and fn → f on [0, 1],

where f ≡ 0, and∫ 10 fn(x)dx = 1− (1/en)− (n/en) → 1, which

is not equal to∫ 10 f(x)dx = 0.

2. Uniform Convergence of a Sequence

In an attempt to obtain affirmative answers to the questions posed atthe beginning of the previous section, we introduce a stronger concept ofconvergence.

UNIFORM CONVERGENCE 3

Let E be a set and consider functions fn : E → R for n = 1, 2, . . . . Wesay that the sequence (fn) of functions converges uniformly on E if thereis a function f : E → R such that for every ε > 0, there is n0 ∈ N satisfying

n ≥ n0, p ∈ E =⇒ |fn(p) − f(p)| < ε.

Note that the natural number n0 mentioned in the above definition maydepend upon the given sequence (fn) of functions and on the given positivenumber ε, but it is independent of p ∈ E. Clearly, such a function f isunique and it is called the uniform limit of (fn) on E. We then writefn ⇒ f on E. Obviously, fn ⇒ f on E =⇒ fn → f on E, but the converseis not true : Let E := (0, 1] and define fn(x) := 1/(nx +1) for 0 < x ≤ 1. Iff(x) := 0 for x ∈ (0, 1], then fn → f on (0, 1], but fn 6⇒ f on (0, 1]. To seethis, let ε := 1/2, note that there is no n0 ∈ N satisfying

|fn(x) − f(x)| =1

nx + 1<

1

2for all n ≥ n0 and for all x ∈ (0, 1],

since 1/(nx + 1) = 1/2 when x = 1/n, n ∈ N.A sequence (fn) of real-valued functions defined on a set E is said to be

uniformly Cauchy on E if for every ε > 0, there is n0 ∈ N satisfying

m,n ≥ n0, p ∈ E =⇒ |fm(p) − fn(p)| < ε.

Proposition 2.1. (Cauchy Criterion for Uniform Convergence of aSequence) Let (fn) be a sequence of real-valued functions defined on a setE. Then (fn) is uniformly convergent on E if and only if (fn) is uniformlyCauchy on E.

Proof. =⇒) Let fn ⇒ f . For all m,n ∈ N and p ∈ E, we have

|fm(p) − fn(p)| ≤ |fm(p) − f(p)| + |f(p) − fn(p)|.⇐=) For each p ∈ E, (fn(p)) is a Cauchy sequence in R, and so it convergesto a real number which we denote by f(p). Let ε > 0. There is n0 ∈ N

satisfyingm,n ≥ n0, p ∈ E =⇒ |fm(p) − fn(p)| < ε.

For any m ≥ n0 and p ∈ E, letting n → ∞, we have |fm(p) − f(p)| ≤ ε. �

We have the following useful test for checking the uniform convergence of(fn) when its pointwise limit is known.

Proposition 2.2. (Test for Uniform Convergence of a Sequence)Let fn and f be real-valued functions defined on a set E. If fn → f onE, and if there is a sequence (an) of real numbers such that an → 0 and|fn(p) − f(p)| ≤ an for all p ∈ E, then fn ⇒ f on E.

Proof. Let ε > 0. Since an → 0, there is n0 ∈ N such that n ≥ n0 =⇒ an < ε,and so |fn(p) − f(p)| < ε for all p ∈ E. �

Example 2.3. Let 0 < r < 1 and fn(x) := xn for x ∈ [−r, r]. Thenfn(x) →0 for each x ∈ [−r, r]. Since rn → 0 and |fn(x) − 0| ≤ rn for all x ∈ [−r, r],(fn) is uniformly convergent on [−r, r].

Let us now pose the four questions stated in the last section with ‘con-vergence’ replaced by ‘uniform convergence’. We shall answer them one byone, but not necessarily in the same order.


Uniform Convergence and Boundedness.

Proposition 2.4. Let fn and f be real-valued functions defined on a set E.If fn ⇒ f on E and each fn is bounded on E, then f bounded on E.

Proof. There is n0 ∈ N such that n ≥ n0, p ∈ E =⇒ |fn(p)−f(p)| < 1. Also,since fn0

is bounded on E, there is α0 such that p ∈ E =⇒ |fn0(p)| ≤ α0.

Hence p ∈ E =⇒ |f(p)| ≤ |f(p) − fn0(p)| + |fn0

(p)| < 1 + α0. �

The converse of the above result is not true, that is, each fn as wellas f bounded on E and fn → f 6=⇒ fn ⇒ f on E. For example, letE := (0, 1], fn(x) := 1/(nx + 1) and f ≡ 0.

Given a set E, let B(E) denote the set of all real-valued bounded functionsdefined on E. For f, g in B(E), define d(f, g) := sup{|f(p)− g(p)| : p ∈ E}.Then it is easy to see that d is a metric on B(E), known as the sup-metricon B(E). Also, by Proposition 2.2, for fn and f in B(E), we have fn ⇒ f onE if and only if d(fn, f) → 0, that is, (fn) converges to f in the sup-metric onB(E). Similarly, (fn) is uniformly Cauchy on E if and only if d(fn, fm) → 0as n,m → ∞, that is, (fn) is a Cauchy sequence in the sup-metric on B(E).Thus Propositions 2.1 and 2.4 show that B(E) is a complete metric space.Also, under the hypotheses of Proposition 2.4, we have

∣

∣

∣supp∈E

|fn(p)| − supp∈E

|f(p)|∣

∣

∣≤ d(fn, f) → 0,

and so, supp∈E |fn(p)| → supp∈E |f(p)|.

Uniform Convergence and Integration.

Proposition 2.5. Let (fn) be a sequence of real-valued functions defined on[a, b]. If fn ⇒ f on [a, b] and each fn is Riemann integrable on [a, b], then

f is Riemann integrable on [a, b] and∫ b

afn(x)dx →

∫ b

af(x)dx.

Proof. Since fn ⇒ f and each fn is bounded, we see that f is bounded on[a, b] by Proposition 2.4. For n ∈ N, let αn := d(fn, f), where d denotes thesup-metric on B([a, b]). For each n ∈ N and x ∈ [a, b], we have |fn(x) −f(x)| ≤ αn, that is, fn(x) − αn ≤ f(x) ≤ fn(x) + αn, and so

L(fn) − αn(b − a) ≤ L(f) ≤ U(f) ≤ U(fn) + αn(b − a).

But since fn is Riemann integrable, we have L(fn) = U(fn), and hence0 ≤ U(f) − L(f) ≤ 2αn(b − a) → 0 as n → ∞. Thus L(f) = U(f), that is,f is Riemann integrable on [a, b]. Also,

∫ b

a

fn(x)dx − αn(b − a) ≤∫ b

a

f(x)dx ≤∫ b

a

fn(x)dx + αn(b − a),

that is,∣

∣

∣

∫ b

afn(x)dx −

∫ b

af(x)dx

∣

∣

∣≤ αn(b − a) → 0 as n → ∞. �

The converse of the above result is not true, that is, each fn as well

as f Riemann integrable on [a, b], fn → f on [a, b] and∫ b

afn(x)dx →

∫ b

af(x)dx 6=⇒ fn ⇒ f . For example, if fn(x) := 1/(nx + 1) for x ∈ [0, 1],


f(0) := 1, f(x) := 0 for x ∈ (0, 1], then fn 6⇒ f on [0, 1], but f is integrableand

∫ 1

0fn(x)dx =

ln(nx + 1)

n

∣

∣

∣

1

0=

ln(1 + n)

n→ 0 =

∫ 1

0f(x)dx.

Uniform Convergence and Continuity.

Proposition 2.6. Let (fn) be a sequence of real-valued functions defined ona metric space E. If fn ⇒ f on E and each fn is continuous on E, then fis continuous on E.

Proof. Let ε > 0. There is n0 ∈ N such that p ∈ E =⇒ |fn0(p)−f(p)| < ε/3.

Consider p0 ∈ E. Since fn0is continuous at p0, there is δ > 0 such that

p ∈ E, d(p, p0) < δ =⇒ |fn0(p) − fn0

(p0)| < ε/3. and hence

|f(p) − f(p0)| ≤ |f(p) − fn0(p)| + |fn0

(p) − fn0(p0)| + |fn0

(p0) − f(p0)| < ε,

establishing the continuity of f at p0 ∈ E. �

The converse of the above result is not true, that is, each fn as well as fcontinuous on a metric space E, fn → f on E 6=⇒ fn ⇒ f . For example,let fn(x) := nxe−nx and f(x) := 0 for x ∈ [0, 1]. Since fn(0) = 0 and forx ∈ (0, 1], fn(x) → 0 as n → ∞ by L’Hopital’s Rule, we see that fn → f .But there is no n0 ∈ N such that n ≥ n0, x ∈ [0, 1] =⇒ |nxe−nx − 0| < 1,since |nxe−nx| = e−1 for x = 1/n, n ∈ N. However, the following partialconverse holds.

Proposition 2.7. (Dini’s Theorem) Let (fn) be a sequence of real-valuedfunctions defined on a compact metric space E. If fn → f on E, each fn andf are continuous on E, and (fn) is a monotonic sequence (that is, fn ≤ fn+1

for all n ∈ N, or fn ≥ fn+1 for all n ∈ N), then fn ⇒ f on E.

For a proof, see Theorem 7.13 of [3].

The following examples show that neither the compactness of the metricspace E nor the continuity of the function f can be dropped from Dini’sTheorem: (i) E := (0, 1] and fn(x) := 1/(nx+1), x ∈ E, (ii) E := [0, 1] andfn(x) := xn, x ∈ E.

Uniform Convergence and Differentiation. Answers to the questionsregarding differentiation posed in the last section are not affirmative evenwhen fn ⇒ f on an interval of R.

(a) Let fn(x) :=√

x2 + (1/n2) and f(x) := |x| for x ∈ [−1, 1]. Since

|fn(x) − f(x)| =∣

∣

∣

√

x2 + (1/n2) −√

x2∣

∣

∣≤

∣

∣

∣

√

x2 + (1/n2) − x2∣

∣

∣=

1

n

for all n ∈ N and x ∈ [−1, 1], Proposition 2.2 shows that fn ⇒ f on [−1, 1].Although each fn is differentiable on [−1, 1], the limit function f is not.

(b) In Example 1.1 (iii) (b), fn ⇒ f on R, each fn differentiable, but (f ′

n)does not converge pointwise.

(c) In Example 1.1 (iii) (c), fn ⇒ f on (−1, 1) and each fn as well as f isdifferentiable on (−1, 1), and f ′

n → g on (−1, 1), where g 6= f ′.However, if we assume the uniform convergence of the ‘derived sequence’

(f ′

n) along with the convergence of the sequence (fn) at only one point ofthe interval, we have a satisfactory answer.


Proposition 2.8. Let (fn) be a sequence of real-valued functions definedon [a, b]. If (fn) converges at one point of [a, b], each fn is continuouslydifferentiable on [a, b] and (f ′

n) converges uniformly on [a, b], then there isf : [a, b] → R such that f is continuously differentiable on [a, b], f ′

n → f ′ on[a, b] and in fact, fn ⇒ f on [a, b].

Proof. Let x0 ∈ [a, b] and c0 ∈ R be such that fn(x0) → c0. Also, let eachfn be continuously differentiable and g : [a, b] → R be such that f ′

n ⇒ gon [a, b]. By Proposition 2.6, the function g is continuous on [a, b]. Definef : [a, b] → R by

f(x) := c0 +

∫ x

x0

g(t)dt for x ∈ [a, b].

By part (ii) of the Fundamental Theorem of Calculus (FTC), f ′ exists on[a, b] and f ′(x) = g(x) for x ∈ [a, b]. Thus f is continuously differentiableon [a, b] and f ′

n → g = f ′. Also, by part (i) of the FTC, we have

fn(x) = fn(x0) +

∫ x

x0

f ′

n(t)dt for x ∈ [a, b].

Hence for n ∈ N and x ∈ [a, b], we obtain

|fn(x) − f(x)| ≤ |fn(x0) − c0| +∣

∣

∣

∫ x

x0

(

f ′

n(t) − g(t))

dt∣

∣

∣

≤ |fn(x0) − c0| + |x − x0| supt∈[a,b]

|f ′

n(t) − g(t)|

≤ |fn(x0) − c0| + (b − a)d(f ′

n, g).

Thus fn ⇒ f on [a, b] by Proposition 2.2. �

The converse of the above result is not true, that is, each fn as wellas f continuously differentiable on [a, b], fn ⇒ f on [a, b], f ′

n → f ′ on[a, b] 6=⇒ f ′

n ⇒ f ′. For example, let fn(x) := (nx + 1)e−nx/n and f(x) := 0for x ∈ [0, 1]. Since f ′

n(x) = −nxe−nx for each n ∈ N and all x ∈ [0, 1],each fn is monotonically decreasing on [0, 1]. As fn(0) = 1/n, we obtain|fn(x)−f(x)| ≤ 1/n for all x ∈ [0, 1], and so fn ⇒ f on [0, 1]. Also, we haveseen after the proof of Proposition 2.6 that f ′

n → f ′, but f ′

n 6⇒ f ′ on [0, 1].

Remark 2.9. Proposition 2.8 holds if we drop the word ‘continuously’ ap-pearing (two times) in its statement, but then the proof is much more in-volved. See Theorem 7.17 of [3].

The results in Propositions 2.4, 2.5, 2.6 and 2.8 are summarized in thefollowing theorem.

Theorem 2.10. (i) The uniform limit of a sequence of real-valuedbounded functions defined on a set is bounded.

(ii) The uniform limit of a sequence of Riemann integrable functionsdefined on [a, b] is Riemann integrable, and its Riemann integral isthe limit of the sequence of termwise Riemann integrals, that is, if(fn) is uniformly convergent to f on [a, b] and each fn is Riemannintegrable on [a, b], then the function f is Riemann integrable on

[a, b] and∫ b

af(x)dx = limn→∞

∫ b

afn(x)dx.


(iii) The uniform limit of a sequence of continuous functions defined ona metric space is continuous.

(vi) If a sequence of continuously differentiable functions defined on [a, b]is convergent at one point of [a, b] and if the ‘derived’ sequence isuniformly convergent on [a, b], then the given sequence convergesuniformly on [a, b], the uniform limit is continuously differentiableon [a, b] and its derivative is the limit of the sequence of termwisederivatives, that is, if (fn) converges at one point of [a, b], each fn iscontinuously differentiable on [a, b] and (f ′

n) is uniformly convergenton [a, b], then (fn) converges uniformly to a continuously differen-tiable function f on [a, b], and f ′(x) = limn→∞ f ′

n(x) for all x ∈[a, b].

3. Uniform Convergence of a series

The reader is assumed to be familiar with the elementary theory of seriesof real numbers. (See, for example, Chapter 9 of [1], or Chapter 3 of [3].)

Let (fk) be a sequence of real-valued functions defined on a set E. Con-sider the sequence (sn) of real-valued functions on E defined by

sn := f1 + · · · + fn =

n∑

k=1

fk.

Note: Just as the sequence (fk) determines the sequence (sn), so does (sn)determine (fk): If we let s0 = 0, then we have fk = sk − sk−1 for all k ∈ N.

We say that the series∑

∞

k=1 fk converges pointwise on E if the se-quence (sn) converges pointwise on E, and we say that the series

∑

∞

k=1 fk

converges uniformly on E if the sequence (sn) converges uniformly onE. For n ∈ N, the function sn is called the nth partial sum of the series∑

∞

k=1 fk and if sn → s, then the function s is called its sum.Results about convergence / uniform convergence of sequences of func-

tions carry over to corresponding results about convergence / uniform con-vergence of series of functions.

Proposition 3.1. (Cauchy Criterion for Uniform Convergence of aSeries) Let (fk) be a sequence of real-valued functions defined on a set E.Then the series

∑

∞

k=1 fk converges uniformly on E if and only if for everyε > 0, there is n0 ∈ N such that

m ≥ n ≥ n0, p ∈ E =⇒∣

∣

∣

m∑

k=n

fk(p)∣

∣

∣< ε.

Proof. Use Proposition 2.1 for the sequence (sn) of partial sums. �

Proposition 3.2. (Weierstrass M-Test for Uniform Absolute Con-vergence of a Series) Let (fk) be a sequence of real-valued functionsdefined on a set E. Suppose there is a sequence (Mk) in R such that|fk(p)| ≤ Mk for all k ∈ N and all p ∈ E. If

∑

∞

k=1 Mk is convergent,then

∑

∞

k=1 fk converges uniformly and absolutely on E.


Proof. Note that∣

∣

∣

m∑

k=n

fk(p)∣

∣

∣≤

m∑

k=n

|fk(p)| ≤m

∑

k=n

Mk for all m ≥ n,

and use Proposition 3.1. �

Examples 3.3. (i) Consider the series∑

∞

k=0 xk, where x ∈ (−1, 1). Ifr < 1, then the series converges uniformly on {x ∈ R : |x| ≤ r} sincethe series

∑

∞

k=0 Mk is convergent, where Mk := rk for k = 0, 1, . . .

(ii) For k ∈ N, let fk(x) := (−1)k(x + k)/k2, where x ∈ [0, 1]. We showthat the series

∑

∞

k=1 fk converges uniformly on [0, 1]. For k ∈ N,

let gk(x) := (−1)kx/k2, where x ∈ [0, 1]. Letting Mk := 1/k2 fork ∈ N, we observe that the series

∑

∞

k=0 Mk is convergent. Hencethe series

∑

∞

k=1 gk(x) converges uniformly on [0, 1]. Also, the se-

ries∑

∞

k=1(−1)k/k converges uniformly on [0, 1], being a convergent

series of constants. Since fk(x) = gk(x) + (−1)k/k for k ∈ N andx ∈ [0, 1], the series

∑

∞

k=1 fk(x) converges uniformly on [0, 1].This example also shows that the converse of Weierstrass’ M -test

does not hold: If Mk := supx∈[0,1] |fk(x)| = (1 + k)/k2 for k ∈ N,

then∑

∞

k=1 Mk does not converge, since∑

∞

k=1 1/k2 converges, but∑

∞

k=1 1/k diverges.

Proposition 3.4. (Dirichlet’s Test for Uniform Conditional Con-vergence of a Series) Let (fk) be a monotonic sequence of real-valuedfunctions defined on a set E such that fk ⇒ 0 on E. If (gk) is a sequenceof real-valued functions defined on E such that the partial sums of the series∑

∞

k=1 gk are uniformly bounded on E, then the series∑

∞

k=1 fkgk converges

uniformly on E. In particular, the series∑

∞

k=1(−1)kfk converges uniformlyon E.

Proof. For each p ∈ E, the series∑

∞

k=1 fk(p)gk(p) converges in R by Dirich-let’s Test for conditional convergence of a series of real numbers. (See,for example, Proposition 9.20 of [1], or Theorem 3.42 of [3].) For p ∈ E,let H(p) :=

∑

∞

k=1 fk(p)gk(p). Also, for n ∈ N, let Gn :=∑n

k=1 gk andHn :=

∑nk=1 fkgk. Further, let β ∈ R be such that |Gn(p)| ≤ β for all n ∈ N

and all p ∈ E. Then by using the partial summation formula

n∑

k=1

fkgk =n−1∑

k=1

(fk − fk+1)Gk + fnGn for all n ≥ 2,

we have |H(p) − Hn(p)| ≤ 2β|fn+1(p)| for all p ∈ E. Since fn+1 ⇒ 0 onE, it follows that Hn ⇒ H on E, that is, the series

∑

∞

k=1 fkgk convergesuniformly on E.

In particular, letting gk(p) := (−1)k for all k ∈ N and p ∈ E, and notingthat |Gn(p)| ≤ 1 for all n ∈ N and all p ∈ E, we obtain the uniformconvergence of the series

∑

∞

k=1(−1)kfk on E. �

Example 3.5. Let E := [0, 1] and fk(x) := xk/k for k ∈ N and x ∈ [0, 1].Then (fk) is a momotonically decreasing sequence and since |fk(x)| ≤ 1/kfor k ∈ N and x ∈ [0, 1], we see that fk ⇒ 0 on [0, 1] by Proposition 2.2.Hence the series

∑

∞

k=1(−1)kxk/k converges uniformly on [0, 1].


Results regarding the boundedness, Riemann integrability, continuity anddifferentiability of the sum function of a convergent series of functions can beeasily deduced from the corresponding results for the sequence of its partialsums.

Theorem 3.6. (i) The sum function of a uniformly convergent seriesof real-valued bounded functions defined on a set is bounded.

(ii) The sum function of a uniformly convergent series of Riemann in-tegrable functions defined on [a, b] is Riemann integrable, and theseries can be integrated term by term, that is, if

∑

∞

k=1 fk is uni-formly convergent on [a, b] and each fk is Riemann integrable on[a, b], then the function

∑

∞

k=1 fk is Riemann integrable on [a, b] and

∫ b

a

(

∞∑

k=1

fk(x))

dx =

∞∑

k=1

∫ b

a

fk(x)dx.

(iii) The sum function of a uniformly convergent series of real-valuedcontinuous functions defined on a metric space is continuous.

(vi) If a series of continuously differentiable functions defined on [a, b]is convergent at one point of [a, b] and if the ‘derived’ series isuniformly convergent on [a, b], then the given series converges uni-formly on [a, b], the sum function is continuously differentiable on[a, b] and the series can be differentiated term by term, that is, if∑

∞

k=1 fk converges at one point of [a, b], each fk is continuously dif-ferentiable on [a, b] and

∑

∞

k=1 f ′

k is uniformly convergent on [a, b],then

∑

∞

k=1 fk converges uniformly to a continuously differentiablefunction, and

(

∞∑

k=1

fk(x))

′

=

∞∑

k=1

f ′

k(x) for all x ∈ [a, b].

Proof. The results follow by applying Theorem 2.10 to the sequence of par-tial sums of the given series. �

4. Two Celebrated Theorems on Uniform Approximation

We have seen in Proposition 2.6 that a uniform limit of a sequence ofcontinuous functions on a metric space is continuous. In this section, wereverse the procedure and ask whether every continuous function on a closedand bounded interval of R is the uniform limit of a sequence of some ‘special’continuous functions.

For a function f : [0, 1] → R and n ∈ N, we define the nth Bernsteinpolynomial of f by

Bn(f) :=

n∑

k=0

(

nk

)

f(k

n

)

xk(1 − x)n−k.

Theorem 4.1. (Polynomial Approximation Theorem of Weierstrass)If f : [0, 1] → R is continuous, then Bn(f) ⇒ f on [0, 1]. Consequently,every real-valued continuous function on [0, 1] is the uniform limit of a se-quence of real-valued polynomial functions.


For a proof, see Theorem 7.26 of [3], or Corollary 3.12 of [2].

Remark 4.2. Theorem 4.1 can be used to prove that every real-valuedcontinuous function on any closed and bounded interval [a, b] is the uniformlimit of a sequence of real-valued polynomial functions. Let φ : [0, 1] → [a, b]be defined by φ(x) := (1 − x)a + xb for x ∈ [0, 1]. Then φ is a bijectivecontinuous function and its continuous inverse φ−1 : [a, b] → [0, 1] is givenby φ−1(t) = (t − a)/(b − a) for t ∈ [a, b]. Given a continuous real-valuedfunction g on [a, b], consider the continuous function f := g ◦ φ defined on[0, 1]. If (Pn) is a sequence of polynomial functions such that Pn ⇒ f on[0, 1], and if we let Qn := Pn ◦ φ−1, then since Qn(t) = Pn

(

(t − a)/(b − a))

for t ∈ [a, b], each Qn is a polynomial function, and Qn ⇒ f ◦ φ−1 = g on[a, b].

Instead of polynomials, let us now consider trigonometric polynomials forapproximating a function. They are given by

a0 +

n∑

k=1

(ak cos kx + bk sin kx) for n ∈ N,

where a0, a1, a2, . . . , b1, b2, . . . are real numbers. For a Riemann integrablefunction f on [−π, π], we define the Fourier coefficients of f by

a0(f) :=1

2π

∫ π

−π

f(t)dt, and for k ∈ N,

ak(f) :=1

π

∫ π

−π

f(t) cos kt dt, bk(f) :=1

π

∫ π

−π

f(t) sin kt dt.

The series a0(f)+∑

∞

k=1

(

ak(f) cos kx+ bk(f) sin kx)

of functions definedon [−π, π] is called the Fourier series of the function f . For n = 0, 1, 2, . . .,let sn(f) denote the nth partial sum of this series, and consider the arith-metic means of these partial sums given by

σn(f) :=s0(f) + s1(f) · · · + sn(f)

n + 1for n = 0, 1, 2 . . .

Theorem 4.3. (Trigonometric Polynomial Approximation Theo-rem of Fejer) If f : [−π, π] → R is continuous and f(−π) = f(π), thenσn(f) ⇒ f on [−π, π]. Consequently, every real-valued continuous functionon [−π, π] having the same value at −π and π is the uniform limit of asequence of real-valued trigonometric polynomial functions.

For a proof, see Theorem 8.15 and Exercise 8.15 of [3], or Theorem 3.13of [2].

References

[1] S. R. Ghorpade and B. V. Limaye, A Course in Calculus and Real Analysis,Springer International Ed., New Delhi, 2006.

[2] B. V. Limaye, Functional Analysis, New Age International, 2nd Ed., New Delhi,1996.

[3] W. Rudin, Principles of Mathematical Analysis, 3rd Ed., McGraw Hill, New Delhi,1976.

uniform convergence

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