44 CHAPTER 3. INTEGRALS, FUNCTIONS OF ONE VARIABLE

Definition 3.2.4 Let { fn} be a sequence of continuous functions defined on [a,b]. It is saidto be uniformly Cauchy if for every ε > 0 there exists nε such that if m,k > nε

∥ fm− fk∥< ε

Corollary 3.2.5 Suppose { fn} is a uniformly Cauchy sequence of functions defined on D.Then there exists a unique continuous function f such that limn→∞ ∥ fn− f∥= 0. If each fnis uniformly continuous, then so is f .

Proof: The hypothesis implies that { fn (x)} is a Cauchy sequence in R for each x. Bycompleteness of R, this sequence converges for each x. Let f (x)≡ limn→∞ fn (x). Then bycontinuity of y→ |y− fn (x)| , for each x,

| f (x)− fn (x)|= limm→∞| fm (x)− fn (x)| ≤ lim inf

m→∞∥ fm− fn∥< ε

provided n is sufficiently large. Since x is arbitrary, this shows that

∥ f − fn∥ ≡ supx∈[a,b]

| f (x)− fn (x)| ≤ ε

if n is large enough. this says limn→∞ ∥ fn− f∥= 0. Now the continuity of f follows fromTheorem 3.2.3. How many such functions f are there? There can be only one because f (x)must equal the limit of fn (x). ■

3.3 Uniform Convergence And The IntegralIt turns out that uniform convergence is very agreeable in terms of the integral. The follow-ing is the main result.

Theorem 3.3.1 Let fn be continuous and converging uniformly to f on [a,b]. Then itfollows f is also continuous and ∫ b

af dx = lim

n→∞

∫ b

afndx

Proof: The uniform convergence implies f is also continuous. See Theorem 3.2.3.Therefore,

∫ ba f dx exists. Using the triangle inequality and definition of ∥·∥ described ear-

lier in conjunction with this theorem,∣∣∣∣∫ b

af (x)dx−

∫ b

afn (x)dx

∣∣∣∣ =

∣∣∣∣∫ b

a( f (x)− fn (x))dx

∣∣∣∣≤

∫ b

a| f (x)− fn (x)|dx≤

∫ b

a∥ f − fn∥dx

≤ ∥ f − fn∥(b−a)

which is given to converge to 0 as n→ ∞. ■

44 CHAPTER 3. INTEGRALS, FUNCTIONS OF ONE VARIABLEDefinition 3.2.4 Let {f,,} be a sequence of continuous functions defined on |a,b]. It is saidto be uniformly Cauchy if for every € > 0 there exists ng such that if m,k > negIlfm — full <€Corollary 3.2.5 Suppose {f,} is a uniformly Cauchy sequence of functions defined on D.Then there exists a unique continuous function f such that limp ||fn—f || = 0. If each fnis uniformly continuous, then so is f.Proof: The hypothesis implies that {f,, (x)} is a Cauchy sequence in R for each x. Bycompleteness of R, this sequence converges for each x. Let f (x) = limy oo fy (x). Then bycontinuity of y > |y— f, (x)|, for each x,Lf () — fn ()| = lim | fin 8) — fn (&)| Slim inf | fn — fall < €m—eo m—ooprovided n is sufficiently large. Since x is arbitrary, this shows thatIf — rll = sup |f (x) — Sn (x)| <Ex€[a,b]if n is large enough. this says limy_,. || fn — f|| = 0. Now the continuity of f follows fromTheorem 3.2.3. How many such functions f are there? There can be only one because f (x)must equal the limit of f, (x). I3.3. Uniform Convergence And The IntegralIt turns out that uniform convergence is very agreeable in terms of the integral. The follow-ing is the main result.Theorem 3.3.1 Let f,, be continuous and converging uniformly to f on [a,b]. Then itfollows f is also continuous andb b| fdx= lim | f,dxa no JaqProof: The uniform convergence implies f is also continuous. See Theorem 3.2.3.Therefore, ” fdx exists. Using the triangle inequality and definition of ||-|| described ear-lier in conjunction with this theorem,[re feoyas[rear [intgarIAb b[f@)-felaes [it fallaeIf full (0a)which is given to converge to 0 as n > oo.lA