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5.9: Convergence Theorems in Differentiation and Integration

This page is a draft and is under active development. 

( \newcommand{\kernel}{\mathrm{null}\,}\)

Given

Fn=fn or Fn=fn,n=1,2,,

what can one say about limfn or (limFn) if the limits exist? Below we give some answers, for complete range spaces E (such as En). Roughly, we have limFn=(limFn) on IQ if

(a) {Fn(p)} converges for at least one pI and

(b) {Fn} converges uniformly.

Here I is a finite or infinite interval in E1 and Q is countable. We include in Q the endpoints of I (if any), so IQI0(= interior of I).

Theorem 5.9.1

Let Fn:E1E (n=1,2,) be finite and relatively continuous on I and differentiable on IQ. Suppose that

(a) limnFn(p) exists for some pI;

(b) Fnf±(uniformly) on JQ for each finite subinterval JI;

(c) E is complete.

Then

(i) limnFn=F exists uniformly on each finite subinterval JI;

(ii) F=f on I; and

(iii) (limFn)=F=f=limnFn on IQ.

Proof

Fix ε>0 and any subinterval JI of length δ<, with pJ (p as in (a)). By (b), Fnf (uniformly) on JQ, so there is a k such that for m,n>k,

|Fn(t)f(t)|<ε2,tJQ;

hence

suptJQ|Fm(t)Fn(t)|ε.(Why?)

Now apply Corollary 1 in §4 to the function h=FmFn on J. Then for each xJ,|h(x)h(p)|M|xp|, where

MsuptJQ|h(t)|ε

by (2). Hence for m,n>k,xJ and

|Fm(x)Fn(x)Fm(p)+Fn(p)|ε|xp|εδ.

As ε is arbitrary, this shows that the sequence

{FnFn(p)}

satisfies the uniform Cauchy criterion (Chapter 4, §12, Theorem 3). Thus as E is complete, {FnFn(p)} converges uniformly on J. So does {Fn}, for {Fn(p)} converges, by (a). Thus we may set

F=limFn (uniformly) on J,

proving assertion (i).

Here by Theorem 2 of Chapter 4, §12, F is relatively continuous on each such JI, hence on all of I. Also, letting m+ (with n fixed), we have FmF in (3), and it follows that for n>k and xGp(δ)I.

|F(x)Fn(x)F(p)+Fn(p)|ε|xp|εδ.

Having proved (i), we may now treat p as just any point in I. Thus formula (4) holds for any globe Gp(δ),pI. We now show that

F=f on IQ; i.e., F=f on I.

Indeed, if pIQ, each Fn is differentiable at p (by assumption), and pI0 (since IQI0 by our convention). Thus for each n, there is δn>0 such that

|ΔFnΔxFn(p)|=|Fn(x)Fn(p)xpFn(p)|<ε2

for all xG¬p(δn);Gp(δn)I.

By assumption (b) and by (4), we can fix n so large that

|Fn(p)f(p)|<ε2

and so that (4) holds for δ=δn. Then, dividing by |Δx|=|xp|, we have

|ΔFΔxΔFnΔx|ε.

Combining with (5), we infer that for each ε>0, there is δ>0 such that

|ΔFΔxf(p)||ΔFΔxΔFnΔx|+|ΔFnΔxFn(p)|+|Fn(p)f(p)|<2ε,xGp(δ).

This shows that

limxpΔFΔx=f(p) for pIQ,

i.e., F=f on IQ, with f finite by assumption, and F finite by (4). As F is also relatively continuous on I, assertion (ii) is proved, and (iii) follows since F=limFn and f=limFn.

Note 1. The same proof also shows that FnF (uniformly) on each closed subinterval JI if Fnf (uniformly) on JQ for all such J (with the other assumptions unchanged). In any case, we then have FnF (pointwise) on all of I.

We now apply Theorem 1 to antiderivatives.

Theorem 5.9.2

Let the functions fn:E1E,n=1,2,, have antiderivatives, Fn=fn, on I. Suppose E is complete and fnf (uniformly) on each finite subinterval JI, with f finite there. Then f exists on I, and

xpf=xplimnfn=limnxpfn for any p,xI.

Proof

Fix any pI. By Note 2 in §5, we may choose

Fn(x)=xpfn for xI.

Then Fn(p)=ppfn=0, and so limnFn(p)=0 exists, as required in Theorem 1(a).

Also, by definition, each Fn is relatively continuous and finite on I and differentiable, with Fn=fn, on IQn. The countable sets Qn need not be the same, so we replace them by

Q=n=1Qn

(including in Q also the endpoints of I, if any. Then Q is countable (see Chapter 1, §9, Theorem 2), and IQIQn, so all Fn are differentiable on IQ, with Fn=fn there.

Additionally, by assumption,

fnf (uniformly)

on finite subintervals JI. Hence

Fnf (uniformly) on JQ

for all such J, and so the conditions of Theorem 1 are satisfied.

By that theorem, then,

F=f=limFn exists on I

and, recalling that

Fn(x)=xpfn,

we obtain for xI

xpf=F(x)F(p)=limFn(x)limFn(p)=limFn(x)0=limxpfn.

As pI was arbitrary, and f=limfn (by assumption), all is proved.

Note 2. By Theorem 1, the convergence

xpfnxpf( i.e. ,FnF)

is uniform in x (with p fixed), on each finite subinterval JI.

We now "translate" Theorems 1 and 2 into the language of series.

Corollary 5.9.1

Let E and the functions Fn:E1E be as in Theorem 1. Suppose the series

Fn(p)

converges for some pI, and

Fn

converges uniformly on JQ, for each finite subinterval JI.

Then Fn converges uniformly on each such J, and

F=n=1Fn

is differentiable on IQ, with

F=(n=1Fn)=n1Fn there.

In other words, the series can be differentiated termwise.

Proof

Let

sn=nk=1Fk,n=1,2,,

be the partial sums of Fn. From our assumptions, it then follows that the sn satisfy all conditions of Theorem 1. (Verify!) Thus the conclusions of Theorem 1 hold, with Fn replaced by sn.

We have F=limsn and F=(limsn)=limsn, whence (7) follows.

Corollary 5.9.2

If E and the fn are as in Theorem 2 and if fn converges uniformly to f on each finite interval JI, then f exists on I, and

xpf=xpn=1fn=n=1xpfn for any p,xI.

Briefly, a uniformly convergent series can be integrated termwise.

Theorem 5.9.3 (Power Series)

Let r be the convergence radius of

an(xp)n,anE,pE1.

Suppose E is complete. Set

f(x)=n=0an(xp)n on I=(pr,p+r).

Then the following are true:

(i) f is differentiable and has an exact antiderivative on I.

(ii) f(x)=n=1nan(xp)n1 and xpf=n=0ann+1(xp)n+1,xI.

(iii) r is also the convergence radius of the two series in (ii).

(iv) n=0an(xp)n is exactly the Taylor series for f(x) on I about p.

Proof

We prove (iii) first.

By Theorem 6 of Chapter 4, §13, r=1/d, where

d=¯limnan.

Let r be the convergence radius of nan(xp)n1, so

r=1d with d=¯limnnan.

However, lim nn=1 (see §3, Example (e)). It easily follows that

d=¯limnnan=1¯limnan=d2.

Hence r=1/d=1/d=r.

The convergence radius of ann+1(xp)n+1 is determined similarly. Thus (iii) is proved.

Next, let

fn(x)=an(xp)n and Fn(x)=ann+1(xp)n+1,n=0,1,2,.

Then the Fn are differentiable on I, with Fn=fn there. Also, by Theorems 6 and 7 of Chapter 4, §13, the series

Fn=an(xp)n

converges uniformly on each closed subinterval JI=(pr,p+r). Thus the functions Fn satisfy all conditions of Corollary 1, with Q=, and the fn satisfy Corollary 2. By Corollary 1, then,

F=n=1Fn

is differentiable on I, with

F(x)=n=1Fn(x)=n=1an(xp)n=f(x)

for all xI. Hence F is an exact antiderivative of f on I, and (8) yields the second formula in (ii).

Quite similarly, replacing Fn and F by fn and f, one shows that f is differentiable on I, and the first formula in (ii) follows. This proves (i) as well.

Finally, to prove (iv), we apply (i)-(iii) to the consecutive derivatives of f and obtain

f(k)(x)=n=kn(n1)(nk+1)an(xp)nk

for each xI and kN.

If x=p, all terms vanish except the first term (n=k), i.e., k!ak. Thus f(k)(p)=k!ak,kN. We may rewrite it as

an=f(n)(p)n!,n=0,1,2,,

since f(0)(p)=f(p)=a0. Assertion (iv) now follows since

f(x)=n=0an(xp)n=n=0f(n)(p)n!(xp)n,xI, as required. 

Note 3. If an(xp)n converges also for x=pr or x=p+r, so does the integrated series

an(xp)n+1n+1

since we may include such x in I. However, the derived series nan(xp)n1 need not converge at such x. (Why?) For example (see §6, Problem 9), the expansion

ln(1+x)=xx22+x33

converges for x=1 but the derived series

1x+x2

does not.

In this respect, there is the following useful rule for functions f:E1Em (*Cm).

Corollary 5.9.3

Let a function f:E1Em(Cm) be relatively continuous on [p,x0] (or [x0,p]), x0p. If

f(x)=n=0an(xp)n for px<x0 (respectively, x0<xp),

and if an(x0p)n converges, then necessarily

f(x0)=n=0an(x0p)n.

Proof

The proof is sketched in Problems 4 and 5.

Thus in the above example, f(x)=ln(1+x) defines a continuous function on [0,1], with

f(x)=n=1(1)n1xnn on [0,1].

We gave a direct proof in §6, Problem 9. However, by Corollary 3, it suffices to prove this for [0,1), which is much easier. Then the convergence of

n=1(1)n1n

yields the result for x=1 as well.


This page titled 5.9: Convergence Theorems in Differentiation and Integration is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform.

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