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 F′n=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 limF′n=(limFn)′ on I−Q if
(a) {Fn(p)} converges for at least one p∈I and
(b) {F′n} 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 I−Q⊆I0(= interior of I).
Let Fn:E1→E (n=1,2,…) be finite and relatively continuous on I and differentiable on I−Q. Suppose that
(a) limn→∞Fn(p) exists for some p∈I;
(b) F′n→f≠±∞(uniformly) on J−Q for each finite subinterval J⊆I;
(c) E is complete.
Then
(i) limn→∞Fn=F exists uniformly on each finite subinterval J⊆I;
(ii) F=∫f on I; and
(iii) (limFn)′=F′=f=limn→∞F′n on I−Q.
- Proof
-
Fix ε>0 and any subinterval J⊆I of length δ<∞, with p∈J (p as in (a)). By (b), F′n→f (uniformly) on J−Q, so there is a k such that for m,n>k,
|F′n(t)−f(t)|<ε2,t∈J−Q;
hence
supt∈J−Q|F′m(t)−F′n(t)|≤ε.(Why?)
Now apply Corollary 1 in §4 to the function h=Fm−Fn on J. Then for each x∈J,|h(x)−h(p)|≤M|x−p|, where
M≤supt∈J−Q|h′(t)|≤ε
by (2). Hence for m,n>k,x∈J and
|Fm(x)−Fn(x)−Fm(p)+Fn(p)|≤ε|x−p|≤εδ.
As ε is arbitrary, this shows that the sequence
{Fn−Fn(p)}
satisfies the uniform Cauchy criterion (Chapter 4, §12, Theorem 3). Thus as E is complete, {Fn−Fn(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 J⊆I, hence on all of I. Also, letting m→+∞ (with n fixed), we have Fm→F in (3), and it follows that for n>k and x∈Gp(δ)∩I.
|F(x)−Fn(x)−F(p)+Fn(p)|≤ε|x−p|≤εδ.
Having proved (i), we may now treat p as just any point in I. Thus formula (4) holds for any globe Gp(δ),p∈I. We now show that
F′=f on I−Q; i.e., F=∫f on I.
Indeed, if p∈I−Q, each Fn is differentiable at p (by assumption), and p∈I0 (since I−Q⊆I0 by our convention). Thus for each n, there is δn>0 such that
|ΔFnΔx−F′n(p)|=|Fn(x)−Fn(p)x−p−F′n(p)|<ε2
for all x∈G¬p(δn);Gp(δn)⊆I.
By assumption (b) and by (4), we can fix n so large that
|F′n(p)−f(p)|<ε2
and so that (4) holds for δ=δn. Then, dividing by |Δx|=|x−p|, we have
|ΔFΔx−ΔFnΔx|≤ε.
Combining with (5), we infer that for each ε>0, there is δ>0 such that
|ΔFΔx−f(p)|≤|ΔFΔx−ΔFnΔx|+|ΔFnΔx−F′n(p)|+|F′n(p)−f(p)|<2ε,x∈Gp(δ).
This shows that
limx→pΔFΔx=f(p) for p∈I−Q,
i.e., F′=f on I−Q, 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=limF′n.◻
Note 1. The same proof also shows that Fn→F (uniformly) on each closed subinterval J⊆I if F′n→f (uniformly) on J−Q for all such J (with the other assumptions unchanged). In any case, we then have Fn→F (pointwise) on all of I.
We now apply Theorem 1 to antiderivatives.
Let the functions fn:E1→E,n=1,2,…, have antiderivatives, Fn=∫fn, on I. Suppose E is complete and fn→f (uniformly) on each finite subinterval J⊆I, with f finite there. Then ∫f exists on I, and
∫xpf=∫xplimn→∞fn=limn→∞∫xpfn for any p,x∈I.
- Proof
-
Fix any p∈I. By Note 2 in §5, we may choose
Fn(x)=∫xpfn for x∈I.
Then Fn(p)=∫ppfn=0, and so limn→∞Fn(p)=0 exists, as required in Theorem 1(a).
Also, by definition, each Fn is relatively continuous and finite on I and differentiable, with F′n=fn, on I−Qn. 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 I−Q⊆I−Qn, so all Fn are differentiable on I−Q, with F′n=fn there.
Additionally, by assumption,
fn→f (uniformly)
on finite subintervals J⊆I. Hence
F′n→f (uniformly) on J−Q
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 x∈I
∫xpf=F(x)−F(p)=limFn(x)−limFn(p)=limFn(x)−0=lim∫xpfn.
As p∈I was arbitrary, and f=limfn (by assumption), all is proved. ◻
Note 2. By Theorem 1, the convergence
∫xpfn→∫xpf( i.e. ,Fn→F)
is uniform in x (with p fixed), on each finite subinterval J⊆I.
We now "translate" Theorems 1 and 2 into the language of series.
Let E and the functions Fn:E1→E be as in Theorem 1. Suppose the series
∑Fn(p)
converges for some p∈I, and
∑F′n
converges uniformly on J−Q, for each finite subinterval J⊆I.
Then ∑Fn converges uniformly on each such J, and
F=∞∑n=1Fn
is differentiable on I−Q, with
F′=(∞∑n=1Fn)′=∞∑n−1F′n there.
In other words, the series can be differentiated termwise.
- Proof
-
Let
sn=n∑k=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)′=lims′n, whence (7) follows. ◻
If E and the fn are as in Theorem 2 and if ∑fn converges uniformly to f on each finite interval J⊆I, then ∫f exists on I, and
∫xpf=∫xp∞∑n=1fn=∞∑n=1∫xpfn for any p,x∈I.
Briefly, a uniformly convergent series can be integrated termwise.
Let r be the convergence radius of
∑an(x−p)n,an∈E,p∈E1.
Suppose E is complete. Set
f(x)=∞∑n=0an(x−p)n on I=(p−r,p+r).
Then the following are true:
(i) f is differentiable and has an exact antiderivative on I.
(ii) f′(x)=∑∞n=1nan(x−p)n−1 and ∫xpf=∑∞n=0ann+1(x−p)n+1,x∈I.
(iii) r is also the convergence radius of the two series in (ii).
(iv) ∑∞n=0an(x−p)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=¯limn√an.
Let r′ be the convergence radius of ∑nan(x−p)n−1, so
r′=1d′ with d′=¯limn√nan.
However, lim n√n=1 (see §3, Example (e)). It easily follows that
d′=¯limn√nan=1⋅¯limn√an=d2.
Hence r′=1/d′=1/d=r.
The convergence radius of ∑ann+1(x−p)n+1 is determined similarly. Thus (iii) is proved.
Next, let
fn(x)=an(x−p)n and Fn(x)=ann+1(x−p)n+1,n=0,1,2,….
Then the Fn are differentiable on I, with F′n=fn there. Also, by Theorems 6 and 7 of Chapter 4, §13, the series
∑F′n=∑an(x−p)n
converges uniformly on each closed subinterval J⊆I=(p−r,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=1F′n(x)=∞∑n=1an(x−p)n=f(x)
for all x∈I. 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(n−1)⋯(n−k+1)an(x−p)n−k
for each x∈I and k∈N.
If x=p, all terms vanish except the first term (n=k), i.e., k!ak. Thus f(k)(p)=k!ak,k∈N. 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(x−p)n=∞∑n=0f(n)(p)n!(x−p)n,x∈I, as required. ◻
Note 3. If ∑an(x−p)n converges also for x=p−r or x=p+r, so does the integrated series
∑an(x−p)n+1n+1
since we may include such x in I. However, the derived series ∑nan(x−p)n−1 need not converge at such x. (Why?) For example (see §6, Problem 9), the expansion
ln(1+x)=x−x22+x33−⋯
converges for x=1 but the derived series
1−x+x2−⋯
does not.
In this respect, there is the following useful rule for functions f:E1→Em (*Cm).
Let a function f:E1→Em(∗Cm) be relatively continuous on [p,x0] (or [x0,p]), x0≠p. If
f(x)=∞∑n=0an(x−p)n for p≤x<x0 (respectively, x0<x≤p),
and if ∑an(x0−p)n converges, then necessarily
f(x0)=∞∑n=0an(x0−p)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)n−1xnn 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)n−1n
yields the result for x=1 as well.