5.9: Convergence Theorems in Differentiation and Integration
This page is a draft and is under active development.
Given
\[F_{n}=\int f_{n} \text { or } F_{n}^{\prime}=f_{n}, \quad n=1,2, \ldots,\]
what can one say about \(\int \lim f_{n}\) or \(\left(\lim F_{n}\right)^{\prime}\) if the limits exist? Below we give some answers, for complete range spaces \(E\) (such as \(E^{n}\)). Roughly, we have \(\lim F_{n}^{\prime}=\left(\lim F_{n}\right)^{\prime}\) on \(I-Q\) if
(a) \(\left\{F_{n}(p)\right\}\) converges for at least one \(p \in I\) and
(b) \(\left\{F_{n}^{\prime}\right\}\) converges uniformly.
Here \(I\) is a finite or infinite interval in \(E^{1}\) and \(Q\) is countable. We include in \(Q\) the endpoints of \(I\) (if any), so \(I-Q \subseteq I^{0}(=\text { interior of } I).\)
Let \(F_{n} : E^{1} \rightarrow E \text{ } (n=1,2, \ldots)\) be finite and relatively continuous on \(I\) and differentiable on \(I-Q.\) Suppose that
(a) \(\lim _{n \rightarrow \infty} F_{n}(p)\) exists for some \(p \in I\);
(b) \(F_{n}^{\prime} \rightarrow f \neq \pm \infty(\text {uniformly})\) on \(J-Q\) for each finite subinterval \(J \subseteq I\);
(c) \(E\) is complete.
Then
(i) \(\lim _{n \rightarrow \infty} F_{n}=F\) exists uniformly on each finite subinterval \(J \subseteq I\);
(ii) \(F=\int f\) on \(I ;\) and
(iii) \((\lim F_{n})^{\prime}=F^{\prime}=f=\lim _{n \rightarrow \infty} F_{n}^{\prime}\) on \(I-Q\).
- Proof
-
Fix \(\varepsilon>0\) and any subinterval \(J \subseteq I\) of length \(\delta<\infty,\) with \(p \in J\) (\(p\) as in (a)). By (b), \(F_{n}^{\prime} \rightarrow f\) (uniformly) on \(J-Q,\) so there is a \(k\) such that for \(m, n>k\),
\[\left|F_{n}^{\prime}(t)-f(t)\right|<\frac{\varepsilon}{2}, \quad t \in J-Q;\]
hence
\[\sup _{t \in J-Q}\left|F_{m}^{\prime}(t)-F_{n}^{\prime}(t)\right| \leq \varepsilon. \quad(\text {Why?})\]
Now apply Corollary 1 in §4 to the function \(h=F_{m}-F_{n}\) on \(J.\) Then for each \(x \in J,|h(x)-h(p)| \leq M|x-p|,\) where
\[M \leq \sup _{t \in J-Q}\left|h^{\prime}(t)\right| \leq \varepsilon\]
by (2). Hence for \(m, n>k, x \in J\) and
\[\left|F_{m}(x)-F_{n}(x)-F_{m}(p)+F_{n}(p)\right| \leq \varepsilon|x-p| \leq \varepsilon \delta.\]
As \(\varepsilon\) is arbitrary, this shows that the sequence
\[\left\{F_{n}-F_{n}(p)\right\}\]
satisfies the uniform Cauchy criterion (Chapter 4, §12, Theorem 3). Thus as \(E\) is complete, \(\left\{F_{n}-F_{n}(p)\right\}\) converges uniformly on \(J.\) So does \(\left\{F_{n}\right\},\) for \(\left\{F_{n}(p)\right\}\) converges, by (a). Thus we may set
\[F=\lim F_{n}\text { (uniformly) on } J,\]
proving assertion (i).
Here by Theorem 2 of Chapter 4, §12, \(F\) is relatively continuous on each such \(J \subseteq I,\) hence on all of \(I\). Also, letting \(m \rightarrow+\infty\) (with \(n\) fixed), we have \(F_{m} \rightarrow F\) in (3), and it follows that for \(n>k\) and \(x \in G_{p}(\delta) \cap I.\)
\[\left|F(x)-F_{n}(x)-F(p)+F_{n}(p)\right| \leq \varepsilon|x-p| \leq \varepsilon \delta.\]
Having proved (i), we may now treat \(p\) as just any point in \(I\). Thus formula (4) holds for any globe \(G_{p}(\delta), p \in I.\) We now show that
\[F^{\prime}=f \text { on } I-Q ; \text { i.e., } F=\int f \text { on } I.\]
Indeed, if \(p \in I-Q,\) each \(F_{n}\) is differentiable at \(p\) (by assumption), and \(p \in I^{0}\) (since \(I-Q \subseteq I^{0}\) by our convention). Thus for each \(n,\) there is \(\delta_{n}>0\) such that
\[\left|\frac{\Delta F_{n}}{\Delta x}-F_{n}^{\prime}(p)\right|=\left|\frac{F_{n}(x)-F_{n}(p)}{x-p}-F_{n}^{\prime}(p)\right|<\frac{\varepsilon}{2}\]
for all \(x \in G_{\neg p}\left(\delta_{n}\right); G_{p}\left(\delta_{n}\right) \subseteq I\).
By assumption (b) and by (4), we can fix \(n\) so large that
\[\left|F_{n}^{\prime}(p)-f(p)\right|<\frac{\varepsilon}{2}\]
and so that (4) holds for \(\delta=\delta_{n}.\) Then, dividing by \(|\Delta x|=|x-p|,\) we have
\[\left|\frac{\Delta F}{\Delta x}-\frac{\Delta F_{n}}{\Delta x}\right| \leq \varepsilon.\]
Combining with (5), we infer that for each \(\varepsilon>0,\) there is \(\delta>0\) such that
\[\left|\frac{\Delta F}{\Delta x}-f(p)\right| \leq\left|\frac{\Delta F}{\Delta x}-\frac{\Delta F_{n}}{\Delta x}\right|+\left|\frac{\Delta F_{n}}{\Delta x}-F_{n}^{\prime}(p)\right|+\left|F_{n}^{\prime}(p)-f(p)\right|<2 \varepsilon, x \in G_{p}(\delta).\]
This shows that
\[\lim _{x \rightarrow p} \frac{\Delta F}{\Delta x}=f(p) \text { for } p \in I-Q,\]
i.e., \(F^{\prime}=f\) 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 \(F=\lim F_{n}\) and \(f=\lim F_{n}^{\prime}. \quad \square\)
Note 1. The same proof also shows that \(F_{n} \rightarrow F\) (uniformly) on each closed subinterval \(J \subseteq I\) if \(F_{n}^{\prime} \rightarrow f\) (uniformly) on \(J-Q\) for all such \(J\) (with the other assumptions unchanged). In any case, we then have \(F_{n} \rightarrow F\) (pointwise) on all of \(I.\)
We now apply Theorem 1 to antiderivatives.
Let the functions \(f_{n} : E^{1} \rightarrow E, n=1,2, \ldots,\) have antiderivatives, \(F_{n}=\int f_{n},\) on \(I.\) Suppose \(E\) is complete and \(f_{n} \rightarrow f\) (uniformly) on each finite subinterval \(J \subseteq I,\) with \(f\) finite there. Then \(\int f\) exists on \(I,\) and
\[\int_{p}^{x} f=\int_{p}^{x} \lim _{n \rightarrow \infty} f_{n}=\lim _{n \rightarrow \infty} \int_{p}^{x} f_{n} \text { for any } p, x \in I.\]
- Proof
-
Fix any \(p \in I.\) By Note 2 in §5, we may choose
\[F_{n}(x)=\int_{p}^{x} f_{n} \text { for } x \in I.\]
Then \(F_{n}(p)=\int_{p}^{p} f_{n}=0,\) and so \(\lim _{n \rightarrow \infty} F_{n}(p)=0\) exists, as required in Theorem 1(a).
Also, by definition, each \(F_{n}\) is relatively continuous and finite on \(I\) and differentiable, with \(F_{n}^{\prime}=f_{n},\) on \(I-Q_{n}.\) The countable sets \(Q_{n}\) need not be the same, so we replace them by
\[Q=\bigcup_{n=1}^{\infty} Q_{n}\]
(including in \(Q\) also the endpoints of \(I,\) if any. Then \(Q\) is countable (see Chapter 1, §9, Theorem 2), and \(I-Q \subseteq I-Q_{n},\) so all \(F_{n}\) are differentiable on \(I-Q,\) with \(F_{n}^{\prime}=f_{n}\) there.
Additionally, by assumption,
\[f_{n} \rightarrow f \text { (uniformly)}\]
on finite subintervals \(J \subseteq I.\) Hence
\[F_{n}^{\prime} \rightarrow f \text { (uniformly) on } J-Q\]
for all such \(J,\) and so the conditions of Theorem 1 are satisfied.
By that theorem, then,
\[F=\int f=\lim F_{n} \text { exists on } I\]
and, recalling that
\[F_{n}(x)=\int_{p}^{x} f_{n},\]
we obtain for \(x \in I\)
\[\int_{p}^{x} f=F(x)-F(p)=\lim F_{n}(x)-\lim F_{n}(p)=\lim F_{n}(x)-0=\lim \int_{p}^{x} f_{n}.\]
As \(p \in I\) was arbitrary, and \(f=\lim f_{n}\) (by assumption), all is proved. \(\quad \square\)
Note 2. By Theorem 1, the convergence
\[\int_{p}^{x} f_{n} \rightarrow \int_{p}^{x} f \quad\left(\text { i.e. }, F_{n} \rightarrow F\right)\]
is uniform in \(x\) (with \(p\) fixed), on each finite subinterval \(J \subseteq I\).
We now "translate" Theorems 1 and 2 into the language of series.
Let \(E\) and the functions \(F_{n} : E^{1} \rightarrow E\) be as in Theorem 1. Suppose the series
\[\sum F_{n}(p)\]
converges for some \(p \in I,\) and
\[\sum F_{n}^{\prime}\]
converges uniformly on \(J-Q,\) for each finite subinterval \(J \subseteq I\).
Then \(\sum F_{n}\) converges uniformly on each such \(J,\) and
\[F=\sum_{n=1}^{\infty} F_{n}\]
is differentiable on \(I-Q,\) with
\[F^{\prime}=\left(\sum_{n=1}^{\infty} F_{n}\right)^{\prime}=\sum_{n-1}^{\infty} F_{n}^{\prime} \text { there.}\]
In other words, the series can be differentiated termwise.
- Proof
-
Let
\[s_{n}=\sum_{k=1}^{n} F_{k}, \quad n=1,2, \ldots,\]
be the partial sums of \(\sum F_{n}.\) From our assumptions, it then follows that the \(s_{n}\) satisfy all conditions of Theorem 1. (Verify!) Thus the conclusions of Theorem 1 hold, with \(F_{n}\) replaced by \(s_{n}\).
We have \(F=\operatorname{lims}_{n}\) and \(F^{\prime}=\left(\lim s_{n}\right)^{\prime}=\lim s_{n}^{\prime},\) whence (7) follows. \(\quad \square\)
If \(E\) and the \(f_{n}\) are as in Theorem 2 and if \(\sum f_{n}\) converges uniformly to \(f\) on each finite interval \(J \subseteq I,\) then \(\int f\) exists on \(I,\) and
\[\int_{p}^{x} f=\int_{p}^{x} \sum_{n=1}^{\infty} f_{n}=\sum_{n=1}^{\infty} \int_{p}^{x} f_{n} \text { for any } p, x \in I.\]
Briefly, a uniformly convergent series can be integrated termwise.
Let \(r\) be the convergence radius of
\[\sum a_{n}(x-p)^{n}, \quad a_{n} \in E, p \in E^{1}.\]
Suppose \(E\) is complete. Set
\[f(x)=\sum_{n=0}^{\infty} a_{n}(x-p)^{n} \quad \text { on } I=(p-r, p+r).\]
Then the following are true:
(i) \(f\) is differentiable and has an exact antiderivative on \(I.\)
(ii) \(f^{\prime}(x)=\sum_{n=1}^{\infty} n a_{n}(x-p)^{n-1}\) and \(\int_{p}^{x} f=\sum_{n=0}^{\infty} \frac{a_{n}}{n+1}(x-p)^{n+1}, x \in I\).
(iii) \(r\) is also the convergence radius of the two series in (ii).
(iv) \(\sum_{n=0}^{\infty} a_{n}(x-p)^{n}\) 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
\[d=\overline{\lim } \sqrt[n]{a_{n}}.\]
Let \(r^{\prime}\) be the convergence radius of \(\sum n a_{n}(x-p)^{n-1},\) so
\[r^{\prime}=\frac{1}{d^{\prime}} \text { with } d^{\prime}=\overline{\lim } \sqrt[n]{n a_{n}}.\]
However, lim \(\sqrt[n]{n}=1\) (see §3, Example (e)). It easily follows that
\[d^{\prime}=\overline{\lim } \sqrt[n]{n a_{n}}=1 \cdot \overline{\lim } \sqrt[n]{a_{n}}=d^{2}.\]
Hence \(r^{\prime}=1 / d^{\prime}=1 / d=r\).
The convergence radius of \(\sum \frac{a_{n}}{n+1}(x-p)^{n+1}\) is determined similarly. Thus (iii) is proved.
Next, let
\[f_{n}(x)=a_{n}(x-p)^{n} \text { and } F_{n}(x)=\frac{a_{n}}{n+1}(x-p)^{n+1}, n=0,1,2, \ldots.\]
Then the \(F_{n}\) are differentiable on \(I,\) with \(F_{n}^{\prime}=f_{n}\) there. Also, by Theorems 6 and 7 of Chapter 4, §13, the series
\[\sum F_{n}^{\prime}=\sum a_{n}(x-p)^{n}\]
converges uniformly on each closed subinterval \(J \subseteq I=(p-r, p+r).\) Thus the functions \(F_{n}\) satisfy all conditions of Corollary 1, with \(Q=\emptyset,\) and the \(f_{n}\) satisfy Corollary 2. By Corollary 1, then,
\[F=\sum_{n=1}^{\infty} F_{n}\]
is differentiable on \(I,\) with
\[F^{\prime}(x)=\sum_{n=1}^{\infty} F_{n}^{\prime}(x)=\sum_{n=1}^{\infty} a_{n}(x-p)^{n}=f(x)\]
for all \(x \in I.\) Hence \(F\) is an exact antiderivative of \(f\) on \(I,\) and (8) yields the second formula in (ii).
Quite similarly, replacing \(F_{n}\) and \(F\) by \(f_{n}\) 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
\[f^{(k)}(x)=\sum_{n=k}^{\infty} n(n-1) \cdots(n-k+1) a_{n}(x-p)^{n-k}\]
for each \(x \in I\) and \(k \in N\).
If \(x=p,\) all terms vanish except the first term \((n=k),\) i.e., \(k ! a_{k}.\) Thus \(f^{(k)}(p)=k ! a_{k}, k \in N.\) We may rewrite it as
\[a_{n}=\frac{f^{(n)}(p)}{n !}, \quad n=0,1,2, \ldots,\]
since \(f^{(0)}(p)=f(p)=a_{0}.\) Assertion (iv) now follows since
\[f(x)=\sum_{n=0}^{\infty} a_{n}(x-p)^{n}=\sum_{n=0}^{\infty} \frac{f^{(n)}(p)}{n !}(x-p)^{n}, \quad x \in I, \text { as required. } \quad \square\]
Note 3. If \(\sum a_{n}(x-p)^{n}\) converges also for \(x=p-r\) or \(x=p+r,\) so does the integrated series
\[\sum a_{n} \frac{(x-p)^{n+1}}{n+1}\]
since we may include such \(x\) in \(I.\) However, the derived series \(\sum n a_{n}(x-p)^{n-1}\) need not converge at such \(x.\) (Why?) For example (see §6, Problem 9), the expansion
\[\ln (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\cdots\]
converges for \(x=1\) but the derived series
\[1-x+x^{2}-\cdots\]
does not.
In this respect, there is the following useful rule for functions \(f : E^{1} \rightarrow E^{m}\) (*\(C^{m}\)).
Let a function \(f : E^{1} \rightarrow E^{m}\left(^{*} C^{m}\right)\) be relatively continuous on \(\left[p, x_{0}\right]\) (or \([x_{0}, p])\), \(x_{0} \neq p.\) If
\[f(x)=\sum_{n=0}^{\infty} a_{n}(x-p)^{n} \text { for } p \leq x<x_{0} \text { (respectively, } x_{0}<x \leq p),\]
and if \(\sum a_{n}\left(x_{0}-p\right)^{n}\) converges, then necessarily
\[f\left(x_{0}\right)=\sum_{n=0}^{\infty} a_{n}\left(x_{0}-p\right)^{n}.\]
- 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
\[f(x)=\sum_{n=1}^{\infty}(-1)^{n-1} \frac{x^{n}}{n} \text { on }[0,1].\]
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
\[\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\]
yields the result for \(x=1\) as well.