5.9: Convergence Theorems in Differentiation and Integration

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 $I-Q \subseteq I^{0}(=\text { interior of } I).$

Theorem $\PageIndex{1}$

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.

Theorem $\PageIndex{2}$

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.

Corollary $\PageIndex{1}$

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. $\quad \square$

Corollary $\PageIndex{2}$

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.

Theorem $\PageIndex{3}$ (Power Series)

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 $f : E^{1} \rightarrow E^{m}$ (*$C^{m}$).

Corollary $\PageIndex{3}$

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.