9.4: Convergence of Parametrized Integrals and Functions
I. We now consider C-integrals of the form
\[C \int f(t, u) d m(t),\]
where \(m\) is Lebesgue or LS measure in \(E^{1}.\) Here the variable \(u,\) called a parameter, remains fixed in the process of integration; but the end result depends on \(u,\) of course.
We assume \(f : E^{2} \rightarrow E\) (\(E\) complete) even if not stated explicitly. As before, we give our definitions and theorems for the case
\[C \int_{a}^{\infty}.\]
The other cases \(\left(C \int_{-\infty}^{a}, C \int_{a}^{b-}, \text { etc. }\right)\) are analogous; they are treated in Problems 2 and 3. We assume
\[a, b, c, x, t, u, v \in E^{1}\]
throughout, and write "\(d t\)" for "\(d m(t)\)" iff \(m\) is Lebesgue measure.
If
\[C \int_{a}^{\infty} f(t, u) d m(t)\]
converges for each \(u\) in a set \(B \subseteq E^{1},\) we can define a map \(F : B \rightarrow E\) by
\[F(u)=C \int_{a}^{\infty} f(t, u) d m(t)=\lim _{x \rightarrow \infty} \int_{a}^{x} f(t, u) d m(t).\]
This means that
\[(\forall u \in B)(\forall \varepsilon>0)(\exists b>a)(\forall x \geq b) \quad\left|\int_{a}^{x} f(t, u) d m(t)-F(u)\right|<\varepsilon,\]
so \(|F|<\infty\) on \(B\).
Here \(b\) depends on both \(\varepsilon\) and \(u\) (convergence is "pointwise"). However, it may occur that one and the same \(b\) fits all \(u \in B,\) so that \(b\) depends on \(\varepsilon\) alone. We then say that
\[C \int_{a}^{\infty} f(t, u) d m(t)\]
converges uniformly on \(B\) (i.e., for \(u \in B\)), and write
\[F(u)=C \int_{a}^{\infty} f(t, u) d m(t) \text { (uniformly) on } B.\]
Explicitly, this means that
\[(\forall \varepsilon>0)(\exists b>a)(\forall u \in B)(\forall x \geq b) \quad\left|\int_{a}^{x} f(t, u) d m(t)-F(u)\right|<\varepsilon.\]
Clearly, this implies (1), but not conversely. We now obtain the following.
Suppose
\[\int_{a}^{x} f(t, u) d m(t)\]
exists for \(x \geq a\) and \(u \in B \subseteq E^{1}.\) (This is automatic if \(E \subseteq E^{*};\) see Chapter 8, §5.)
Then
\[C \int_{a}^{\infty} f(t, u) d m(t)\]
converges uniformly on \(B\) iff for every \(\varepsilon>0,\) there is \(b>a\) such that
\[(\forall v, x \in[b, \infty))(\forall u \in B) \quad\left|\int_{v}^{x} f(t, u) d m(t)\right|<\varepsilon,\]
and
\[\left|\int_{a}^{b} f(t, u) d m(t)\right|<\infty.\]
- Proof
-
The necessity of (3) follows as in Theorem 2 of §3. (Verify!)
To prove sufficiency, suppose the desired \(b\) exists for every \(\varepsilon>0.\) Then for each (fixed) \(u \in B\),
\[C \int_{a}^{\infty} f(t, u) d m(t)\]
satisfies Theorem 2 of §3. Hence
\[F(u)=\lim _{x \rightarrow \infty} \int_{a}^{x} f(t, u) d m(t) \neq \pm \infty\]
exists for every \(u \in B\) (pointwise). Now, from (3), writing briefly \(\int f\) for \(\int f(t, u) d m(t),\) we obtain
\[\left|\int_{v}^{x} f\right|=\left|\int_{a}^{x} f-\int_{a}^{v} f\right|<\varepsilon\]
for all \(u \in B\) and all \(x>v \geq b\).
Making \(x \rightarrow \infty\) (with \(u\) and \(v\) temporarily fixed), we have by (4) that
\[\left|F(u)-\int_{a}^{v} f\right| \leq \varepsilon\]
whenever \(v \geq b\).
But by our assumption, \(b\) depends on \(\varepsilon\) alone (not on \(u\)). Thus unfixing \(u\), we see that (5) establishes the uniform convergence of
\[\int_{a}^{\infty} f,\]
as required.\(\quad \square\)
Under the assumptions of Theorem 1,
\[C \int_{a}^{\infty} f(t, u) d m(t)\]
converges uniformly on \(B\) if
\[C \int_{a}^{\infty}|f(t, u)| d m(t)\]
does.
Indeed,
\[\left|\int_{v}^{x} f\right| \leq \int_{v}^{x}|f|<\varepsilon.\]
Let \(f : E^{2} \rightarrow E\) and \(M : E^{2} \rightarrow E^{*}\) satisfy
\[|f(t, u)| \leq M(t, u)\]
for \(u \in B \subseteq E^{1}\) and \(t \geq a\).
Then
\[C \int_{a}^{\infty}|f(t, u)| d m(t)\]
converges uniformly on \(B\) if
\[C \int_{a}^{\infty} M(t, u) d m(t)\]
does.
Indeed, Theorem 1 applies, with
\[\left|\int_{v}^{x} f\right| \leq \int_{v}^{x} M<\varepsilon.\]
Hence we have the following corollary.
Let \(f : E^{2} \rightarrow E\) and \(M : E^{1} \rightarrow E^{*}\) satisfy
\[|f(t, u)| \leq M(t)\]
for \(u \in B \subseteq E^{1}\) and \(t \geq a.\) Suppose
\[C \int_{a}^{\infty} M(t) d m(t)\]
converges. Then
\[C \int_{a}^{\infty}|f(t, u)| d m(t)\]
converges (uniformly) on \(B.\) So does
\[C \int_{a}^{\infty} f(t, u) d m(t)\]
by Corollary 1.
- Proof
-
Set
\[h(t, u)=M(t) \geq|f(t, u)|.\]
Then Corollary 2 applies (with \(M\) replaced by \(h\) there). Indeed, the convergence of
\[C \int h=C \int M\]
is trivially "uniform" for \(u \in B,\) since \(M\) does not depend on \(u\) at all.\(\quad \square\)
Note 1. Observe also that, if \(h(t, u)\) does not depend on \(u,\) then the (pointwise) and (uniform) convergence of \(C \int h\) are trivially equivalent.
We also have the following result.
Suppose
\[C \int_{a}^{\infty} f(t, u) d m(t)\]
converges (pointwise) on \(B \subseteq E^{1}.\) Then this convergence is uniform iff
\[\lim _{\nu \rightarrow \infty} C \int_{v}^{\infty} f(t, u) d m(t)=0 \text { (uniformly) on } B,\]
i.e., iff
\[(\forall \varepsilon>0)(\exists b>a)(\forall u \in B)(\forall v \geq b) \quad\left|C \int_{v}^{\infty} f(t, u) d m(t)\right|<\varepsilon.\]
- Proof
-
The proof (based on Theorem 1) is left to the reader, along with that of the following corollary.
Suppose
\[\int_{a}^{b} f(t, u) d m(t) \neq \pm \infty\]
exists for each \(u \in B \subseteq E^{1}\).
Then
\[C \int_{a}^{\infty} f(t, u) d m(t)\]
converges (uniformly) on \(B\) iff
\[C \int_{b}^{\infty} f(t, u) d m(t)\]
does.
II. The Abel-Dirichlet tests for uniform convergence of series (Problems 9 and 11 in Chapter 4, §13) have various analogues for C-integrals. We give two of them, using the second law of the mean (Corollary 5 in §1).
First, however, we generalize our definitions, "unstarring" some ideas of Chapter 4, §11. Specifically, given
\[H : E^{2} \rightarrow E \text{ (} E \text { complete),}\]
we say that \(H(x, y)\) converges to \(F(y),\) uniformly on \(B,\) as \(x \rightarrow q\left(q \in E^{*}\right),\) and write
\[\lim _{x \rightarrow q} H(x, y)=F(y) \text { (uniformly) on } B\]
iff we have
\[(\forall \varepsilon>0)\left(\exists G_{\neg q}\right)(\forall y \in B)\left(\forall x \in G_{\neg q}\right) \quad|H(x, y)-F(y)|<\varepsilon;\]
hence \(|F|<\infty\) on \(B\).
If here \(q=\infty,\) the deleted globe \(G_{\neg q}\) has the form \((b, \infty).\) Thus if
\[H(x, u)=\int_{a}^{x} f(t, u) d t,\]
(6) turns into (2) as a special case. If (6) holds with \("\left(\exists G_{\neg q}\right) "\) and \("(\forall y \in B)"\) interchanged, as in (1), convergence is pointwise only.
As in Chapter 8, §8, we denote by \(f(\cdot, y),\) or \(f^{y},\) the function of \(x\) alone (on \(E^{1}\)) given by
\[f^{y}(x)=f(x, y).\]
Similarly,
\[f_{x}(y)=f(x, y).\]
Of course, we may replace \(f(x, y)\) by \(f(t, u)\) or \(H(t, u),\) etc.
We use Lebesgue measure in Theorems 2 and 3 below.
Assume \(f, g : E^{2} \rightarrow E^{1}\) satisfy
(i) \(C \int_{a}^{\infty} g(t, u) d t\) converges (uniformly) on \(B\);
(ii) each \(g^{u}(u \in B)\) is \(L\)-measurable on \(A=[a, \infty)\);
(iii) each \(f^{u}(u \in B)\) is monotone \((\downarrow\) or \(\uparrow)\) on \(A;\) and
(iv) \(|f|<K \in E^{1}\) (bounded) on \(A \times B\).
Then
\[C \int_{a}^{\infty} f(t, u) g(t, u) d t\]
converges uniformly on \(B\).
- Proof
-
Given \(\varepsilon>0,\) use assumption (i) and Theorem 1 to choose \(b>a\) so that
\[\left|L \int_{v}^{x} g(t, u) d t\right|<\frac{\varepsilon}{2 K},\]
written briefly as
\[\left|L \int_{v}^{x} g^{u}\right|<\frac{\varepsilon}{2 K},\]
for all \(u \in B\) and \(x>v \geq b,\) with \(K\) as in (iv).
Hence by (ii), each \(g^{u}(u \in B)\) is \(L\)-integrable on any interval \([v, x] \subset A\), with \(x>v \geq b.\) Thus given such \(u\) and \([v, x],\) we can use (iii) and Corollary 5 from §1 to find that
\[L \int_{v}^{x} f^{u} g^{u}=f^{u}(v) L \int_{v}^{c} g^{u}+f^{u}(x) L \int_{c}^{x} g^{u}\]
for some \(c \in[v, x]\).
Combining with (7) and using (iv), we easily obtain
\[\left|L \int_{v}^{x} f(t, u) g(t, u) d t\right|<\varepsilon\]
whenever \(u \in B\) and \(x>v \geq b.\) (Verify!)
Our assertion now follows by Theorem 1.\(\quad \square\)
Let \(f, g : E^{2} \rightarrow E^{*}\) satisfy
(a) \(\lim _{t \rightarrow \infty} f(t, u)=0\) (uniformly) for \(u \in B\);
(b) each \(f^{u}(u \in B)\) is nonincreasing \((\downarrow)\) on \(A=[0, \infty)\);
(c) each \(g^{u}(u \in B)\) is \(L\)-measurable on \(A;\) and
(d) \(\left(\exists K \in E^{1}\right)(\forall x \in A)(\forall u \in B)\left|L \int_{a}^{x} g(t, u) d t\right|<K\).
Then
\[C \int_{a}^{\infty} f(t, u) g(t, u) d t\]
converges uniformly on \(B\).
- Proof Outline
-
Argue as in Problem 13 of §3, replacing Theorem 2 in §3 by Theorem 1 of the present section.
By Lemma 2 in §1, obtain
\[\left|L \int_{v}^{x} f^{u} g^{u}\right|=\left|f^{u}(v) L \int_{a}^{x} g^{u}\right| \leq K f(v, u)\]
for \(u \in B\) and \(x>v \geq a\).
Then use assumption (a) to fix \(k\) so that
\[|f(t, u)|<\frac{\varepsilon}{2 K}\]
for \(t>k\) and \(u \in B. \quad \square\)
Note 2. Via components, Theorems 2 and 3 extend to the case \(g : E^{2} \rightarrow\) \(E^{n}\left(C^{n}\right).\)
Note 3. While Corollaries 2 and 3 apply to absolute convergence only, Theorems 2 and 3 cover conditional convergence, too (a great advantage!). The theorems also apply if \(f\) or \(g\) is independent of \(u\) (see Note 1). This supersedes Problems 13 and 14 in §3.
(A) The integral
\[\int_{0}^{\infty} \frac{\sin t u}{t} d t\]
converges uniformly on \(B_{\delta}=[\delta, \infty)\) if \(\delta>0,\) and pointwise on \(B=[0, \infty)\).
Indeed, we can use Theorem 3, with
\[g(t, u)=\sin t u\]
and
\[f(t, u)=\frac{1}{t}, f(0, u)=1,\]
say. Then the limit
\[\lim _{t \rightarrow \infty} \frac{1}{t}=0\]
is trivially uniform for \(u \in B_{\delta},\) as \(f\) is independent of \(u.\) Thus assumption (a) is satisfied. So is (d) because
\[\left|\int_{0}^{x} \sin t u d t\right|=\left|\frac{1}{u} \int_{0}^{x u} \sin \theta d \theta\right| \leq \frac{1}{\delta} \cdot 2.\]
(Explain!) The rest is easy.
Note that Theorem 2 fails here since assumption (i) is not satisfied.
(B) The integral
\[\int_{0}^{\infty} \frac{1}{t} e^{-t u} \sin a t d t\]
converges uniformly on \(B=[0, \infty).\) It does so absolutely on \(B_{\delta}=[\delta, \infty),\) if \(\delta>0.\)
Here we shall use Theorem 2 (though Theorem 3 works, too). Set
\[f(t, u)=e^{-t u}\]
and
\[g(t, u)=\frac{\sin a t}{t}, g(0, u)=a.\]
Then
\[\int_{0}^{\infty} g(t, u) d t\]
converges (substitute \(x=a t\) in Problem 8 or 15 in §3). Convergence is trivially uniform, by Note 1. Thus assumption (i) holds, and so do the other assumptions. Hence the result.
For absolute convergence on \(B_{\delta},\) use Corollary 3 with
\[M(t)=e^{-\delta t},\]
so \(M \geq|f g|\).
Note that, quite similarly, one treats C-integrals of the form
\[\int_{a}^{\infty} e^{-t u} g(t) d t, \int_{a}^{\infty} e^{-t^{2} u} g(t) d t, \text { etc.,}\]
provided
\[\int_{a}^{\infty} g(t) d t\]
converges \((a \geq 0)\).
In fact, Theorem 2 states (roughly) that the uniform convergence of \(C \int g\) implies that of \(C \int f g,\) provided \(f\) is monotone (in \(t\)) and bounded.
III. We conclude with some theorems on uniform convergence of functions \(H : E^{2} \rightarrow E\) (see (6)). In Theorem 4, \(m\) is again an LS (or Lebesgue) measure in \(E^{1};\) the deleted globe \(G_{\neg q}^{*}\) is fixed.
Suppose
\[\lim _{x \rightarrow q} H(x, y)=F(y) \text { (uniformly)}\]
for \(y \in B \subseteq E^{1}.\) Then we have the following:
(i) If all \(H_{x}\left(x \in G_{\neg q}^{*}\right)\) are continuous or \(m\)-measurable on \(B,\) so also is \(F\).
(ii) The same applies to \(m\)-integrability on \(B,\) provided \(m B<\infty;\) and then
\[\lim _{x \rightarrow q} \int_{B}\left|H_{x}-F\right|=0;\]
hence
\[\lim _{x \rightarrow q} \int_{B} H_{x}=\int_{B} F=\int_{B}\left(\lim _{x \rightarrow q} H_{x}\right).\]
Formula (8') is known as the rule of passage to the limit under the integral sign.
- Proof
-
(i) Fix a sequence \(x_{k} \rightarrow q\) \((x_{k}\) in the deleted globe \(G_{\neg q}^{*}),\) and set
\[H_{k}=H_{x_{k}} \quad(k=1,2, \ldots).\]
The uniform convergence
\[H(x, y) \rightarrow F(y)\]
is preserved as \(x\) runs over that sequence (see Problem 4). Hence if all \(H_{k}\) are continuous or measurable, so is \(F\) (Theorem 2 in Chapter 4, §12 and Theorem 4 in Chapter 8, §1. Thus clause (i) is proved.
(ii) Now let all \(H_{x}\) be \(m\)-integrable on \(B;\) let
\[m B<\infty.\]
Then the \(H_{k}\) are \(m\)-measurable on \(B,\) and so is \(F,\) by (i). Also, by (6),
\[(\forall \varepsilon>0)\left(\exists G_{\neg q}\right)\left(\forall x \in G_{\neg q}\right) \quad \int_{B}\left|H_{x}-F\right| \leq \int_{B}(\varepsilon)=\varepsilon m B<\infty,\]
proving (8). Moreover, as
\[\int_{B}\left|H_{x}-F\right|<\infty,\]
\(H_{x}-F\) is \(m\)-integrable on \(B,\) and so is
\[F=H_{x}-\left(H_{x}-F\right).\]
Hence
\[\left|\int_{B} H_{x}-\int_{B} F\right|=\left|\int_{B}\left(H_{x}-F\right)\right| \leq \int_{B}\left|H_{x}-F\right| \rightarrow 0,\]
as \(x \rightarrow q,\) by (8). Thus (8') is proved, too.\(\quad \square\)
Quite similarly (keeping \(E\) complete and using sequences), we obtain the following result.
Suppose that
(i) all \(H_{x}\left(x \in G_{-q}^{*}\right)\) are continuous and finite on a finite interval \(B \subset E^{1}\), and differentiable on \(B-Q,\) for a fixed countable set \(Q\);
(ii) \(\lim _{x \rightarrow q} H\left(x, y_{0}\right) \neq \pm \infty\) exists for some \(y_{0} \in B;\) and
(iii) \(\lim _{x \rightarrow q} D_{2} H(x, y)=f(y)\) (uniformly) exists on \(B-Q\).
Then \(f,\) so defined, has a primitive \(F\) on \(B,\) exact on \(B-Q\) (so \(F^{\prime}=f\) on \(B-Q);\) moreover,
\[F(y)=\lim _{x \rightarrow y} H(x, y) \text { (uniformly) for } y \in B.\]
- Outline of Proof
-
Note that
\[D_{2} H(x, y)=\frac{d}{d y} H_{x}(y).\]
Use Theorem 1 of Chapter 5, §9, with \(F_{n}=H_{x_{n}}, x_{n} \rightarrow q. \quad \square\)
Note 4. If \(x \rightarrow q\) over a path \(P\) (clustering at \(q\)), one must replace \(G_{\neg q}\) and \(G_{\neg q}^{*}\) by \(P \cap G_{\neg q}\) and \(P \cap G_{\neg q}^{*}\) in (6) and in Theorems 4 and 5.