5.4: Complex and Vector-Valued Functions on \(E^{1}\)
The theorems of §§2-3 fail for complex and vector-valued functions (see Problem 3 below and Problem 2 in §3). However, some analogues hold. In a sense, they even are stronger, for, unlike the previous theorems, they do not require the existence of a derivative on an entire interval \(I \subseteq E^{1},\) but only on \(I-Q\), where \(Q\) is a countable set, one contained in the range of a sequence, \(Q \subseteq\left\{p_{m}\right\}.\) (We henceforth presuppose §9 of Chapter 1.)
In the following theorem, due to N. Bourbaki, \(g : E^{1} \rightarrow E^{*}\) is extended real while \(f\) may also be complex or vector valued. We call it the finite increments law since it deals with "finite increments" \(f(b)-f(a)\) and \(g(b)-g(a) .\) Roughly, it states that \(\left|f^{\prime}\right| \leq g^{\prime}\) implies a similar inequality for increments.
Let \(f : E^{1} \rightarrow E\) and \(g : E^{1} \rightarrow E^{*}\) be relatively continuous and finite on a closed interval \(I=[a, b] \subseteq E^{1},\) and have derivatives with \(\left|f^{\prime}\right| \leq g^{\prime},\) on \(I-Q\) where \(Q \subseteq\left\{p_{1}, p_{2}, \ldots, p_{m}, \ldots\right\}.\) Then
\[|f(b)-f(a)| \leq g(b)-g(a).\]
The proof is somewhat laborious, but worthwhile. (At a first reading, one may omit it, however.) We outline some preliminary ideas.
Given any \(x \in I,\) suppose first that \(x>p_{m}\) for at least one \(p_{m} \in Q.\) In this case, we put
\[Q(x)=\sum_{p_{m}<x} 2^{-m};\]
here the summation is only over those \(m\) for which \(p_{m}<x.\) If, however, there are no \(p_{m} \in Q\) with \(p_{m}<x,\) we put \(Q(x)=0.\) Thus \(Q(x)\) is defined for all \(x \in I.\) It gives an idea as to "how many" \(p_{m}\) (at which \(f\) may have no derivative) precede \(x.\) Note that \(x<y\) implies \(Q(x) \leq Q(y)\). (Why?) Also,
\[Q(x) \leq \sum_{m=1}^{\infty} 2^{-m}=1.\]
Our plan is as follows. To prove (1), it suffices to show that for some fixed \(K \in E^{1},\) we have
\[(\forall \varepsilon>0) \quad|f(b)-f(a)| \leq g(b)-g(a)+K \varepsilon,\]
for then, letting \(\varepsilon \rightarrow 0,\) we obtain (1). We choose
\[K=b-a+Q(b), \text { with } Q(x) \text { as above. }\]
Temporarily fixing \(\varepsilon>0,\) let us call a point \(r \in I\) "good" iff
\[|f(r)-f(a)| \leq g(r)-g(a)+[r-a+Q(r)] \varepsilon\]
and "bad" otherwise. We shall show that \(b\) is "good." First, we prove a lemma.
Every "good" point \(r \in I(r<b)\) is followed by a whole interval \((r, s), r<s \leq b,\) consisting of "good" points only.
- Proof
-
First let \(r \notin Q,\) so by assumption, \(f\) and \(g\) have derivatives at \(r,\) with
\[\left|f^{\prime}(r)\right| \leq g^{\prime}(r).\]
Suppose \(g^{\prime}(r)<+\infty.\) Then (treating \(g^{\prime}\) as a right derivative) we can find \(s>r\) \((s \leq b)\) such that, for all \(x\) in the interval \((r, s),\),
\[\left|\frac{g(x)-g(r)}{x-r}-g^{\prime}(r)\right|<\frac{\varepsilon}{2} \quad\text {(why?);}\]
similarly for \(f.\) Multiplying by \(x-r,\) we get
\[\begin{aligned}\left|f(x)-f(r)-f^{\prime}(r)(x-r)\right|<(x-r) \frac{\varepsilon}{2} \text { and } \\ \left|g(x)-g(r)-g^{\prime}(r)(x-r)\right|<(x-r) \frac{\varepsilon}{2}, \end{aligned}\]
and hence by the triangle inequality (explain!),
\[|f(x)-f(r)| \leq\left|f^{\prime}(r)\right|(x-r)+(x-r) \frac{\varepsilon}{2}\]
and
\[g^{\prime}(r)(x-r)+(x-r) \frac{\varepsilon}{2}<g(x)-g(r)+(x-r) \varepsilon.\]
Combining this with \(\left|f^{\prime}(r)\right| \leq g^{\prime}(r),\) we obtain
\[|f(x)-f(r)| \leq g(x)-g(r)+(x-r) \varepsilon \text { whenever } r<x<s.\]
Now as \(r\) is "good," it satisfies \((2) ;\) hence, certainly, as \(Q(r) \leq Q(x)\),
\[|f(r)-f(a)| \leq g(r)-g(a)+(r-a) \varepsilon+Q(x) \varepsilon \text { whenever } r<x<s.\]
Adding this to (3) and using the triangle inequality again, we have
\[|f(x)-f(a)| \leq g(x)-g(a)+[x-a+Q(x)] \varepsilon \text { for all } x \in(r, s).\]
By definition, this shows that each \(x \in(r, s)\) is "good," as claimed. Thus the lemma is proved for the case \(r \in I-Q,\) with \(g^{\prime}(r)<+\infty\).
The cases \(g^{\prime}(r)=+\infty\) and \(r \in Q\) are left as Problems 1 and 2. \(\quad \square\)
We now return to Theorem 1.
Proof of Theorem 1. Seeking a contradiction, suppose \(b\) is "bad," and let \(B \neq \emptyset\) be the set of all "bad" points in \([a, b].\) Let
\[r=\inf B, \quad r \in[a, b].\]
Then the interval \([a, r)\) can contain only "good" points, i.e., points \(x\) such that
\[|f(x)-f(a)| \leq g(x)-g(a)+[x-a+Q(x)] \varepsilon.\]
As \(x<r\) implies \(Q(x) \leq Q(r),\) we have
\[|f(x)-f(a)| \leq g(x)-g(a)+[x-a+Q(r)] \varepsilon \text { for all } x \in[a, r).\]
Note that \([a, r) \neq \emptyset,\) for by (2), \(a\) is certainly "good' (why?), and so Lemma 1 yields a whole interval \([a, s)\) of "good" points contained in \([a, r).\)
Letting \(x \rightarrow r\) in (4) and using the continuity of \(f\) at \(r,\) we obtain (2). Thus \(r\) is "good" itself. Then, however, Lemma 1 yields a new interval \((r, q)\) of "good" points. Hence \([a, q)\) has no "bad" points, and so \(q\) is a lower bound of the set \(B\) of "bad" points in \(I\), contrary to \(q>r=\operatorname{glb} B\). This contradiction shows that \(b\) must be "good," i.e.,
\[|f(b)-f(a)| \leq g(b)-g(a)+[b-a+Q(b)] \varepsilon.\]
Now, letting \(\varepsilon \rightarrow 0,\) we obtain formula (1), and all is proved. \(\quad \square\)
If \(f : E^{1} \rightarrow E\) is relatively continuous and finite on \(I=[a, b] \subseteq\) \(E^{1},\) and has a derivative on \(I-Q,\) then there is a real \(M\) such that
\[|f(b)-f(a)| \leq M(b-a) \text { and } M \leq \sup _{t \in I-Q}\left|f^{\prime}(t)\right|.\]
- Proof
-
Let
\[M_{0}=\sup _{t \in I-Q}\left|f^{\prime}(t)\right|.\]
If \(M_{0}<+\infty,\) put \(M=M_{0} \geq\left|f^{\prime}\right|\) on \(I-Q,\) and take \(g(x)=M x\) in Theorem 1. Then \(g^{\prime}=M \geq\left|f^{\prime}\right|\) on \(I-Q,\) so formula (1) yields (5) since
\[g(b)-g(a)=M b-M a=M(b-a).\]
If, however, \(M_{0}=+\infty,\) let
\[M=\left|\frac{f(b)-f(a)}{b-a}\right|<M_{0}.\]
Then (5) clearly is true. Thus the required \(M\) exists always. \(\quad \square\)
Let \(f\) be as in Corollary 1. Then \(f\) is constant on \(I\) iff \(f^{\prime}=0\) on \(I-Q.\)
- Proof
-
If \(f^{\prime}=0\) on \(I-Q,\) then \(M=0\) in Corollary 1, so Corollary 1 yields, for any subinterval \([a, x](x \in I),|f(x)-f(a)| \leq 0;\) i.e., \(f(x)=f(a)\) for all \(x \in I.\) Thus \(f\) is constant on \(I.\)
Conversely, if so, then \(f^{\prime}=0,\) even on all of \(I. \quad \square\)
Let \(f, g : E^{1} \rightarrow E\) be relatively continuous and finite on \(I=[a, b],\) and differentiable on \(I-Q.\) Then \(f-g\) is constant on \(I\) iff \(f^{\prime}=g^{\prime}\) on \(I-Q.\)
- Proof
-
Apply Corollary 2 to the function \(f-g. \quad \square\)
We can now also strengthen parts (ii) and (iii) of Corollary 4 in §2.
Let \(f\) be real and have the properties stated in Corollary 1. Then
(i) \(f \uparrow\) on \(I=[a, b]\) iff \(f^{\prime} \geq 0\) on \(I-Q ;\) and
(ii) \(f \downarrow\) on \(I\) iff \(f^{\prime} \leq 0\) on \(I-Q\).
- Proof
-
Let \(f^{\prime} \geq 0\) on \(I-Q.\) Fix any \(x, y \in I(x<y)\) and define \(g(t)=0\) on \(E^{1}.\) Then \(\left|g^{\prime}\right|=0 \leq f^{\prime}\) on \(I-Q .\) Thus \(g\) and \(f\) satisfy Theorem 1 (with their roles reversed on \(I,\) and certainly on the subinterval \([x, y].\) Thus we have
\[f(y)-f(x) \geq|g(y)-g(x)|=0, \text { i.e., } f(y) \geq f(x) \text { whenever } y>x \text { in } I,\]
so \(f \uparrow\) on \(I\).
Conversely, if \(f \uparrow\) on \(I,\) then for every \(p \in I,\) we must have \(f^{\prime}(p) \geq 0,\) for otherwise, by Lemma 1 of §2, \(f\) would decrease at \(p.\) Thus \(f^{\prime} \geq 0\), even on all of \(I,\) and (i) is proved. Assertion (ii) is proved similarly. \(\quad \square\)