Uniform Calculus and the Law of Bounded Change

Introduction

In a recent exchange about the role of the mean value theorem in the theory of the calculus, T. Tucker notes that ``the origin of the Mean Value Theorem in the structure of the real numbers" is much too difficult for a standard course [6]. He shows how the increasing function theorem (a function with positive derivative is increasing) serves very nicely in place of the mean value theorem, and sketches a proof of it from the nested interval property of the real number system.

In support of the mean value theorem, H. Swann recalls its derivation from the extreme value theorem (a continuous function on a closed interval has a maximum value) via Rolle's theorem and remarks that ``such a sequence of arguments reveals the charm and power of mathematics, for we prove that a questionable complicated result must be true if we assume other simpler results that are less questionable'' [5 ].

We agree with Swann about the charm and power of mathematics and with Tucker about the ability of the increasing function theorem to play a role traditionally accorded the mean value theorem. In fact, we give several examples that support Tucker's claim. But Tucker and Swann work with pointwise continuity and differentiability, weak notions that make proving statements like the increasing function theorem more difficult. On closed finite intervals, uniform continuity and differentiability are as easy to verify, and using them as starting points permits a natural development of the calculus in which such difficulties do not arise.

Our treatment of continuity and differentiation is from our forthcoming book, A New Course of Analysis, where it is expressed in terms of a theory of real numbers based on interval order and arithmetic. We offer no theory of real numbers in this article but we use repeatedly the fact that each real number can be approximated by rationals to arbitrary accuracy.

Uniform Calculus

Continuity. Uniform continuity of a one variable function $f$ is a condition on its variation, $f(y)-f(x)$. The condition, written MATH as $y-x\rightarrow 0$ for any two-variable function $\phi $, is that for each $\epsilon >0$, there is a $\delta >0$ such that MATH if $|y-x|\leq \delta $. When MATH, we also write: MATH as $y\rightarrow x$.




Example. The relationship
MATH
shows that MATH on any interval of the form $[-C,C]$ and, hence, that $x^{n}$ is uniformly continuous on each finite interval. A proof of pointwise continuity could hardly be simpler.




Example. Using MATH with $p=x^{1/n}$ and $q=(y-x)^{1/n}$, we have
MATH
It follows that for each positive integer $n$, $x^{1/n}$ is uniformly continuous on $[0,\infty )$.




Proposition

A composition of uniformly continuous functions is uniformly continuous.

Proposition

A uniformly continuous function $f$ on a finite interval $I $ is bounded.

Proof

For $\epsilon >0$, let $\delta >0$ be given by uniform continuity. Because $I $ is finite, we can find finitely many points such that every $p\in I$ is within $\delta $ of at least one of them. Hence, $f$ is bounded by $\epsilon $ plus the maximum of its values at these finitely many points.




Differentiability. Uniform differentiability of a function $f$ also is a condition on its variation: it factors as MATH, where MATH as $y\rightarrow x$. If $f$ is uniformly differentiable, its derivative is the function MATH. Thus, MATH, for $y$ different from $x$, and MATH.

Because the difference quotient converges to the derivative as $y\rightarrow x$, the derivative is unique on any domain $S$ for which each $x$ in $S$ is approximable to arbitrary accuracy by points $y$ in $S$ different from $x$.

Example. For all positive integers $n$, using the factorization of $y^{n}-x^{n}$ and the arithmetic of convergence (see Lemma ), it follows that on each finite interval, $x^{n}$ is uniformly differentiable with derivative $nx^{n-1}$.

Example. Because MATH and $y+x\rightarrow 2x$ on $\QTR{Bbb}{R}$ as $x\rightarrow 2x$, $x^{2}$ is differentiable on $\QTR{Bbb}{R}$ with derivative $2x$.




Proposition

If $f$ is uniformly differentiable, then $f~^{\prime }$ is uniformly continuous.

Proof

Because $F$ is symmetric, if $x$ and $y$ are close enough, both $f\ ^{\prime }(x)$ and $f\ ^{\prime }(y)$ are within $\epsilon $ of $F(x,y)=F(y,x)$ and hence within $2\epsilon $ of each other.




Corollary

On finite intervals, $f\ ^{\prime }$ is bounded.

Proof

See Proposition and proposition.




Proposition

If $f\ ^{\prime }$ is bounded, then $f$ is uniformly continuous.

Proof

When $f\ ^{\prime }$ is bounded, so is $F(x,y)$, say by $C$, for $|y-x|$ sufficiently small. Hence, MATH as $y-x\rightarrow 0$.




Theorem

(Fundamental Theorem of the Calculus) If $g$ is uniformly continuous on $[a,b]$, then MATH is uniformly differentiable on $[a,b]$ with $G^{\prime }=g$.

Proof

$G(y)-G(x)$ equals the integral of $g$ from $x$ to $y$, which equals $y-x$ times a limit of averages of values of $g$ at points in $[x,y]$. (To see this, approximate the integral by Riemann sums with equal spacing.) Also, for each $\epsilon >0$, if $|y-x|$ is small enough, every value of $g$ at a point in $[x,y]$ is within $\epsilon $ of $g(x)$. But then, also, any limit of averages of values of $g$ at points in $[x,y]$ is within $\epsilon $ of $g(x)$, so we are done.




The Arithmetic of Uniform Continuity

The arithmetic of uniform continuity is very simple. If both $f$ and $g$ are uniformly continuous, so is $f + g$. If also $f$ and $g$ are bounded, then $fg$ is uniformly continuous. Finally, if $1/f$ is defined and bounded, it too is uniformly continuous.

These statements can be verified by first relating the variations of the sum, product and reciprocal to those $f$ and $g$. Simple algebra shows that var($f + g$) = var($f$) + var($g$), var($fg) = g(y)$var($f) + f(x)$var($g$) and var($1/f$) = $-$var($f$)/($f(x)f(y)$).

Because each expression is a sum of expressions of the form $B(x,y)\phi (x,y) $, where $B(x,y)$ is bounded and MATH as $y-x\rightarrow 0$, it suffices to verify that each such sum again converges to $0$ as $y-x\rightarrow 0$. We omit the simple proof of this.




The Arithmetic of Uniform Differentiability

If $f$ and $g$ are uniformly differentiable, derivatives for their arithmetic combinations are given by the following rules.




Sums. $f+g$ is uniformly differentiable with MATH.

Products. If $f$, $g$, and their derivatives are bounded, e.g., if their domain is a finite interval, then $fg $ is uniformly differentiable with MATH.

Reciprocals. If $1/f$ is defined and bounded, and $f^{\prime }$ also is bounded, then $1/f$ is uniformly differentiable with MATH.




To prove these assertions, we begin by substituting $F(x,y)(y-x)$ and $G(x,y)(y-x)$ for var($f$) and var($g$) in our expressions for var($f+g$), var($fg$), and var($1/f$). For the sum, we get $F(x,y)+G(x,y)$, for the product, MATH. and for the reciprocal, MATH, each multiplied by $y-x$.

For $y=x$, these expressions become MATH, MATH, and MATH.

The case of the sum is clear. Both MATH and MATH converge to $0$ as $y-x\rightarrow 0$, hence so does the sum. For the product and reciprocal, multiplications are involved. The following lemma gives us what we need to deal with them.




Lemma

Suppose that $u$ and $v$ are bounded. If MATH and MATH as $y\rightarrow x$, then for $y-x$ sufficiently small, $U$ and $V$ are bounded and MATH as $y\rightarrow x$.

Proof

Write MATH and note that, because $u$ and $V$ are bounded for $y-x$ small enough, each summand converges to $0$ as $y-x\rightarrow 0$.




The next lemma is used to prove Proposition about the differentiability of an inverse function.




Lemma

Suppose that $1/u$ is defined and bounded. If MATH, then for $y-x$ sufficiently small, $1/U$ is defined and bounded, and MATH as $y\rightarrow x$.

Proof

We prove only the second part. Write $1/u-1/U=(U-u)/uU$ and note that $1/uU$ is bounded for $y-x$ sufficiently small.




For the product rule, we reason as follows. By assumption, $f$ is bounded and MATH as $y\rightarrow x$. Thus, MATH as $y\rightarrow x$. Because limits add, it suffices to prove that MATH as $y\rightarrow x$. But, also by assumption, MATH as $y\rightarrow x$, and $f^{\prime }$ and $g$ are bounded. Hence, if MATH as $y\rightarrow x$, we can apply Lemma . It therefore suffices to note that $g$ is uniformly continuous because $g^{\prime }$ is bounded.

Similarly, for the reciprocal rule, $f$ is uniformly continuous because $f^{\prime }$ is bounded, and because $1/f$ is bounded, it too is uniformly continuous. Hence, MATH as $y\rightarrow x$. Multiplying by $1/f(x)$, we see that MATH as $y\rightarrow x$. Because MATH as $y\rightarrow x$, and the limit functions $1/f^{2}$ and $-f^{\prime }$ are bounded, the product converges to the product by Lemma .







The Chain Rule

Proposition

If $f$ and $g$ are uniformly differentiable, and if $f^{\prime}$ and $g^{\prime}$ are bounded, then $f(g)$ is uniformly differentiable with derivative MATH.

Proof

Because MATH, which in turn equals MATH, the candidate for the derivative of $f(g)$ is indeed MATH. Because $g^{\prime }$ is bounded, $g$ is uniformly continuous. Hence, MATH as $y\rightarrow x$. Because MATH as $y\rightarrow x$ and $f^{\prime }(g)$ is bounded, an application of Lemma gives us the desired result.




Differentiability of the Inverse

Proposition

If $h$ is a uniformly continuous inverse for $f$, and if $1/f^{\prime}$ is defined and bounded, then $h$ is uniformly differentiable with MATH.

Proof

Because $h$ is an inverse for $f$, we can factor the variation of the identity function as
MATH
This shows that $1/F(h(x),h(y))$ is equal to the difference quotient for $h$ when $|y-x|>0$. Because $h$ is uniformly continuous, MATH as $y-x\rightarrow 0$. Therefore, because MATH is defined and bounded, we can apply Lemma to conclude that MATH as $y\rightarrow x$.




The Law of Bounded Change

Theorem

If $f$ is uniformly differentiable and MATH on $[a,b]$, then MATH.

This is the law of bounded change. It says that bounds for the derivative are bounds for the difference quotient. Notice that the increasing function theorem is just the law of bounded change for $A=0$ (and we don't care about $B$) and the law of bounded change is the increasing function theorem applied to the functions $Bx-f(x)$ and $f(x)-Ax$.




Proof

It suffices to prove that for all $\epsilon >0$, the conclusion holds with $A $ and $B$ replaced by $A-\epsilon $ and $B+\epsilon $. The justification for this is the general truth that if $p<q+\epsilon $ for all $\epsilon >0$, then $p\leq q$. That this holds for reals follows by rational approximation from the fact that it holds for rationals.

Since MATH as $v \rightarrow u$, for each $\epsilon >0$ there is a $\delta > 0$ such that MATH for $0\le v-u < \delta$. But MATH, so MATH lies between MATH and MATH.

Hence, if we express $f(b)-f(a)$ as a telescoping sum of $n$ differences MATH, where $u_{0}=a$ and each MATH, we have that MATH.




We now draw several useful and easy consequences of the law of bounded change.




Corollary

$f$ is constant on any interval on which $f^{\prime }=0$.

Proof

This is just the law of bounded change with $A$ and $B$ equal to $0$.

Is there any simpler or essentially different way to prove this deceptively obvious-looking fact?

Corollary

MATH

Proof

By the fundamental theorem of the calculus, the two sides of the equation have the same derivative. Hence, by previous Corollary, they differ by a constant. But they agree at $x=a$, so they agree everywhere.

(Alternatively, we can observe that in the proof of the law of bounded change, we in effect approximate $f(x)-f(a)$ to arbitrary accuracy by Riemann sums for the integral of $f^{\prime }$ from $a$ to $x$. Because these sums also approximate the integral, the two must be equal.)

Corollary

If MATH on $[a,b]$ and $f(h(u)) = u$ for all $u $ in $[f(a),f(b)]$, then $h$ is uniformly continuous.

Proof

By the law of bounded change, if $h(u)<h(v)$, then MATH. So MATH as $v-u\rightarrow 0$.

By the inverse function theorem, whenever MATH on $[a,b]$, there is a function $h$ as in the statement of the previous Corollary.

Corollary

If MATH on $[a,b]$, then
MATH

Proof

We apply the law of bounded change on $[x,y]$. Because the values of $f^{\prime }$ are in $[A,B]$, so is the difference quotient $(f(y)-f(x))/(y-x) $, which therefore cannot differ from any value of $f^{\prime }$ by more than $B-A$.

Corollary

If $f$ is uniformly differentiable on all sufficiently small subintervals of an interval $I$ and if $f^{\prime }$ is uniformly continuous on $I$, then $f$ is uniformly differentiable on $I$.

Proof

For $\epsilon >0$, the values of $f^{\prime }$ lie between MATH and MATH on each sufficiently small $[x,y]$ in $I$. Therefore, if MATH, the previous Corollary shows that MATH for $|y-x|$ sufficiently small.




The next consequence of the law of bounded change is needed for L'Hôpital's Rule. In it, $A$ and $B$ are constants, and $f$ and $g$ are uniformly differentiable on $[a,b]$.

Corollary

(Generalized Law of Bounded Change) If MATH on $[a,b]$, then MATH.

Proof

Apply the increasing function theorem to $Bg-f$ and $f-Ag$, and rearrange the resulting inequalities.




Application: L'Hôpital's Rule

We now present a few examples in support of Tucker's contention that the increasing function theorem serves nicely to prove major theorems of the calculus that traditionally are derived from the mean value theorem [6]. We begin with L'Hôpital's Rule; see also [2]. There are two cases. In both, we assume that $f$ and $g$ are defined on a semi-infinite interval $[c,\infty )$ and are uniformly differentiable on each finite subinterval. We assume also that $g$ and $g^{\prime }$ are positive.




Proposition

If $f(x)$ and $g(x)\rightarrow 0$ and MATH as MATH, then also MATH as MATH.

Proof

For $\epsilon >0$, MATH on $[p,\infty )$ if $p$ is large enough. In that case, if $p\leq x\leq y$, the generalized law of bounded change ensures that
MATH
Because weak inequalities are preserved in the limit, if we let MATH and divide by $g(x)>0$, we obtain MATH for all $x\geq p$.

In the second case of L'Hôpital's Rule, it is common to assume also that MATH, but there is no need to do so.




Proposition

If MATH and MATH as MATH, then also MATH as MATH.

Here too, the generalized law of bounded change is used only once. We note that if MATH lies between $L-\epsilon /2$ and $L+\epsilon /2$ for $x\geq p$, then so does MATH. But to complete the argument, one has to be more artful than in the first case.




Application: Differentiation Under the Integral Sign

Definition

A two-variable function $f$ is uniformly continuous if for each $\epsilon> 0$, there is a $\delta>0$ such that MATH whenever both $|x^{\prime}- x|$ and $|y^{\prime}- y|$ are smaller than $\delta$.

Remark

If the second coordinates in the previous Definition are equal, then the condition that guarantees that MATH involves only the first coordinates: MATH. That is, for each $\epsilon >0$, one $\delta >0$ works for all horizontal lines $y$ = constant. This simple observation is a key to our proof of the next Theorem.




Theorem

(Differentiation Under the Integral Sign) Let $f$ be defined on MATH and uniformly continuous on MATH. If $f$ is uniformly differentiable on each $[a,b]\times \{y\}$ and its partial derivative $f_{x}$ is uniformly continuous on $Q$, then $f$ is uniformly continuous on $Q$ and the integral of $f_{x}$ over $[c,d]$ is a derivative for the integral of $f$ over $[c,d]$.

Proof

We assume the uniform continuity of $f$ on $Q$. It can be be proved fairly easily using Corollary but we prefer to focus here on the second part of the argument, which employs a less familiar application of the law of bounded change. Integrating MATH over $[c,d]$, we see that to complete the proof, it suffices to demonstrate that the integral of MATH over $[c,d]$ converges to $0$ as $y-x\rightarrow 0$. To this end, it suffices to show that MATH can be made less than any $\epsilon >0$ by making $|y-x|$ less than some $\delta >0$, independent of $t$.

By Corollary , MATH is bounded for each $t$ by any bound for MATH, for all $u$ and $v$ in $[x,y]$. By Remark, for any $\epsilon >0$, there is a $\delta >0$ such that if $|x-y|\leq \delta $, then, for all $t$ in $[c,d]$, MATH for all $u$ and $v$ in $[x,y]$. This is precisely what we need.




Fubini's Theorem also follows easily from reversal of order of integration and Corollary. Because reversal of order of integration is a simple consequence of the existence of the double integral of a uniformly continuous function, this provides a proof of Fubini's Theorem that uses the law of bounded change in a more familiar way.




Higher Dimensions

In higher dimensions, there is no obvious counterpart to the increasing function theorem, and the mean value theorem is false even for a mapping from an interval to ${}\QTR{Bbb}{R}^{2}$. Yet the law of bounded change generalizes almost without alteration if we regard convex sets as higher dimensional counterparts to intervals and read the proof as showing that if $f$ is defined on an interval $[a,b]$, then any closed interval that contains MATH also contains $f(b)-f(a)$.

Definition

A map $f$ from a subset $U$ of one normed linear space $X$ to another $Y$ is uniformly differentiable if there is a map $Df$ from $U$ to the set of bounded linear transformations from $X$ to $Y$ such that for each $\epsilon>0$, MATH if $\|q - p\|_X$ is sufficiently small.

Proposition

Using the notation in the previous definition, if $U$ is convex then, for each $p$ and $q$ in $U$, $f(q)-f(p)$ belongs to every convex subset of $Y$ that contains $Df(u)(q-p)$ for all $u$ in $U$. Hence, if each MATH is bounded by $K$ on the unit sphere of $X$, then MATH.

Proof

If $p$ and $q$ are in $U$, so is the line segment joining them. Hence, using a telescoping sum as in the proof of the Law of Bounded Change, and approximating each summand by the value of $Df$ at a point on the segment, applied to $(q-p)/n$, we can approximate $f(q)-f(p)$ to arbitary accuracy by an average of finitely many values of $Df$ at points along the segment, applied to $q-p$.




Afterword

We believe that this development, which is in the constructivist manner of Errett Bishop and L. E. J. Brouwer [4], produces proofs that are shorter and more transparent than those encountered in classical treatments. The idea of working with uniform rather than pointwise notions is a hallmark of the constructivist tradition.

For the one-dimensional case, our definition of differentiable function is a uniform version of a definition of Carath\eodory. See [3] and the references therein. For a definition of this kind in higher dimensions, see [1].




References

1. Acosta, E. and Delgado, C., Fréchet vs. Carathéodory, Amer. Math. Monthly 101 (1994) 332-338.

2. Boas, R.P., L'Hôpital's Rule without Mean Values, Amer. Math. Monthly 76 (1969) 1051-1053.

3. Kuhn, Stephen, The Derivative á la Carathéodory, Amer. Math. Monthly 98 (1991) 40-44.

4. Stolzenberg, Gabriel, review of Foundations of Constructive Analysis by Errett Bishop, Bull. Amer. Math. Soc. 76 (1970) 301-323.

5. Swann, Howard, Commentary on Rethinking Rigor in Calculus: The Role of the Mean Value Theorem, Amer. Math. Monthly 104 (1997) 241-245.

6. Tucker, Thomas, Rethinking Rigor in Calculus: The Role of the Mean Value Theorem, Amer. Math. Monthly 104 (1997) 231-240.







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