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Cauchy sum theorem

The Cauchy sum theorem

Overview

In his influential 1821 textbook Cours d'Analyse, Augustin Cauchy states a theorem that is now widely regarded as false, attributed to a confusion between pointwise convergence and uniform convergence. This mistake —if indeed it is a mistake— is of both pedagogical and philosophical-historical interest.

Counterexamples (specific Fourier series) were known already to Joseph Fourier?, and Niels Abel specifically pointed them out as counterexamples to Cauchy's claim. However, Cauchy denied that these were counterexamples, on the grounds that the series did not converge everywhere (which is a hypothesis in the theorem). Imre Lakatos? has argued that the confusion rests on different conceptions of the continuum, so that Cauchy's notion of convergence everywhere is really more like Weierstrass's notion of uniform convergence than pointwise convergence, and the theorem as he intended it is true.

Although Cauchy's original formulation was about the sum of an infinite series, we will consider it in the slightly more elementary context of the limit of an infinite sequence; the modern isomorphism between these was already well established and used explicitly by Cauchy.

Statements

Non-theorem (attributed to Cauchy, 1821)

Let f=(f 1,f 2,)f = (f_1, f_2, \ldots) be an infinite sequence of continuous functions from the real line to itself. Suppose that, for every real number xx, the sequence (f 1(x),f 2(x),)(f_1(x), f_2(x), \ldots) converges to some (necessarily unique) real number f (x)f_\infty(x), defining a function f f_\infty; in other words, the sequence ff converges pointwise to f f_\infty. Then f f_\infty is also continuous.

Theorem (attributed to Seidel, 1847)

Let f=(f 1,f 2,)f = (f_1, f_2, \ldots) be an infinite sequence of continuous functions from the real line to itself. Suppose that the sequence ff converges uniformly to a function f f_\infty. Then f f_\infty is also continuous.

Theorem (attributed to ???)

Let f=(f 1,f 2,)f = (f_1, f_2, \ldots) be an equicontinuous sequence of (necessarily continuous) functions from the real line to itself. Suppose that, for every real number xx, the sequence (f 1(x),f 2(x),)(f_1(x), f_2(x), \ldots) converges to some (necessarily unique) real number f (x)f_\infty(x), defining a function f f_\infty; in other words, the sequence ff converges pointwise to f f_\infty. Then f f_\infty is also continuous.

Counterexamples

The first counterexamples to Non-Theorem 1 arose as Fourier series. The sawtooth wave

k=1 sin(kx)k=π2xmod2π2 \sum_{k = 1}^\infty \frac{\sin(k x)}{k} = \frac{\pi}{2} - \frac{x \mathbin{mod} 2 \pi}{2}

may be the simplest. Each partial sum of this infinite series is continuous; the sum converges pointwise as indicated for xx not a multiple of 2π2 \pi and to 00 (which is the average of the limits ±π/2\pm\pi/2 on either side) for xx a multiple of 2π2 \pi. However, the sequence of partial sums is not equicontinuous, nor does it converge uniformly. And indeed, the sum is not continuous at multiples of 2π2 \pi.

A simple counterexample from outside of this context is

f n(x)x n f_n(x) \coloneqq x^n

on [0,1][0,1]. (We can make this a function on all of \mathbb{R} by letting f n(x)f_n(x) be f n(0)=0f_n(0) = 0 for x<0x \lt 0 and letting f n(x)f_n(x) be f n(1)=1f_n(1) = 1 for x>1x \gt 1.) Each f nf_n is continuous; the pointwise limit is f (x)=0f_\infty(x) = 0 for x<1x \lt 1 but f (x)=1f_\infty(x) = 1 for x1x \geq 1, which is not continuous. Again, the sequence is not equicontinuous, and its convergence is not uniform.

Another counterexample is

f n(x)exp(n|x|). f_n(x) \coloneqq \exp ({-n {|x|}}) .

(The absolute value is here only to handle negative values of xx; the example is analytic on ]0,[]0,\infty[.) Now f (0)=1f_\infty(0) = 1, while f (x)=0f_\infty(x) = 0 for x0x \ne 0.

Proofs

Here is Cauchy's argument:

Proof?

At any xx, we have

|f (x+h)f (x)|=|f (x+h)f n(x+h)+f n(x+h)f n(x)+f n(x)f (x)||f (x+h)f n(x+h)|+|f n(x+h)f n(x)|+|f n(x)f (x)|. {|f_\infty(x + h) - f_\infty(x)|} = {|f_\infty(x + h) - f_n(x + h) + f_n(x + h) - f_n(x) + f_n(x) - f_\infty(x)|} \leq {|f_\infty(x + h) - f_n(x + h)|} + {|f_n(x + h) - f_n(x)|} + {|f_n(x) - f_\infty(x)|} .

As nn becomes infinite and hh becomes infinitesimal, all of the terms on the right become infinitesimal (respectively, because ff converges to f f_\infty at x+hx + h, because f nf_n is continuous at xx, and because ff converges to f f_\infty at xx). Thus, the expression on the left also becomes infinitesimal, proving that f f_\infty is continuous at xx.

Nowadays, we'd say that a variable approaches zero rather than that it becomes infinitesimal. If we interpret this argument in modern analysis, then the proof is flawed, because (the lower bound for) nn depends on hh in the first term, while (the upper bound for the absolute value of) hh depends on nn in the second term; they cannot be chosen simultaneously.

Analysing this argument, Philipp von Seidel (first, and others afterwards) realised that it could be fixed if ff converges to f f_\infty uniformly, so that nn is independent of hh. Another fix is to make hh independent of nn, by requiring the family ff to be equicontinuous, although this idea came later.

In epsilontics

Writing Cauchy's argument in the epsilontic language developed by Karl Weierstrass, we have:

Non-proof

(Non-theorem 1).

Let ϵ\epsilon be a positive number, and consider ϵ/3\epsilon/3. Because ff converges to f f_\infty at xx, there is some natural number NN such that |f n(x)f (x)|<ϵ/3{|f_n(x) - f_\infty(x)|} \lt \epsilon/3 whenever nNn \geq N. Because f nf_n is continuous at xx for each nn, there is some positive number δ\delta such that |f n(x+h)f n(x)|<ϵ/3{|f_n(x + h) - f_n(x)|} \lt \epsilon/3 whenever h<δh \lt \delta. Because ff converges to f f_\infty at x+hx + h for each hh, there is some natural number NN such that |f (x+h)f n(x+h)|<ϵ/3{|f_\infty(x + h) - f_n(x + h)|} \lt \epsilon/3 whenever nNn \geq N. Therefore, there are NN and δ\delta such that

|f (x+h)f (x)||f (x+h)f n(x+h)|+|f n(x+h)f n(x)|+|f n(x)f (x)|<ϵ/3+ϵ/3+ϵ/3=ϵ {|f_\infty(x + h) - f_\infty(x)|} \leq {|f_\infty(x + h) - f_n(x + h)|} + {|f_n(x + h) - f_n(x)|} + {|f_n(x) - f_\infty(x)|} \lt \epsilon/3 + \epsilon/3 + \epsilon/3 = \epsilon

whenever nNn \geq N and h<δh \lt \delta. Fixing any such nn, f f_\infty is continuous.

We have used the variable names nn and hh in two different contexts each, then pretended that they arose in a single context for the final inequality. The error can be made more explicit by using different variable names:

Non-proof

(Non-theorem 1).

Let ϵ\epsilon be a positive number, and consider ϵ/3\epsilon/3. Because ff converges to f f_\infty at xx, there is some natural number NN such that |f n(x)f (x)|<ϵ/3{|f_n(x) - f_\infty(x)|} \lt \epsilon/3 whenever nNn \geq N. Because f nf_{n'} is continuous at xx for each nn', there is some positive number δ\delta such that |f n(x+h)f n(x)|<ϵ/3{|f_{n'}(x + h) - f_{n'}(x)|} \lt \epsilon/3 whenever h<δh \lt \delta. Because ff converges to f f_\infty at x+hx + h' for each hh', there is some natural number NN' such that |f (x+h)f n(x+h)|<ϵ/3{|f_\infty(x + h') - f_{n''}(x + h')|} \lt \epsilon/3 whenever nNn'' \geq N'. Therefore, there are N,NN, N' and δ\delta such that

|f (x+h)f (x)||f (x+h)f n(x+h)|+|f n(x+h)f n(x)|+|f n(x)f (x)|<ϵ/3+ϵ/3+ϵ/3=ϵ {|f_\infty(x + h) - f_\infty(x)|} \leq {|f_\infty(x + h') - f_{n''}(x + h')|} + {|f_{n'}(x + h) - f_{n'}(x)|} + {|f_n(x) - f_\infty(x)|} \lt \epsilon/3 + \epsilon/3 + \epsilon/3 = \epsilon

whenever n,n,nmax(N,N)n, n', n'' \geq max(N, N') and h,h<δh, h' \lt \delta. Fixing any such n,n,nn, n', n'', f f_\infty is continuous.

The final inequality is now clearly spurious, since n,n,nn, n', n'' need not be equal, nor h,hh, h'.

This can be fixed up to a point. We may let nn' be nn and let hh' be hh, but then we have no control over nn''. Conversely, we may let n,n,nn, n', n'' all be (anything bounded below by) max(N,N)max(N,N'), but then we have no control over hh'. Indeed, Non-theorem 1 is false, as the counterexamples show.

However, the weaker theorems with strengthened hypotheses are valid, by similar proofs:

Proof

(Theorem 2).

Let ϵ\epsilon be a positive number, and consider ϵ/3\epsilon/3. Because ff converges uniformly to f f_\infty, there is some natural number NN such that |f n(x+h)f (x+h)|<ϵ/3{|f_n(x + h) - f_\infty(x + h)|} \lt \epsilon/3 for every hh (including h0h \coloneqq 0) whenever nNn \geq N. Because f nf_n is continuous at xx for each such nn, there is some positive number δ\delta such that |f n(x+h)f n(x)|<ϵ/3{|f_n(x + h) - f_n(x)|} \lt \epsilon/3 whenever h<δh \lt \delta. Therefore, there is NN such that, whenever nNn \geq N, there is δ\delta such that

|f (x+h)f (x)||f (x+h)f n(x+h)|+|f n(x+h)f n(x)|+|f n(x)f (x)|<ϵ/3+ϵ/3+ϵ/3=ϵ {|f_\infty(x + h) - f_\infty(x)|} \leq {|f_\infty(x + h) - f_n(x + h)|} + {|f_n(x + h) - f_n(x)|} + {|f_n(x) - f_\infty(x)|} \lt \epsilon/3 + \epsilon/3 + \epsilon/3 = \epsilon

whenever h<δh \lt \delta. Fixing any such nn, f f_\infty is continuous.

Proof

(Theorem 3).

Let ϵ\epsilon be a positive number, and consider ϵ/3\epsilon/3. Because ff converges to f f_\infty at xx, there is some natural number NN such that |f n(x)f (x)|<ϵ/3{|f_n(x) - f_\infty(x)|} \lt \epsilon/3 whenever nNn \geq N. Because ff is equicontinuous at xx, there is some positive number δ\delta such that |f n(x+h)f n(x)|<ϵ/3{|f_n(x + h) - f_n(x)|} \lt \epsilon/3 for every nn whenever h<δh \lt \delta. Because ff converges to f f_\infty at x+hx + h for each such hh, there is some natural number NN' such that |f (x+h)f n(x+h)|<ϵ/3{|f_\infty(x + h) - f_n(x + h)|} \lt \epsilon/3 whenever nNn \geq N'. Therefore, there are NN and δ\delta such that, whenever h<δh \lt \delta, there is NN' such that

|f (x+h)f (x)||f (x+h)f n(x+h)|+|f n(x+h)f n(x)|+|f n(x)f (x)|<ϵ/3+ϵ/3+ϵ/3=ϵ {|f_\infty(x + h) - f_\infty(x)|} \leq {|f_\infty(x + h) - f_n(x + h)|} + {|f_n(x + h) - f_n(x)|} + {|f_n(x) - f_\infty(x)|} \lt \epsilon/3 + \epsilon/3 + \epsilon/3 = \epsilon

whenever nmax(N,N)n \geq max(N,N'). Fixing any such nn, f f_\infty is continuous.

In nonstandard analysis

Writing Cauchy's argument in the language of nonstandard analysis developed by Abraham Robinson, we have:

Non-proof

(Non-theorem 1).

Because ff converges to f f_\infty at xx, |f n(x)f (x)|{|f_n(x) - f_\infty(x)|} is infinitesimal whenever nn is infinite?. Because f nf_n is continuous at xx for each finite nn, it is continuous at xx for each infinite nn (by the transfer principle), so |f n(x+h)f n(x)|{|f_n(x + h) - f_n(x)|} is infinitesimal whenever hh is infinitesimal. Because ff converges to f f_\infty at each standard point?, ff converges to f f_\infty at the nonstandard point x+hx + h (by the transfer principle), so |f (x+h)f n(x+h)|{|f_\infty(x + h) - f_n(x + h)|} is infinitesimal whenever nn is infinite. Therefore,

|f (x+h)f (x)||f (x+h)f n(x+h)|+|f n(x+h)f n(x)|+|f n(x)f (x)| {|f_\infty(x + h) - f_\infty(x)|} \leq {|f_\infty(x + h) - f_n(x + h)|} + {|f_n(x + h) - f_n(x)|} + {|f_n(x) - f_\infty(x)|}

is also infinitesimal whenever nn is infinite and hh is infinitesimal. Fixing any such nn, f f_\infty is continuous.

Both uses of the transfer principle are illegitimate, since the statements to which they're applied are not first-order in the language of the real numbers. Indeed, consider the counterexample f n(x)=exp(n|x|)f_n(x) = \exp ({-n {|x|}}) near x=1x = 1. Then |f n(0+h)f n(0)|=exp(n|h|)1{|f_n(0 + h) - f_n(0)|} = \exp ({-n {|h|}}) - 1 is infinitesimal as claimed iff nhn h is infinitesimal, which may fail. Conversely, |f (0+h)f n(0+h)|=exp(n|h|){|f_\infty(0 + h) - f_n(0 + h)|} = \exp ({-n {|h|}}) (for h0h \ne 0) is infinitesimal iff nhn h is infinite, which may also fail. In between, when nhn h is finite and finitesimal, both fail!

We can rewrite this proof more carefully, putting down what really follows in place of the misuse of the transfer principle:

Non-proof

(Non-theorem 1).

Because ff converges to f f_\infty at xx, |f n(x)f (x)|{|f_n(x) - f_\infty(x)|} is infinitesimal whenever nn is infinite?. Because f nf_n is continuous at xx for each finite nn, there is some infinite NN such that f nf_n is continuous at xx for all nNn \leq N, so then |f n(x+h)f n(x)|{|f_n(x + h) - f_n(x)|} is infinitesimal whenever hh is infinitesimal. Because ff converges to f f_\infty at each standard point? and x+hx + h is nearstandard?, there is an infinite number NN' such that |f (x+h)f n(x+h)|{|f_\infty(x + h) - f_n(x + h)|} is infinitesimal whenever nNn \geq N'. Therefore,

|f (x+h)f (x)||f (x+h)f n(x+h)|+|f n(x+h)f n(x)|+|f n(x)f (x)| {|f_\infty(x + h) - f_\infty(x)|} \leq {|f_\infty(x + h) - f_n(x + h)|} + {|f_n(x + h) - f_n(x)|} + {|f_n(x) - f_\infty(x)|}

is also infinitesimal whenever nNn \leq N, nNn \geq N', and hh is infinitesimal. Fixing any such nn, f f_\infty is continuous.

This is more clearly wrong, since there may be no such nn in the last step (as NN' may be larger than NN).

However, we can modify this to produce correct proofs of the weaker theorems:

Proof

(Theorem 2).

Because ff converges to f f_\infty at xx, |f n(x)f (x)|{|f_n(x) - f_\infty(x)|} is infinitesimal whenever nn is infinite?. Because f nf_n is continuous at xx for each finite nn, there is some infinite NN such that f nf_n is continuous at xx for all nNn \leq N, so then |f n(x+h)f n(x)|{|f_n(x + h) - f_n(x)|} is infinitesimal whenever hh is infinitesimal. Because ff converges to f f_\infty uniformly, ff converges to f f_\infty at the nonstandard point x+hx + h, so |f (x+h)f n(x+h)|{|f_\infty(x + h) - f_n(x + h)|} is infinitesimal whenever nn is infinite. Therefore,

|f (x+h)f (x)||f (x+h)f n(x+h)|+|f n(x+h)f n(x)|+|f n(x)f (x)| {|f_\infty(x + h) - f_\infty(x)|} \leq {|f_\infty(x + h) - f_n(x + h)|} + {|f_n(x + h) - f_n(x)|} + {|f_n(x) - f_\infty(x)|}

is also infinitesimal whenever nNn \leq N is infinite and hh is infinitesimal. Fixing any such nn, f f_\infty is continuous.

Proof

(Theorem 3).

Because ff converges to f f_\infty at xx, |f n(x)f (x)|{|f_n(x) - f_\infty(x)|} is infinitesimal whenever nn is infinite?. Because ff is equicontinuous at xx, f nf_n is continuous at xx even for infinite nn, so |f n(x+h)f n(x)|{|f_n(x + h) - f_n(x)|} is infinitesimal whenever hh is infinitesimal. Because ff converges to f f_\infty at each standard point? and x+hx + h is nearstandard?, there is an infinite number NN such that |f (x+h)f n(x+h)|{|f_\infty(x + h) - f_n(x + h)|} is infinitesimal whenever nNn \geq N. Therefore,

|f (x+h)f (x)||f (x+h)f n(x+h)|+|f n(x+h)f n(x)|+|f n(x)f (x)| {|f_\infty(x + h) - f_\infty(x)|} \leq {|f_\infty(x + h) - f_n(x + h)|} + {|f_n(x + h) - f_n(x)|} + {|f_n(x) - f_\infty(x)|}

is also infinitesimal whenever nNn \geq N and hh is infinitesimal. Fixing any such nn, f f_\infty is continuous.

(To check that these are valid, one must know the nonstandard formulations of uniform convergence and equicontinuity that are used; but they are correct. Specifically, a sequence ff of functions converges uniformly to a function f f_\infty iff, for every hyperreal number xx, the sequence f(x)=(f n(x)) nf(x) = (f_n(x))_n of hyperreal numbers converges to the hyperreal number f (x)f_\infty(x); and ff is equicontinuous at a real number xx iff, for every hyperinteger nn, the hyperfunction f nf_n is continuous at xx.)

On the other hand, in the light of nonstandard analysis, we can also reevaluate Cauchy's original claim. Cauchy himself, when confronted with counterexamples such as the Fourier series, denied that the sequence of functions converged everywhere. One interpretation of this is that it fails to converge at some nonstandard points. By this analysis, Cauchy's notion of convergence everywhere is our modern notion of uniform convergence, not pointwise convergence, and his theorem is true (and his proof correct, even if less detailed than one might like).

Lakatos's view

In his experimental textbook Proofs and Refutations?, Imre Lakatos? used Cauchy's sum theorem to motivate the concept of uniform convergence. Two years later, he partially reevaluated his discussion in light of the question of what Cauchy's conception of the continuum was.

According to Lakatos, it is ahistorical to interpret Cauchy's 1821 result using either Weierstrass's epsilontics or Robinson's nonstandard analysis. Cauchy did not mean that f n(x)f_n(x) converges (in nn) for each fixed standard real number xx, nor that it converges for each fixed hyperreal number xx; rather, he said that it converges for each variable real number. In particular, when discussing the Fourier series

k=1 sin(kx)k, \sum_{k = 1}^\infty \frac{\sin(k x)}{k} ,

Cauchy states (in 1853) that the sequence

(1) k=1 nsin(kx)k \sum_{k = 1}^n \frac{\sin(k x)}{k}

of partial sums fails to converge when x=1/nx = 1/n; that is, xx is a variable whose value depends on the position in the sequence! Such a claim is hard to interpret in an epsilontic framework; but in a nonstandard framework, it simply means that for some particular infinite values of nn and infinitesimal values of xx (namely those for which x=1/nx = 1/n), this partial sum is not infinitely close to 00 (which is the value of the partial sum at x=lim n(1/n)=0x = \lim_n (1/n) = 0), demonstrating that the sequence is not equicontinuous.

If we (still arguably ahistorically) interpret convergence to mean to that f n(x n)f_n(x_n) converges (in nn) for every convergent sequence (x n) n(x_n)_n of real numbers, then this is equivalent (given that each f nf_n is continuous) to uniform convergence, and Cauchy's result would again be true and his argument would be (when formalised with either epsilontics or nonstandard analysis) valid.

On the other hand (and this goes beyond what Lakatos wrote), one way to interpret Cauchy's claim that (1) fails to converge when x=1/nx = 1/n is, using nonstandard analysis, simply that for any particular infinite value of nn and using the infinitesimal value x=1/nx = 1/n, this partial sum is not infinitely close to 00 (which is the value of the infinite sum at x=lim n(1/n)=0x = \lim_n (1/n) = 0). This suggests interpreting convergence to mean that f n(c+h)f_n(c+h) is infinitely close to f (c)f_\infty(c) whenever nn is infinite and hh is infinitesimal (for cc real), which again is equivalent (given continuity of f nf_n) to uniform convergence, again justifying Cauchy's proof. However, proving that uniform convergence is equivalent to this is basically the same concept as the nonstandard proof of Theorem 2 anyway.

References

The original sum theorem is in

Cauchy's most thorough defence of the theorem against critics is in

  • Augustin Cauchy (1853). Note sur les séries convergentes dont les divers termes sont des fonctions continues d'une variable réelle ou imaginaire, entre des limites données.

The reconstruction in terms of nonstandard analysis is in the historical appendix to

Lakatos's discussion forms Chapter 3 of

  • Imre Lakatos? (1978). Mathematics, Science, and Epistemology.

Revised on May 3, 2017 13:46:07 by Anonymous (2.40.107.129)