Todd Trimble
Continuity of the exponential function

Contents

Introduction

This little note was motivated by a discussion at the nForum on alternate approaches to the teaching of calculus. We first present several ways of seeing that continuous exponential functions have derivatives and are in fact infinitely differentiable; the third proof we give can in principle be explained to students who are just beginning calculus, and without taking any shortcuts such as integration theory, or the assumption of major theorems like the intermediate value theorem or extreme value theorem. We then carry out a similar analysis on trigonometric functions (we do appeal to complex numbers for the sake of conceptual clarity and brevity, but of course it is a simple matter to rewrite the arguments so as to circumvent complex numbers).

Exponential functions

Proposition

Suppose E:E: \mathbb{R} \to \mathbb{R} is continuous and satisfies E(x+y)=E(x)E(y)E(x+y)=E(x)E(y). Then EE is infinitely differentiable.

To get something out of the way: if E(a)=0E(a) = 0 for any aa, then E(x)=E(xa)E(a)=0E(x) = E(x-a)E(a) = 0 for all xx. Otherwise, we have E(x)>0E(x) \gt 0 for all xx, since after all E(x)=E(x/2) 2E(x) = E(x/2)^2.

One can dream up various proofs of this proposition. But the ultimate goal is to try to keep it simple.

Proof 1

Construct the natural logarithm first as log(x)= 1 xdtt\log(x) = \int_1^x \frac{d t}{t} and define exp\exp as its inverse. These give a C C^\infty diffeomorphism between +\mathbb{R}_+ and \mathbb{R}, so we can use this to reduce the question to the Cauchy functional equation

f(x+y)=f(x)+f(y)f(x+y) = f(x) + f(y)

where continuity of ff implies that ff is given by multiplication by the real scalar k=f(1)k = f(1) (since this is true on the integers, and then on the rationals, and then on the reals by a density argument). Thus EE is of the form E(x)=exp(kx)E(x) = \exp (k x) which is certainly C C^\infty.

Proof 2

This proof also uses integration theory. First one argues that the integral 0 aE(t)dt\int_0^a E(t) d t (which exists since EE is continuous) is nonzero for some aa; let CC denote this quantity. Then we have

CE(x)= 0 aE(t)E(x)dt= 0 aE(x+t)dt= x x+aE(s)dsC E(x) = \int_0^a E(t)E(x) d t = \int_0^a E(x+t) d t = \int_{x}^{x+a} E(s) d s

where the right side is continuously differentiable by the fundamental theorem of calculus. Hence the left side of

E(x)=1C x x+aE(s)dsE(x) = \frac1{C}\int_x^{x+a} E(s) d s

is continuously differentiable. From here, one can apply the usual arguments to show E(x)=E(0)E(x)E'(x) = E'(0)E(x).

However, since the discussion which gave rise to this had to do with the teaching of calculus to beginning students (who might not have any background in integration theory), the question is whether a reasonable proof can be given without using integrals.

Remark

It’s not just a matter of avoiding integrals. More to the point is to avoid hauling out the big guns, such as the intermediate value theorem or the mean value theorem. From the point of view of developing calculus rigorously, such results should not be considered fair play unless one is actually prepared to prove them, a task which is usually left for a first course in real analysis. This makes our job a bit more challenging!

We will however allow ourselves one luxury: we assume the key property of the Dedekind real numbers, that every set of real numbers that admits an upper bound has a least upper bound. (End remark)

It suffices to show that EE has a derivative somewhere (for then it can be shown it has a derivative everywhere, and then one calculates E(x)=E(0)E(x)E'(x) = E'(0)E(x)).

Proof 3

This is based on the convexity of EE. First one proves E(x+y2)E(x)+E(y)2E(\frac{x+y}{2}) \leq \frac{E(x)+E(y)}{2} (easy exercise: it boils down to 2aba 2+b 22 a b \leq a^2 + b^2), so that

E(tx+(1t)y)tE(x)+(1t)E(y)E(t x + (1-t)y) \leq t E(x) + (1-t)E(y)

for all dyadic rationals tt between 00 and 11, and thence for all t[0,1]t \in [0, 1] by virtue of continuity of EE.

This has the consequence that slopes of secant lines through (0,1)(0, 1), namely E(t)1t\frac{E(t)-1}{t} for t0t \neq 0, increase with tt. It follows that the left limit

lim t0E(t)1t\lim_{t \nearrow 0} \frac{E(t)-1}{t}

exists, being a limsup over tt that has an upper bound (given by any E(s)1s\frac{E(s)-1}{s} where s>0s \gt 0). Similarly the right limit exists. We just need to show that the right limit R 0R_0 and left limit L 0L_0 agree. The basic idea is that if they don’t agree (if there is a jump discontinuity at 00, so to speak), then there is a jump discontinuity everywhere, and that spells trouble, mister.

Thus, let us define

L alim h0E(a+h)E(a)h;R alim h0E(a+h)E(a)h.L_a \coloneqq \lim_{h \nearrow 0} \; \frac{E(a+h)-E(a)}{h}; \qquad R_a \coloneqq \lim_{h\searrow 0} \; \frac{E(a+h)-E(a)}{h}.

One easily verifies that R a=E(a)R 0R_a = E(a)R_0 and L a=E(a)L 0L_a = E(a)L_0. Also observe that whenever a<ba \lt b, we have

R a<E(b)E(a)ba<L bR_a \lt \frac{E(b)-E(a)}{b-a} \lt L_b

so that L b/R a>1L_b/R_a \gt 1.

Okay, certainly 0<L 0R 00 \lt L_0 \leq R_0, so cR 0/L 01c \coloneqq R_0/L_0 \geq 1. We have R a/L a=cR_a/L_a = c for any aa. Telescoping, we have for every nn that

L 1L 0 = (R 0L 0L 1/nR 0)(R 1/nL 1/nL 2/nR 1/n)(R (n1)/nL (n1)/nL 1R (n1)/n) > R 0L 0R 1/nL 1/nR (n1)/nL (n1)/n = c n\array{ \frac{L_1}{L_0} & = & (\frac{R_0}{L_0}\frac{L_{1/n}}{R_0})\cdot (\frac{R_{1/n}}{L_{1/n}}\frac{L_{2/n}}{R_{1/n}}) \cdot \ldots \cdot (\frac{R_{(n-1)/n}}{L_{(n-1)/n}} \frac{L_1}{R_{(n-1)/n}}) \\ & \gt & \frac{R_0}{L_0} \frac{R_{1/n}}{L_{1/n}} \ldots \frac{R_{(n-1)/n}}{L_{(n-1)/n}} \\ & = & c^n }

so that 1c<(L 1L 0) 1/n1 \leq c \lt (\frac{L_1}{L_0})^{1/n} for all n1n \geq 1. Thus c=1c = 1, which completes the proof.

Trigonometric functions

Now we set out to prove analogous results for trigonometric functions:

Theorem

Suppose S,C:S, C: \mathbb{R} \to \mathbb{R} are continuous functions satisfying the addition formulas

S(x+y)=S(x)C(y)+C(x)S(y);C(x+y)=C(x)C(y)S(x)S(y).S(x+y) = S(x)C(y) + C(x)S(y); \qquad C(x+y) = C(x)C(y) - S(x)S(y).

Then SS and CC are infinitely differentiable.

Once again, as in Proof 3 above, we will seek to avoid integration theory, including any discussion of arc length along the unit circle (on which the standard treatment of circular functions is based).

Preliminary remarks

Note that if we put E(t)=C(t)+iS(t)E(t) = C(t) + i S(t), then we have a (continuous) exponential function E:E: \mathbb{R} \to \mathbb{C}. Arguing as we did earlier, it is easy to see that EE is either zero nowhere or zero everywhere. We restrict to the “zero nowhere” case, where E(0)=1E(0) = 1 (thus C(0)=1C(0) = 1 and S(0)=0S(0) = 0).

The complex conjugate E¯\bar{E} of EE is a continuous exponential function, as is therefore EE¯=|E| 2:E \bar{E} = {|E|}^2: \mathbb{R} \to \mathbb{R}, and also |E|:{|E|}: \mathbb{R} \to \mathbb{R}. We have already characterized continuous exponential functions of the latter type in the previous section. Therefore we are reduced to understanding continuous exponential functions

E/|E|:S 1E/{|E|}: \mathbb{R} \to S^1

mapping to complex numbers of norm 11.

This reduction justifies adding in the assumption that C(t) 2+S(t) 2=1C(t)^2 + S(t)^2 = 1. From E(t)E(t)=E(t+t)=E(0)=1E(-t)E(t) = E(-t+t) = E(0) = 1, we see that E(t)=E¯(t)E(-t) = \bar{E}(t), so that CC is an even function and SS is an odd function.

Convexity

Lemma

If S(a)<0S(a) \lt 0 (C(a)<0C(a) \lt 0), then there is an interval about aa over which SS (CC) is convex; similarly if S(a)>0S(a) \gt 0 (C(a)>0C(a) \gt 0), then there is an interval about aa over which S-S (C-C) is convex.

Proof

We will just consider one case; the other cases are similarly routine1. Suppose S(a)>0S(a) \gt 0; choose an interval UU about aa where SS is positive. We show that for x,yUx, y \in U that

S(x+y2)S(x)+S(y)2.S(\frac{x+y}{2}) \geq \frac{S(x) + S(y)}{2}.

This is sufficient, since we may argue as we did earlier (viz., that S(tx+(1t)y)tS(x)+(1t)S(y)S(t x + (1-t)y) \geq t S(x) + (1-t)S(y) for all dyadic rationals t[0,1]t \in [0, 1] and thence for all t[0,1]t \in [0, 1] by continuity).

For u=x/2u = x/2 and v=y/2v = y/2, we have S(2u),S(2v)>0S(2 u), S(2 v) \gt 0. Applying the addition formulas, we are to show

S(u)C(v)+C(u)S(v)2S(u)C(u)+2S(v)C(v)2=S(u)C(u)+S(v)C(v)S(u)C(v) + C(u)S(v) \geq \frac{2S(u)C(u) + 2S(v)C(v)}{2} = S(u)C(u) + S(v)C(v)

where S(u)S(u) and C(u)C(u) have the same sign since S(2u)=2S(u)C(u)>0S(2 u) = 2S(u)C(u) \gt 0. Rearranging, we must show

(S(u)S(v))(C(v)C(u))0.(S(u) - S(v))(C(v) - C(u)) \geq 0.

Suppose WLOG that S(u)S(v)S(u) \geq S(v); consider first the case S(u)S(v)>0S(u) \geq S(v) \gt 0. Then S(u) 2S(v) 2S(u)^2 \geq S(v)^2, whence C(u) 2C(v) 2C(u)^2 \leq C(v)^2 (using C(t) 2+S(t) 2=1C(t)^2 + S(t)^2 = 1), whence C(u)C(v)C(u) \leq C(v) (since C(u)>0C(u) \gt 0; remember C(u)C(u) and S(u)S(u) have the same sign). The displayed inequality follows. If 0>S(u)S(v)0 \gt S(u) \geq S(v), then similarly S(u) 2S(v) 2S(u)^2 \leq S(v)^2, whence C(u) 2C(v) 2C(u)^2 \geq C(v)^2, whence 0>C(v)C(u)0 \gt C(v) \geq C(u), and again the displayed inequality follows. Finally, the case where S(u)>0>S(v)S(u) \gt 0 \gt S(v) actually does not occur, because then C(u)>0>C(v)C(u) \gt 0 \gt C(v) and consequently S(x+y2)=S(u)C(v)+C(u)S(v)<0S(\frac{x+y}{2}) = S(u)C(v) + C(u)S(v) \lt 0, contrary to the fact that x+y2\frac{x+y}{2} lies in the interval UU where SS is positive.

Derivative computations

Lemma

If ff is continuous and convex in an open interval UU about a point aa, then lim h0f(a+h)f(a)h\lim_{h \searrow 0} \frac{f(a+h)-f(a)}{h} exists and is finite.

Proof

We have seen this before: by convexity, slopes of secant lines f(a+h)f(a)h\frac{f(a+h)-f(a)}{h} decrease as hh decreases to 00, and yet these slopes have a lower bound given by f(b)f(a)ba\frac{f(b)-f(a)}{b-a} for any bb in UU that is less than aa.

Corollary

The right limit lim h0C(h)1h\lim_{h \searrow 0} \frac{C(h)-1}{h} exists and is finite. (It will be denoted R CR_C below.)

Proof

Since C(0)=1C(0) = 1, lemma 1 implies C-C is convex in a neighborhood of 00. Then apply lemma 2.

Lemma

S(0)S'(0) exists and is finite.

Proof

We divide the proof into the following points.

  • Either S(0)=0S'(0) = 0, or there is a point aa where C(a)C(a) and S(a)S(a) are both nonzero.

For, if δ\delta is sufficiently small, we have C(a)C(a) near 11 if δ<a<δ-\delta \lt a \lt \delta. If S(a)=0S(a) = 0 for all such aa, then obviously S(0)=0S'(0) = 0.

  • If S(a)0S(a) \neq 0, then lim h0S(a+h)S(a)h\lim_{h \searrow 0} \frac{S(a+h)-S(a)}{h} exists and is finite. (We will denote this as R aR_a below.)

By lemma 1, SS or S-S is convex in a neighborhood of aa. Then apply lemma 2.

  • lim h0S(h)h\lim_{h \searrow 0} \frac{S(h)}{h} exists and is finite.

We may assume aa is a point where C(a)C(a) and S(a)S(a) are nonzero. We have

R a=lim h0S(a+h)S(a)h=lim h0S(a)C(h)1h+C(a)S(h)hR_a = \lim_{h \searrow 0} \frac{S(a+h)-S(a)}{h} = \lim_{h \searrow 0} S(a)\frac{C(h)-1}{h} + C(a)\frac{S(h)}{h}

by the addition formula. We then obviously have

lim h0S(h)h=R aS(a)R CC(a).\lim_{h \searrow 0} \frac{S(h)}{h} = \frac{R_a - S(a)R_C}{C(a)}.
  • If lim h0S(h)h\lim_{h \searrow 0} \frac{S(h)}{h} exists and is finite (say RR), then S(0)S'(0) exists and is finite.

For, since SS is odd, we have

lim h0S(h)h=lim |h|0S(|h|)|h|=lim |h|0S(|h|)|h|=R.\lim_{h \nearrow 0} \frac{S(h)}{h} = \lim_{{|h|} \searrow 0} \frac{S(-{|h|})}{-{|h|}} = \lim_{{|h|} \searrow 0} \frac{S({|h|})}{{|h|}} = R.

This completes the proof.

Now the proof of Theorem 1 is routine: from the finiteness of S(0)S'(0) we easily infer that C(0)=0C'(0) = 0, and the computations

S(t)=S(0)C(t),C(t)=S(0)S(t)S'(t) = S'(0)C(t), \qquad C'(t) = -S'(0)S(t)

are easy computations using the addition formulas for SS and CC.


  1. That doesn’t mean that the algebra isn’t annoying. In the case of the function CC, it suffices to prove: If abcd>0a b - c d \gt 0 and a 2c 2>0a^2 - c^2 \gt0 and b 2d 2>0b^2 - d^2 \gt 0 with a 2+c 2=1a^2 + c^2 = 1 and b 2+d 2=1b^2 + d^2 = 1, then 2(abcd)a 2c 2+b 2d 22(a b - c d) \geq a^2 - c^2 + b^2 - d^2. A purely algebraic proof is given at the page “annoying algebra proposition” under my unpublished web.

Revised on September 26, 2013 at 04:34:30 by Todd Trimble