1. Definition of ln
The exponential function f (x) = ex is the function on R defined by the power series f (x) =
∞
X
n=0
xn
n!, x ∈ R.
The radius of convergence of f (x) is R = ∞ and hence it converges everywhere on R.
Moreover, f (x) is smooth and obeys the first order ordinary differential equation f0(x) = f (x), f (0) = 1.
It has the property:
ex+y = ex· ey, x, y ∈ R.
It follows from the definition of f (x) that f (x) ≥ 1 for all x ≥ 0. Using ex· e−x = 1, we find that f (x) > 0 for all x ∈ R. Hence if x > y, then ex−y > 1 and hence ex > ey. In other words, f (x) is an increasing function on R. Since ex> 1 + x for all x > 0, we see that limx→∞ex = ∞. By ex· e−x = 1 again, we find limx→−∞ex = 0. Therefore the image of f is the set of positive real number R>0= (0, ∞). We conclude that
Proposition 1.1. The exponential function f (x) = ex is a smooth increasing function1 onto R>0 such that
(1) f (0) = 1,
(2) f (x + y) = f (x)f (y) for all x, y ∈ R.
Since f is a bijection from R onto R>0, its inverse function exists g : R>0 → R, i.e.
(g ◦ f )(x) = x for all x ∈ R and (f ◦ g)(y) = y for all y > 0. The equation (f ◦ g)(y) = y for y > 0 is equivalent to the equation
eg(y)= y, y > 0 For all x, y > 0, we have
eg(xy) = xy = eg(x)· eg(y) = eg(x)+g(y). Since f (x) = ex is an one-to-one function and e0 = 1,
g(xy) = g(x) + g(y), x, y > 0, and g(1) = 0. We denote g(x) by ln x for x > 0. Then
eln x= x, ∀x > 0.
Proposition 1.2. The function g : R>0 → R is differentiable.
Proof. The proof will be omitted here.
Using eg(x)= x for all x > 0 and the chain rule, we find d
dxeg(x)= eg(x)g0(x) = 1.
Therefore the derivative of g(x) is
g0(x) = 1
x, x > 0.
1If you learn algebra already, R>0is a multiplicative group and f : R → R>0is a continuous group homomorphism.
1
2
By fundamental theorem of calculus, we obtain
(1.1) ln x =
Z x
1
dt t .
Without knowing that f (x) = ex is a bijection from R onto R>0, we can define ln x by the integral (1.1). We define g : R>0→ R by
g(x) = Z x
1
dt t
Proposition 1.3. The function g : R>0 → R is smooth such that (1) g(1) = 0.
(2) g(xy) = g(x) + g(y), for all x, y > 0, (3) eg(x)= x, for all x > 0.
Proof. g(1) = 0 is obvious. For x, y > 0, g(xy) =
Z xy 1
dt t
= Z x
1
dt t +
Z xy x
dt t
= g(x) + Z y
1
dt
t ( Here we use the change of variable u = t/x.)
= g(x) + g(y).
To show that eg(x)= x for all x ∈ R, we only need to show that the function h(x) = xe−g(x) is the constant function h(x) = 1 for x > 0. To do this, we compute
h0(x) = e−g(x)+ x · e−g(x)· −g0(x) = e−g(x)+ x · e−g(x)·−1
x = e−g(x)− e−g(x) = 0.
Then h(x) is a constant function. Since h(1) = 1, h(x) = 1 for all x > 0. This also shows that f (x) = ex is a bijection from R onto R>0.
Now, we are ready to classify all continuous functions f : R → R>02such that f (x + y) = f (x)f (y).
Theorem 1.1. Let f : R → R be a nonzero continuous map such that f (x + y) = f (x)f (y).
Then there is a constant c such that f (x) = ecx for all x ∈ R.
Proof. Let us prove f (0) > 0. We know f (0) = f (0 + 0) = f (0)f (0) = (f (0))2 ≥ 0. If f (0) = 0, then f (x) = f (x + 0) = f (x)f (0) = 0 for all x ∈ R. Since f is nonzero, f (0) 6= 0.
This implies f (0) = (f (0))2 > 0. The equation f (0) = (f (0))2 also implies that f (0) = 1 by f (0) 6= 0. In fact, for all x ∈ R,
f (x) = f
2 ·x
2
=
f
x 2
2
> 0.
Use induction, we can prove f (n) = (f (1))nfor n ≥ 1. Since 1 + (−1) = 0, we know 1 = f (0) = f (1 + (−1)) = f (1)f (−1).
2In group theory, we call such a map a continuous group homomorphism.
3
This implies f (1) and f (−1) are both nonzero and inverse to each other. For a nonnegative integer m, we choose n = −m. Then n + m = 0 and thus
1 = f (0) = f (n + m) = f (n)f (m).
This implies that
f (m) = 1
f (n) = 1
(f (1))n = (f (1))−n= (f (1))m.
We find that the equation f (n) = (f (1))n holds for all integer n. Assume that n ≥ 0. For all n ≥ 1 and all m ∈ Z,
(f (1))m = f (m) = f
n ·m
n
=
f
m n
n
. Taking the n-th root of both side of the equation, we obtain
fm n
= (f (1))mn, m ∈ Z, n ≥ 1.
This shows that the equation f (r) = (f (1))r holds for any rational number r ∈ Q. Since every real number can be approximated by a sequence of rational numbers, for each x, choose {xn} with xn∈ Q so that limn→∞xn= x. By continuity of f,
f (x) = lim
n→∞f (xn) = lim
n→∞(f (1))xn = (f (1))limn→∞xn = (f (1))x. Choose c = ln f (1), we obtain
f (x) = (f (1))x= eln f (1)x = ex ln f (1)= ecx, x > 0.
