1. Definition of ln
The exponential function f (x) = ex is the function on R defined by the power series
f (x) =
∞
X
xn
n=0
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
f 0 (x) = f (x),
f (0) = 1.
ex+y = ex · ey ,
x, y ∈ R.
It has the property:
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 g(x)
e
= eg(x) g 0 (x) = 1.
dx
Therefore the derivative of g(x) is
g 0 (x) =
1
,
x
x > 0.
1If you learn algebra already, R
>0 is a multiplicative group and f : R → R>0 is a continuous group homomorphism.
1
2
By fundamental theorem of calculus, we obtain
Z x
dt
(1.1)
ln x =
.
1 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
Z x
dt
g(x) =
1 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,
Z xy
dt
g(xy) =
t
Z1 x
Z xy
dt
dt
=
+
t
1 t
Z yx
dt
( Here we use the change of variable u = t/x.)
= g(x) +
1 t
= 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
−1
h0 (x) = e−g(x) + x · e−g(x) · −g 0 (x) = e−g(x) + x · e−g(x) ·
= e−g(x) − e−g(x) = 0.
x
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>0 2 such 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,
x x 2
f (x) = f 2 ·
> 0.
= f
2
2
Use induction, we can prove f (n) = (f (1))n for 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
1
= (f (1))−n = (f (1))m .
=
f (n)
(f (1))n
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,
m m n
(f (1))m = f (m) = f n ·
= f
.
n
n
Taking the n-th root of both side of the equation, we obtain
m
m
f
= (f (1)) n , m ∈ Z, n ≥ 1.
n
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 f (xn ) = lim (f (1))xn = (f (1))limn→∞ xn = (f (1))x .
n→∞
n→∞
Choose c = ln f (1), we obtain
x
f (x) = (f (1))x = eln f (1) = ex ln f (1) = ecx ,
x > 0.