§l. Direct construction
337
(6.1)
f(x + y) = f(x)f(y);
this "addition formula", which we have already met a propos the series
expx (Chap. II, nO 22), analogous to those for the trigonometric functions
but simpler, is an example of what is called a functional equation; for the
function f(x) = X S we have an analogous relation
(6.2)
f(xy) = f(x)f(y),
for a linear function f (x) = ax we have
(6.3)
f(x + y) = f(x) + f(y),
for the logarithmic functions we have
(6.4)
f(xy) = f(x) + f(y),
etc. As we shall see, these relations chamcterise the functions in question so
long as one also insists that they be reasonable, i.e. continuous or monotone,
and, if necessary, real-valued. This is already to be found in Cauchy's COUTS
d'analyse.
Theorem 2. Every real-valued function defined on JR which is not identically
zero, satisfies (1), and is monotone OT continuous, is an exponential function.
Since f(x) = f(x/2 + x/2) = f(x/2)2 we have f(x) 2:: 0 for all x. We
even have strict inequality, since if we had f(c) = 0 it would follow that
f(x) = f(c)f(x - c) = 0 for any x E JR. Since f(O +x) = f(O)f(x), we deduce
that f(O) = 1.
Let us put f(l) = a, an excellent idea if we want to prove that f(x) = aX
for some a. Now f(2) = a 2 , then f(3) = f(2 + 1) = a 3 , and more generally
f(n) = an for all n E N. For n negative we write
1 = f(O) = f(n + (-n)) = f(n)f( -n) = f(n)a- n ,
whence again f(n) = an.
Let x = p/q be a rational number, with q > o. The relation (1) shows
that
f(qx) = f(x + ... + x) = f(x) ... f(x) = f(x)q,
whence f(x)q = f(p) = a P • Since f(x) > 0, it follows that
f(x) = (aP )l/q = aP/ q = aX
for all x E Q. If f is continuous or monotone, we can then apply Theorem 1
and conclude that f(x) = aX for all x E JR, qed.
337
(6.1)
f(x + y) = f(x)f(y);
this "addition formula", which we have already met a propos the series
expx (Chap. II, nO 22), analogous to those for the trigonometric functions
but simpler, is an example of what is called a functional equation; for the
function f(x) = X S we have an analogous relation
(6.2)
f(xy) = f(x)f(y),
for a linear function f (x) = ax we have
(6.3)
f(x + y) = f(x) + f(y),
for the logarithmic functions we have
(6.4)
f(xy) = f(x) + f(y),
etc. As we shall see, these relations chamcterise the functions in question so
long as one also insists that they be reasonable, i.e. continuous or monotone,
and, if necessary, real-valued. This is already to be found in Cauchy's COUTS
d'analyse.
Theorem 2. Every real-valued function defined on JR which is not identically
zero, satisfies (1), and is monotone OT continuous, is an exponential function.
Since f(x) = f(x/2 + x/2) = f(x/2)2 we have f(x) 2:: 0 for all x. We
even have strict inequality, since if we had f(c) = 0 it would follow that
f(x) = f(c)f(x - c) = 0 for any x E JR. Since f(O +x) = f(O)f(x), we deduce
that f(O) = 1.
Let us put f(l) = a, an excellent idea if we want to prove that f(x) = aX
for some a. Now f(2) = a 2 , then f(3) = f(2 + 1) = a 3 , and more generally
f(n) = an for all n E N. For n negative we write
1 = f(O) = f(n + (-n)) = f(n)f( -n) = f(n)a- n ,
whence again f(n) = an.
Let x = p/q be a rational number, with q > o. The relation (1) shows
that
f(qx) = f(x + ... + x) = f(x) ... f(x) = f(x)q,
whence f(x)q = f(p) = a P • Since f(x) > 0, it follows that
f(x) = (aP )l/q = aP/ q = aX
for all x E Q. If f is continuous or monotone, we can then apply Theorem 1
and conclude that f(x) = aX for all x E JR, qed.