§l. The intermediate value theorem
195
Direct proof: for any r > 0 there is an r' > 0 such that Ig(y) - g(b)1 < r
for Iy - bl < r', then an r" > 0 such that If(x) - f(a)1 < r' for Ix - al < r".
It is now clear that Ix - al < r" implies Ih(x) - h(a)1 < r, qed.
If for example one knows that the functions sin x and cos x are continuous
on IR one can deduce that the function sin( cos x) is so too.
Example 1. The function f(x) = log[exp(x)] is continuous on R This result,
though trivial, nevertheless has an important consequence. For
f(x + y) = log[exp(x + y)] = log[exp(x) exp(y)] = log[exp(x)] + log[exp(y)]'
in other words f(x + y) = f(x) + f(y). It follows immediately that
f(x + 0) = f(x) + f(O) whence f(O) = 0,
then that 0 = f(x - x) = f(x) + f( -x), whence f( -x) = - f(x). Moreover,
f(nx) = f(x + ... + x) = nf(x) for n EN, then for n E Z since f( -nx) =
-f(nx) = -nf(x). This also shows that f(x/n) = f(x)/n for n =I- 0, whence
f(px/q) = pf(x)/q for p, q E Z, q =I- o. On putting f(l) = a, one thus has
f(x) = ax for all x E Q. But since f is continuous, this relation persists
when x E Q tends to any limit in IR 2. Conclusion: there exists a constant
a = 10g[exp(1)] such that
log[exp(x)] = ax for all x E R
This argument, alas, will not show that a = 1 since it applies equally well to
the function 310g[exp(x)].
But we know that, by definition, logt = limn(tl/n - 1), from which
log[exp(x)] = limn [exp(x)l/n -lJ. Now the addition formula for the exponential function shows that exp(x) = exp(n.x/n) = exp(x/n)n for any x E lR.
and so
log[exp(x)] = limn[exp(x/n) -1].
The power series exp( t) = 1 + t + t 2 /2! + ... shows on the other hand that
exp(t) - 1 rv t when t tends to o. Consequently exp(x/n) - 1 rv x/n; since
the ratio of the two sides of this relation is equal to n[exp(x/n) - l]jx, and
tends to 1, we deduce that
lim n[exp(x/n) - 1] = x,
whence the relation we seek:
2 More generally, let f and 9 be two functions defined and continuous on the
closure X of a set X and suppose that f = 9 on X; then f = 9 on X. For if
Xn E X tends to x E X then f(xn) and g(Xn), which are equal, tend to f(x) and
g(x) respectively.
195
Direct proof: for any r > 0 there is an r' > 0 such that Ig(y) - g(b)1 < r
for Iy - bl < r', then an r" > 0 such that If(x) - f(a)1 < r' for Ix - al < r".
It is now clear that Ix - al < r" implies Ih(x) - h(a)1 < r, qed.
If for example one knows that the functions sin x and cos x are continuous
on IR one can deduce that the function sin( cos x) is so too.
Example 1. The function f(x) = log[exp(x)] is continuous on R This result,
though trivial, nevertheless has an important consequence. For
f(x + y) = log[exp(x + y)] = log[exp(x) exp(y)] = log[exp(x)] + log[exp(y)]'
in other words f(x + y) = f(x) + f(y). It follows immediately that
f(x + 0) = f(x) + f(O) whence f(O) = 0,
then that 0 = f(x - x) = f(x) + f( -x), whence f( -x) = - f(x). Moreover,
f(nx) = f(x + ... + x) = nf(x) for n EN, then for n E Z since f( -nx) =
-f(nx) = -nf(x). This also shows that f(x/n) = f(x)/n for n =I- 0, whence
f(px/q) = pf(x)/q for p, q E Z, q =I- o. On putting f(l) = a, one thus has
f(x) = ax for all x E Q. But since f is continuous, this relation persists
when x E Q tends to any limit in IR 2. Conclusion: there exists a constant
a = 10g[exp(1)] such that
log[exp(x)] = ax for all x E R
This argument, alas, will not show that a = 1 since it applies equally well to
the function 310g[exp(x)].
But we know that, by definition, logt = limn(tl/n - 1), from which
log[exp(x)] = limn [exp(x)l/n -lJ. Now the addition formula for the exponential function shows that exp(x) = exp(n.x/n) = exp(x/n)n for any x E lR.
and so
log[exp(x)] = limn[exp(x/n) -1].
The power series exp( t) = 1 + t + t 2 /2! + ... shows on the other hand that
exp(t) - 1 rv t when t tends to o. Consequently exp(x/n) - 1 rv x/n; since
the ratio of the two sides of this relation is equal to n[exp(x/n) - l]jx, and
tends to 1, we deduce that
lim n[exp(x/n) - 1] = x,
whence the relation we seek:
2 More generally, let f and 9 be two functions defined and continuous on the
closure X of a set X and suppose that f = 9 on X; then f = 9 on X. For if
Xn E X tends to x E X then f(xn) and g(Xn), which are equal, tend to f(x) and
g(x) respectively.