The next big theorem is called the intermediate value theorem and the following picture
illustrates its conclusion.
PICT
It gives the existence of a certain point. You see in the picture there is a horizontal
line, y = c and a continuous function which starts off less than c at the point a and ends
up greater than c at point b. The intermediate value theorem says there is some point
between a and b shown in the picture as z such that the value of the function at this
point equals c.
The theorem is due to Bolzano around 1818. You might think that this is an obvious
theorem but this is not the case. It is not even true if you only had the rational
numbers. Suppose you only consider rational numbers. Then f
(x)
= x^{2}− 2 is
continuous. f
(0)
< 0 and f
(2)
> 0 but the only point between 0 and 2 where this
function is 0 is the point
√-
2
which has been known for thousands of years to be
irrational. You have to use something which rules out holes in the real line.
One can assume a topological concept and say the real line is connected, a
concept not discussed in this book, but the way it is usually done is to use the
completeness axiom. Everything of interest in calculus eventually depends on this
axiom.
Here is a useful lemma.
Lemma 4.4.1If f is continuous on the closed interval
[a,b]
and for somex ∈
[a,b]
, f
(x)
≠c, then there exists δ > 0 such that for y ∈
[a,b]
∩
(x − δ,x + δ)
,the sign of f
(y)
−c is constant. That is, it is either always positive or always negativedepending on the sign of f
(x )
− c.
Proof: Let ε =
|f (x)− c|
> 0 by assumption. Then let δ correspond to this ε in the
definition of continuity.
Case 1: f
(x)
− c > 0. Then for y ∈
[a,b]
∩
(x− δ,x+ δ)
, f
(x)
− f
(y)
< f
(x)
− c
and so 0 < f
(y)
− c.
Case 2: f
(x)
−c < 0. Then for y ∈
[a,b]
∩
(x− δ,x+ δ)
, f
(y)
−f
(x )
< c−f
(x)
and
so 0 > f
(y)
− c. ■
Next here is a proof of the intermediate value theorem.
Theorem 4.4.2Suppose f :
[a,b]
→ ℝ is continuousand suppose f
(a)
<c < f
(b)
. Then there exists x ∈
(a,b)
such that f
(x)
= c.
Proof: Let d =
a+2b
and consider the intervals
[a,d]
and
[d,b]
. If f
(d)
≥ c, then on
[a,d]
, the function is ≤ c at one end point and ≥ c at the other. On the other hand, if
f
(d)
≤ c, then on
[d,b]
, f ≥ 0 at one end point and ≤ 0 at the other. Pick the
interval on which f has values which are at least as large as c and values no
larger than c. Now consider that interval, divide it in half as was done for the
original interval and argue that on one of these smaller intervals, the function has
values at least as large as c and values no larger than c. Continue in this way.
Next apply the nested interval lemma to get x in all these intervals. In the n^{th}
interval, let x_{n},y_{n} be points of this interval such that f
(xn)
≤ c,f
(yn)
≥ c. Now
|xn − x|
≤
(b− a)
2^{−n} and
|yn − x|
≤
(b− a)
2^{−n} and so x_{n}→ x and y_{n}→ x.
Therefore,
f (x)− c = lim (f (xn)− c) ≤ 0
n→ ∞
while
f (x)− c = lim (f (yn)− c) ≥ 0.
n→ ∞
Consequently f
(x)
= c and this proves the theorem. The last step follows from Theorem
4.1.1. ■
Here is another lemma which may seem obvious but when you ask why, you begin
to see that it is not as obvious as you thought. In fact, this is a special case
of a general theory which says that one to one continuous functions from U,
an open set in ℝ^{p} to ℝ^{p} take open sets to open sets. This is a very difficult
result.
Lemma 4.4.3Let ϕ :
[a,b]
→ ℝ be a continuousfunction and suppose ϕ is
1 − 1 on
(a,b)
. Then ϕ is either strictly increasing or strictly decreasing on
[a,b]
.
Proof: First it is shown that ϕ is either strictly increasing or strictly decreasing on
(a,b)
.
If ϕ is not strictly decreasing on
(a,b)
, then there exists x_{1}< y_{1}, x_{1},y_{1}∈
(a,b)
such
that
(ϕ (y1)− ϕ (x1))(y1 − x1) > 0.
If for some other pair of points, x_{2}< y_{2} with x_{2},y_{2}∈
(a,b)
, the above inequality does
not hold, then since ϕ is 1 − 1,
(ϕ (y )− ϕ (x))(y − x ) < 0.
2 2 2 2
Let x_{t}≡ tx_{1} +
(1− t)
x_{2} and y_{t}≡ ty_{1} +
(1− t)
y_{2}. Then x_{t}< y_{t} for all t ∈
[0,1]
because
tx1 ≤ ty1 and (1 − t)x2 ≤ (1 − t)y2
with strict inequality holding for at least one of these inequalities since not both t and
(1 − t)
can equal zero. Now define
h (t) ≡ (ϕ (yt)− ϕ (xt))(yt − xt).
Since h is continuous and h
(0)
< 0, while h
(1)
> 0, there exists t ∈
(0,1)
such that
h
(t)
= 0. Therefore, both x_{t} and y_{t} are points of
(a,b)
and ϕ
(yt)
− ϕ
(xt)
= 0
contradicting the assumption that ϕ is one to one. It follows ϕ is either strictly increasing
or strictly decreasing on
(a,b)
.
This property of being either strictly increasing or strictly decreasing on
(a,b)
carries
over to
[a,b]
by the continuity of ϕ. Suppose ϕ is strictly increasing on
(a,b)
. (A similar
argument holds for ϕ strictly decreasing on
(a,b)
.) If x > a, then let z_{n} be a
decreasing sequence of points of