Advanced Calculus Single Variable

6.4 The Intermediate Value Theorem

The next big theorem is called the intermediate value theorem and the following picture illustrates its conclusion. 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. It may seem this is obvious but without completeness the conclusion of the theorem cannot be drawn. Nevertheless, the above picture makes this theorem very easy to believe. Here is a useful lemma.

Proof: Let ε =

> 0 by assumption. Then let δ correspond to this ε in the definition of continuity.

Case 1: f

−c > 0. Then for y ∈ ∩, f −f < f −c and so 0 < f − c.

Case 2: f

−c < 0. Then for y ∈ ∩, f −f < c−fand so 0 > f − c. ■

Next here is a proof of the intermediate value theorem.

Proof: Let d =

and consider the intervals and . If f ≥ c, then on , the function is ≤ c at one end point and ≥ c at the other. On the other hand, if f ≤ c, then on , 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 ≤ c,f ≥ c. Now ≤2⁻ⁿ and ≤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 = nli→m∞ (f (yn)− c) ≥ 0.

Consequently f

= c and this proves the theorem. The last step follows from Theorem 6.1.1. ■

Here is another proof of this major theorem.

Proof: Since f

< c, the set, S defined as

S ≡ {x ∈ [a,b] : f (t) ≤ c for all t ∈ [a,x]}

is nonempty. In particular a ∈ S. Note that b

S by assumption. Let z ≡ sup. By Lemma 6.4.1, since f < c it follows that z ∈. By the same lemma again, if f≠c, then z≠sup. Hence f = c. ■

Proof: First it is shown that ϕ is either strictly increasing or strictly decreasing on

.

If ϕ is not strictly decreasing on

, then there exists x₁ < y₁, x₁,y₁ ∈ such that

(ϕ(y1)− ϕ (x1))(y1 − x1) > 0.

If for some other pair of points, x₂ < y₂ with x₂,y₂ ∈

, the above inequality does not hold, then since ϕ is 1 − 1,

(ϕ(y2)− ϕ (x2))(y2 − x2) < 0.

Let x_t ≡ tx₁ +

x₂ and y_t ≡ ty₁ + y₂. Then x_t < y_t for all t ∈ 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

can equal zero. Now define

h (t) ≡ (ϕ(yt)− ϕ(xt))(yt − xt).

Since h is continuous and h

< 0, while h > 0, there exists t ∈ such that h = 0. Therefore, both x_t and y_t are points of and ϕ − ϕ = 0 contradicting the assumption that ϕ is one to one. It follows ϕ is either strictly increasing or strictly decreasing on .

This property of being either strictly increasing or strictly decreasing on

carries over to by the continuity of ϕ. Suppose ϕ is strictly increasing on . (A similar argument holds for ϕ strictly decreasing on .) If x > a, then let z_n be a decreasing sequence of points of converging to a. Then by continuity of ϕ at a,

ϕ(a) = lim ϕ(zn) ≤ ϕ(z1) < ϕ (x) . n→ ∞

Therefore, ϕ

< ϕ whenever x ∈. Similarly ϕ > ϕ for all x ∈. ■

Proof 1: Since f is either strictly increasing or strictly decreasing, it follows that f

is an open interval, . Assume f is decreasing. Now let x ∈. Why is f⁻¹ is continuous at f? Since f is decreasing, if f < f, then y ≡ f⁻¹ < x ≡ f⁻¹ and so f⁻¹ is also decreasing. Let ε > 0 be given. Let ε > η > 0 and ⊆. Then f ∈. Let

δ = min (f (x)− f (x + η),f (x− η)− f (x )).

Then if

|f (z)− f (x)| < δ,

it follows

z ≡ f−1(f (z)) ∈ (x− η,x+ η) ⊆ (x − ε,x + ε)

so

| | | | |f−1(f (z))− x| = |f−1(f (z))− f−1 (f (x))| < ε.

This proves the theorem in the case where f is strictly decreasing. The case where f is increasing is similar. ■

Proof 2: Note that since f is either strictly increasing or strictly decreasing, it maps an open interval to an open interval. Now let U be an open set. Thus U = ∪_x∈UI_x where I_x is an open interval. Then

which is a union of open intervals and is therefore, open. ■