Intermediate value theorem

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In Mathematical analysis, the intermediate value theorem is either of two theorems of which an account is given below.

1 Intermediate value theorem
2 Intermediate value theorem of integration
3 Intermediate value theorem of derivatives
4 References
5 External links

[edit] Intermediate value theorem

Intermediate Value Theorem

The intermediate value theorem states the following: If y=f(x) is continuous on [a,b], and N is a number between f(a) and f(b), then there is a c ∈ [a,b] such that f(c) = N.

Suppose that $I$ is an interval [a,b] in the real numbers R and that $f\colon I \rightarrow \mathbf{R}$ is a continuous function. Then the image set $\displaystyle f(I)$ is also an interval, and either it contains [f(a),f(b)], or it contains [f(b),f(a)]; that is,

$\displaystyle f(I) \supseteq [f(a),f(b)]$ ,

$\displaystyle f(I) \supseteq [f(b),f(a)]$ .

It is frequently stated in the following equivalent form: Suppose that $f\colon [a,b] \rightarrow \mathbb{R}$ is continuous and that u is a real number satisfying f(a) < u < f(b) or f(a) > u > f(b). Then for some c ∈ [a,b], f(c) = u.

This captures an intuitive property of continuous functions: given f continuous on [1,2], if f(1) = 3 and f(2) = 5 then f must be equal to 4 somewhere between 1 and 2. It represents the idea that the graph of a continuous function on a closed interval can be drawn without lifting your pencil from the paper.

The theorem depends on the completeness of the real numbers. It is false for the rational numbers Q. For example, the function f(x) = x^2 - 2, x ∈ Q satisfies $f(0) = -2,\, f(2) = 2$ . However there is no rational number $\displaystyle x$ such that $\displaystyle f(x) = 0$ .

[edit] Proof

We shall prove the first case $\displaystyle f(a) < u < f (b)$ ; the second is similar.

Let $\displaystyle \text{S} = \{ x \in [a,b] : f(x) \leq u \}$ . Then $\displaystyle S$ is non-empty (since $\displaystyle a \in \displaystyle S$ ) and bounded above by $\displaystyle b$ . Hence by the completeness property of the real numbers, the supremum $c = \sup \text{S}$ exists. We claim that $\displaystyle f(c) = u$ .

Suppose first that $\displaystyle f(c) > u$ . Then $\displaystyle f(c) - u > 0$ , so there is a $\displaystyle \delta > 0$ such that $\displaystyle |f(x) - f(c)| < f(c) - u$ whenever $\displaystyle |x - c| < \delta$ , since $\displaystyle f$ is continuous. But then $\displaystyle f(x) > f(c) - (f(c) - u) = u$ whenever $\displaystyle |x - c| < \delta$ (i.e. $\displaystyle f(x) > u$ for $\displaystyle x$ in $\displaystyle(c - \delta, c + \delta)$ ). Thus $\displaystyle c - \delta$ is an upper bound for $\displaystyle S$ , a contradiction since we assumed that $\displaystyle c$ was the least upper bound and $\displaystyle c - \delta < \displaystyle c$ .

Suppose next that $\displaystyle f(c) < u$ . Again, by continuity, there is a $\displaystyle \delta > 0$ such that $\displaystyle |f(x) - f(c)| < u - f(c)$ whenever $\displaystyle |x - c| < \delta$ . Then $\displaystyle f(x) < f(c) + (u - f(c)) = u$ for $\displaystyle x$ in $\displaystyle(c - \delta, c + \delta)$ and there are numbers $\displaystyle x$ greater than $\displaystyle c$ for which $\displaystyle f(x) < u$ , again a contradiction to the definition of $\displaystyle c$ .

We deduce that $\displaystyle f(c) = u$ as stated.

[edit] History

For $\displaystyle u = 0$ above, the statement is also known as Bolzano's theorem; this theorem was first stated by Bernard Bolzano (1781–1848), together with a proof which used techniques which were especially rigorous for their time but which are now regarded as non-rigorous.

[edit] Generalization

The intermediate value theorem can be seen as a consequence of the following two statements from topology:

If $\displaystyle X$ and $\displaystyle Y$ are topological spaces, $f\colon X \rightarrow Y$ is continuous, and $\displaystyle X$ is connected, then $\displaystyle f(X)$ is connected.
A subset of $\mathbb{R}$ is connected if and only if it is an interval.

The intermediate value theorem generalizes in a natural way: Suppose that $X$ is a connected topological space and $(Y, < )$ is a totally ordered set equipped with the order topology, and let $f:X\rightarrow Y$ be a continuous map. If $a$ and $b$ are two points of $X$ and $r$ is a point of $Y$ lying between $a$ and $b$ with respect to <, then there exists $c\in X$ such that $\displaystyle f(c)=r$ . The original theorem is recovered by noting that $\mathbb{R}$ is connected and that its natural topology is the order topology.

[edit] Example of use in proof

The theorem is rarely applied with concrete values; instead, it gives some characterization of continuous functions. For example, let $\displaystyle g(x) = f(x) - x$ for $\displaystyle f$ continuous over the real numbers. Also, let $\displaystyle f$ be bounded (above and below). Then we can say $\displaystyle g = 0$ at least once. To see this, consider the following:

Since $\displaystyle f$ is bounded, we can pick $a > \sup(f(x))$ and $b < \inf(f(x))$ . Clearly $\displaystyle g(a) < 0$ and $\displaystyle g(b) > 0$ . If $\displaystyle f$ is continuous, then $\displaystyle g$ is also continuous. Since $\displaystyle g$ is continuous, we can apply the intermediate value theorem and state that $\displaystyle g$ must take on the value of 0 somewhere between $\displaystyle a$ and $\displaystyle b$ . This result proves that any continuous bounded function must cross the function, $\displaystyle x$ .

[edit] Converse is false

Suppose $\displaystyle f$ is a real-valued function defined on some interval $\displaystyle I$ , and for every two elements $\displaystyle a$ and $\displaystyle b$ in $\displaystyle I$ and $\forall\, u \in (f(a),f(b))\, \exists \, c \in (a,b)$ such that $\displaystyle f(c) = u$ . Does $\displaystyle f$ have to be continuous? The answer is no; the converse of the intermediate value theorem fails. As an example, take the function $\displaystyle f(x) = \sin(1/x)$ for $x \neq 0$ , and $\displaystyle f(0) = 0$ . This function is not continuous as the limit when $\displaystyle x$ gets close to 0 does not exist; yet the function has the above intermediate value property. Another, more complicated example is given by the Conway base 13 function.

Historically, this intermediate value property has been suggested as a definition for continuity of real-valued functions; this definition was not adopted.

Darboux's theorem states that all functions that result from the differentiation of some other function on some interval have the intermediate value property (even though they need not be continuous).

[edit] Implications of theorem in real world

The theorem implies that on any great circle around the world, the temperature, pressure, elevation, carbon dioxide concentration, or anything else that varies continuously, there will always exist two antipodal points that share the same value for that variable.

Proof: Take $\displaystyle f$ to be any continuous function on a circle. Draw a line through the center of the circle, intersecting it at two opposite points $\displaystyle A$ and $\displaystyle B$ . Let $\displaystyle d$ be the difference $\displaystyle f(A)-f(B)$ . If the line is rotated 180 degrees, the value $\displaystyle -d$ will be obtained instead. Due to the intermediate value theorem there must be some intermediate rotation angle for which $\displaystyle d=0$ , and as a consequence $\displaystyle f(A)=f(B)$ at this angle.

This is a special case of a more general result called the Borsuk–Ulam theorem.

The theorem also underpins the explanation of why rotating a wobbly table will bring it to stability (subject to certain easily-met constraints).^[1]

[edit] Intermediate value theorem of integration

The intermediate value theorem of integration is derived from the mean value theorem and states:

If $\displaystyle f$ is a continuous function on some interval $\displaystyle [a,b]$ , then the signed area under the function on that interval is equal to the length of the interval $\displaystyle b-a$ multiplied by some function value $\displaystyle f(c)$ such that $\displaystyle a < c < b$ . I.e.,

$\displaystyle \int_{a}^{b}\! f(x)\,dx = (b-a)f(c).$

[edit] Intermediate value theorem of derivatives

If $\displaystyle f$ is a differentiable real-valued function on $\mathbb{R}$ , then the (first order) derivative $\displaystyle f'$ has the intermediate value property, though $\displaystyle f'$ might not be continuous.