# Mean value theorem

(Redirected from Mean Value Theorem)
For the theorem in harmonic function theory, see Harmonic function#The mean value property.
Not to be confused with the Intermediate value theorem.
For any function that is continuous on [a, b] and differentiable on (a, b) there exists some c in the interval (a, b) such that the secant joining the endpoints of the interval [a, b] is parallel to the tangent at c.

In mathematics, the mean value theorem states, roughly: that given a planar arc between two endpoints, there is at least one point at which the tangent to the arc is parallel to the secant through its endpoints.

The theorem is used to prove global statements about a function on an interval starting from local hypotheses about derivatives at points of the interval.

More precisely, if a function f is continuous on the closed interval [a, b], where a < b, and differentiable on the open interval (a, b), then there exists a point c in (a, b) such that

$f'(c) = \frac{f(b) - f(a)}{b-a} \, .$[1]

A special case of this theorem was first described by Parameshvara (1370–1460) from the Kerala school of astronomy and mathematics in his commentaries on Govindasvāmi and Bhaskara II.[2] The mean value theorem in its modern form was later stated by Augustin Louis Cauchy (1789–1857). It is one of the most important results in differential calculus, as well as one of the most important theorems in mathematical analysis, and is useful in proving the fundamental theorem of calculus. The mean value theorem follows from the more specific statement of Rolle's theorem, and can be used to prove the more general statement of Taylor's theorem (with Lagrange form of the remainder term).

## Formal statement

Let f : [a, b] → R be a continuous function on the closed interval [a, b], and differentiable on the open interval (a, b), where a < b. Then there exists some c in (a, b) such that

$f ' (c) = \frac{f(b) - f(a)}{b - a}.$

The mean value theorem is a generalization of Rolle's theorem, which assumes f(a) = f(b), so that the right-hand side above is zero.

The mean value theorem is still valid in a slightly more general setting. One only needs to assume that f : [a, b] → R is continuous on [a, b], and that for every x in (a, b) the limit

$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

exists as a finite number or equals +∞ or −∞. If finite, that limit equals f′(x). An example where this version of the theorem applies is given by the real-valued cube root function mapping x to x1/3, whose derivative tends to infinity at the origin.

Note that the theorem, as stated, is false if a differentiable function is complex-valued instead of real-valued. For example, define f(x) = eix for all real x. Then

f(2π) − f(0) = 0 = 0(2π − 0)

while f′(x) ≠ 0 for any real x.

## Proof

The expression (f(b) − f(a)) / (ba) gives the slope of the line joining the points (a, f(a)) and (b, f(b)), which is a chord of the graph of f, while f '(x) gives the slope of the tangent to the curve at the point (x, f(x)). Thus the Mean value theorem says that given any chord of a smooth curve, we can find a point lying between the end-points of the chord such that the tangent at that point is parallel to the chord. The following proof illustrates this idea.

Define g(x) = f(x) − rx, where r is a constant. Since f is continuous on [a, b] and differentiable on (a, b), the same is true for g. We now want to choose r so that g satisfies the conditions of Rolle's theorem. Namely

\begin{align}g(a)=g(b)&\iff f(a)-ra=f(b)-rb\\ &\iff r(b-a)=f(b)-f(a) \\&\iff r=\frac{f(b)-f(a)}{b-a}\cdot\end{align}

By Rolle's theorem, since g is differentiable and g(a) = g(b), there is some c in (a, b) for which g′(c) = 0, and it follows from the equality g(x) = f(x) − rx that,

$f '(c)=g '(c)+r=0+r=\frac{f(b)-f(a)}{b-a}$

as required.

## A simple application

Assume that f is a continuous, real-valued function, defined on an arbitrary interval I of the real line. If the derivative of f at every interior point of the interval I exists and is zero, then f is constant.

Proof: Assume the derivative of f at every interior point of the interval I exists and is zero. Let (a, b) be an arbitrary open interval in I. By the mean value theorem, there exists a point c in (a,b) such that

$0 = f'(c) = \frac{f(b)-f(a)}{b-a}.$

This implies that f(a) = f(b). Thus, f is constant on the interior of I and thus is constant on I by continuity. (See below for a multivariable version of this result.)

Remarks:

## Cauchy's mean value theorem

Cauchy's mean value theorem, also known as the extended mean value theorem, is a generalization of the mean value theorem. It states: If functions f and g are both continuous on the closed interval [a,b], and differentiable on the open interval (a, b), then there exists some c ∈ (a,b), such that

Geometrical meaning of Cauchy's theorem
$(f(b)-f(a))g\,'(c)=(g(b)-g(a))f\,'(c).\,$

Of course, if g(a) ≠ g(b) and if g′(c) ≠ 0, this is equivalent to:

$\frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}\cdot$

Geometrically, this means that there is some tangent to the graph of the curve

$\begin{array}{ccc}[a,b]&\longrightarrow&\mathbb{R}^2\\t&\mapsto&\bigl(f(t),g(t)\bigr),\end{array}$

which is parallel to the line defined by the points (f(a),g(a)) and (f(b),g(b)). However Cauchy's theorem does not claim the existence of such a tangent in all cases where (f(a),g(a)) and (f(b),g(b)) are distinct points, since it might be satisfied only for some value c with f′(c) = g′(c) = 0, in other words a value for which the mentioned curve is stationary; in such points no tangent to the curve is likely to be defined at all. An example of this situation is the curve given by

$t\mapsto(t^3,1-t^2),$

which on the interval [−1,1] goes from the point (−1,0) to (1,0), yet never has a horizontal tangent; however it has a stationary point (in fact a cusp) at t = 0.

Cauchy's mean value theorem can be used to prove l'Hôpital's rule. The mean value theorem is the special case of Cauchy's mean value theorem when g(t) = t.

### Proof of Cauchy's mean value theorem

The proof of Cauchy's mean value theorem is based on the same idea as the proof of the mean value theorem.

• Suppose that g(a) ≠g(b). Define h(x) = f(x) − rg(x), where r is fixed in such a way that h(a) = h(b), namely
\begin{align}h(a)=h(b)&\iff f(a)-r\,g(a)=f(b)-r\,g(b)\\ &\iff r\,(g(b)-g(a))=f(b)-f(a)\\ &\iff r=\frac{f(b)-f(a)}{g(b)-g(a)}.\end{align}
Since f and g are continuous on [a, b] and differentiable on (a, b), the same is true for h. All in all, h satisfies the conditions of Rolle's theorem: consequently, there is some c in (a, b) for which h′(c) = 0.
From the equality h(x) = f(x) − rg(x) it follows that,
$0=h'(c)=f'(c)-r\, g'(c) \Rightarrow(g(b)-g(a))\,f'(c)=(g(b)-g(a))\,r\,g'(c)=(f(b)-f(a))\,g'(c)$
as required.
• If instead g(a) = g(b), then, applying Rolle's theorem to g, it follows that there exists c in (a, b) for which g′(c) = 0. Using this choice of c, Cauchy's mean value theorem (trivially) holds.

## Generalization for determinants

Assume that $f$, $g$, and $h$ are differentiable functions on $(a,b)$ that are continuous on $[a,b]$. Define

$D(x)=\left|\begin{array}{ccc}f(x) & g(x)& h(x)\\ f(a) & g(a) & h(a)\\ f(b) & g(b)& h(b)\end{array}\right|$

There exists $c\in(a,b)$ such that $D'(c)=0$.

Notice that

$D'(x)=\left|\begin{array}{ccc}f'(x) & g'(x)& h'(x)\\ f(a) & g(a) & h(a)\\ f(b) & g(b)& h(b)\end{array}\right|$

and if we place $h(x)=1$, we get Cauchy's mean value theorem. If we place $h(x)=1$ and $g(x)=x$ we get Lagrange's mean value theorem.

The proof of the generalization is quite simple: each of $D(a)$ and $D(b)$ are determinants with two identical rows, hence $D(a)=D(b)=0$. The Rolle's theorem implies that there exists $c\in (a,b)$ such that $D'(c)=0$.

## Mean value theorem in several variables

The mean value theorem in one variable generalizes to several variables by applying the theorem in one variable via parametrization. Let G be an open connected subset of Rn, and let f : GR be a differentiable function. Fix points x, yG such that the interval x y lies in G, and define g(t) = f((1 − t)x + ty). Since g is a differentiable function in one variable, the mean value theorem gives:

$g(1) - g(0) = g'(c) \!$

for some c between 0 and 1. But since g(1) = f(y) and g(0) = f(x), computing g′(c) explicitly we have:

$f(y) - f(x) = \nabla f ((1- c)x + cy) \cdot (y - x)$

where ∇ denotes a gradient and · a dot product. Note that this is an exact analog of the theorem in one variable (in the case n = 1 this is the theorem in one variable). By the Schwarz inequality, the equation gives the estimate:

$|f(y) - f(x)| \le |\nabla f ((1- c)x + cy)| \, |y - x|.$

In particular, when the partial derivatives of f are bounded, f is Lipschitz continuous (and therefore uniformly continuous). Note that f is not assumed to be continuously differentiable nor continuous on the closure of G. However, in the above, we used the chain rule so the existence of ∇f would not be sufficient.

As an application of the above, we prove that f is constant if G is open and connected and every partial derivative of f is 0. Pick some point x0G, and let g(x) = f(x) − f(x0). We want to show g(x) = 0 for every xG. For that, let E = {xG : g(x) = 0} . Then E is closed and nonempty. It is open too: for every xE,

$|g(y)| = |g(y) - g(x)| \le (0) |y - x| = 0$

for every y in some neighborhood of x. (Here, it is crucial that x and y are sufficiently close to each other.) Since G is connected, we conclude E = G.

Remark that all arguments in the above are made in a coordinate-free manner; hence, they actually generalize to the case when G is a subset of a Banach space.

## Mean value theorem for vector-valued functions

There is no exact analog of the mean value theorem for vector-valued functions. Jean Dieudonné in his classic treatise Foundations of Modern Analysis discards the mean value theorem and replaces it by mean inequality as the proof is not constructive and by no way one can find the mean value. In applications one only needs mean inequality. Serge Lang in Analysis I uses the mean value theorem, in integral form, as an instant reflex but this use requires the continuity of the derivative. If one uses the Henstock-Kurzweil integral one can have the mean value theorem in integral form without the additional assumption that derivative should be continuous as every derivative is Henstock-Kurzweil integrable. The problem is roughly speaking the following: If f : URm is a differentiable function (where URn is open) and if x + th, x, hRn, t ∈ [0, 1] is the line segment in question (lying inside U), then one can apply the above parametrization procedure to each of the component functions fi (i = 1, ..., m) of f (in the above notation set y = x + h). In doing so one finds points x + tih on the line segment satisfying

$f_i(x+h) - f_i(x) = \nabla f_i (x + t_ih) \cdot h.\,$

But generally there will not be a single point x + t*h on the line segment satisfying

$f_i(x+h) - f_i(x) = \nabla f_i (x + t^* h) \cdot h.\,$

for all i simultaneously. (As a counterexample one could take f : [0, 2π] → R2 defined via the component functions f1(x) = cos(x), f2(x) = sin(x). Then f(2π) − f(0) = 0R2, but $\,f_1'(x)=-\sin (x)$ and $\,f_2'(x)=\cos (x)$ are never simultaneously zero as x ranges over [0, 2π].)

However a certain type of generalization of the mean value theorem to vector-valued functions is obtained as follows: Let f be a continuously differentiable real-valued function defined on an open interval I, and let x as well as x + h be points of I. The mean value theorem in one variable tells us that there exists some t* between 0 and 1 such that

$f(x+h)-f(x) = f'(x+t^*h)\cdot h. \,$

On the other hand we have, by the fundamental theorem of calculus followed by a change of variables,

$f(x+h)-f(x) = \int_x^{x+h} f'(u)du = \left(\int_0^1 f'(x+th)\,dt\right)\cdot h.$

Thus, the value f′(x + t*h) at the particular point t* has been replaced by the mean value

$\int_0^1 f'(x+th)\,dt.$

This last version can be generalized to vector valued functions:

Let URn be open, f : URm continuously differentiable, and xU, hRn vectors such that the whole line segment x + th, 0 ≤ t ≤ 1 remains in U. Then we have:

$\text{(*)} \qquad f(x+h)-f(x) = \left(\int_0^1 Df(x+th)\,dt\right)\cdot h,$

where the integral of a matrix is to be understood componentwise. (Df denotes the Jacobian matrix of f.)

From this one can further deduce that if ||Df(x + th)|| is bounded for t between 0 and 1 by some constant M, then

$\text{(**)} \qquad \|f(x+h)-f(x)\| \leq M\|h\|.$

Proof of (*). Write fi (i = 1, ..., m) for the real valued components of f. Define the functions gi: [0, 1] → R by gi(t) := fi(x + th).

Then we have

$f_i(x+h)-f_i(x)\, =\, g_i(1)-g_i(0) =\int_0^1 g_i'(t)dt = \int_0^1 \left(\sum_{j=1}^n \frac{\partial f_i}{\partial x_j} (x+th)h_j\right)\,dt =\sum_{j=1}^n \left(\int_0^1 \frac{\partial f_i}{\partial x_j}(x+th)\,dt\right)h_j.$

The claim follows since Df is the matrix consisting of the components $\frac{\partial f_i}{\partial x_j}$, q.e.d.

Proof of (**). From (*) it follows that

$\|f(x+h)-f(x)\|=\left\|\int_0^1 (Df(x+th)\cdot h)\,dt\right\| \leq \int_0^1 \|Df(x+th)\| \cdot \|h\|\, dt \leq M\| h\|.$

Here we have used the following

Lemma. Let v : [a, b] → Rm be a continuous function defined on the interval [a, b] ⊂ R. Then we have

$\text{(***)}\qquad \left\|\int_a^b v(t)\,dt\right\|\leq \int_a^b \|v(t)\|\,dt.$

Proof of (***). Let u in Rm denote the value of the integral

$u:=\int_a^b v(t)\,dt.$

Now

$\|u\|^2 = \langle u,u \rangle = \left\langle \int_a^b v(t) dt,u \right\rangle = \int_a^b \langle v(t),u \rangle \,dt \leq \int_a^b \| v(t) \|\cdot \|u \|\,dt = \|u\| \int_a^b \|v(t)\|\,dt,$

thus $\| u\| \leq \int_a^b \|v(t)\|\,dt$ as desired. (Note the use of the Cauchy–Schwarz inequality.) This shows (***) and thereby finishes the proof of (**).

## Mean value theorems for integration

### First mean value theorem for integration

The first mean value theorem for integration states

If G : [a, b] → R is a continuous function and $\varphi$ is an integrable function that does not change sign on the interval (a, b), then there exists a number x in [a, b] such that
$\int_a^b G(t)\varphi (t) \, dt=G(x) \int_a^b \varphi (t) \, dt.$

In particular, if φ(t) = 1 for all t in [a, b], then there exists x in (a, b) such that

$\int_a^b G(t) \, dt=\ G(x)(b - a).\,$

More commonly written as:

$\frac{1}{b-a} \int_a^b G(t) \, dt=\ G(x).$

The value G(x) is called the mean value of G(t) on [a, b].

### Proof of the first mean value theorem for integration

Without loss of generality assume the one-signed function $\varphi(t)\ge 0$ for all t (the negative case just changes direction of some inequalities). It follows from the extreme value theorem that the continuous function G has a finite infimum m and a finite supremum M on the interval [a, b]. From the monotonicity of the integral and the fact that mG(t) ≤ M, it follows from the non-negativity of $\varphi(t)$ that

$m I= \int_a^b m\varphi(t)\,dt \le \int^b_aG(t)\varphi(t) \, dt \le \int_a^b M\varphi(t)\,dt = M I,$

where

$I:=\int^b_a\varphi(t) \, dt$

denotes the integral of $\varphi(t)$. Hence, if I = 0, then the claimed equality holds for every x in [a, b]. Therefore, we may assume I > 0 in the following. Dividing through by I we have that

$m \le \frac1I\int^b_aG(t)\varphi(t) \, dt\le M.$

The extreme value theorem tells us more than just that the infimum and supremum of G on [a, b] are finite; it tells us that both are actually attained. Thus we can apply the intermediate value theorem, and conclude that the continuous function G attains every value of the interval [m, M], in particular there exists x in [a, b] such that

$G(x) = \frac1I\int^b_aG(t)\varphi(t) \, dt.$

This completes the proof.

### Second mean value theorem for integration

There are various slightly different theorems called the second mean value theorem for integration. A commonly found version is as follows:

If G : [a, b] → R is a positive monotonically decreasing function and φ : [a, b] → R is an integrable function, then there exists a number x in (a, b] such that
$\int_a^b G(t)\varphi(t)\,dt = G(a+0) \int_a^x \varphi(t)\,dt.$

Here G(a + 0) stands for ${\underset{a_+}{\lim}G}$, the existence of which follows from the conditions. Note that it is essential that the interval (a, b] contains b. A variant not having this requirement is:

If G : [a, b] → R is a monotonic (not necessarily decreasing and positive) function and φ : [a, b] → R is an integrable function, then there exists a number x in (a, b) such that
$\int_a^b G(t)\varphi(t)\,dt = G(a+0) \int_a^x \varphi(t)\,dt + G(b-0) \int_x^b \varphi(t)\,dt.$

This variant was proved by Hiroshi Okamura in 1947.[citation needed][3]

## A probabilistic analogue of the mean value theorem

Let X and Y be non-negative random variables such that E[X] < E[Y] < ∞ and $X\leq_{st} Y$ (i.e. X is smaller than Y in the usual stochastic order). Then there exists an absolutely continuous non-negative random variable Z having probability density function

$f_Z(x)={\Pr(Y>x)-\Pr(X>x)\over {\rm E}[Y]-{\rm E}[X]}\,, \qquad x\geq 0.$

Let g be a measurable and differentiable function such that E[g(X)], E[g(Y)] < ∞, and let its derivative g′ be measurable and Riemann-integrable on the interval [x, y] for all yx ≥ 0. Then, E[g′(Z)] is finite and[4]

${\rm E}[g(Y)]-{\rm E}[g(X)]={\rm E}[g'(Z)]\,[{\rm E}(Y)-{\rm E}(X)].$

## Generalization in complex analysis

As noted above, the theorem does not hold for differentiable complex-valued functions. Instead, a generalization of the theorem is stated such:[5]

Let f : Ω → C be a holomorphic function on the open convex set Ω, and let a and b be distinct points in Ω. Then there exist points u, v on Lab (the line segment from a to b) such that

$\mathrm{Re}(f'(u)) = \mathrm{Re}\left( \frac{f(b)-f(a)}{b-a} \right),$
$\mathrm{Im}(f'(v)) = \mathrm{Im}\left( \frac{f(b)-f(a)}{b-a} \right).$

Where Re() is the Real part and Im() is the Imaginary part of a complex-valued function.