4.3 Consequences of differentiability, the mean value theorem  (Page 2/3)

This proof is tricky. Define a function $h$ on $[a,b]$ by

Clearly, $h$ is continuous on $[a,b]$ and is differentiable at each point $x\in (a,b).$ Indeed,

It follows from this equation that the theorem will be proved if we can show that there exists a point $c\in (a,b)$ for which $h^{\text{'}}\left(c\right)=0.$ Note also that

and

showing that $h\left(a\right)=h\left(b\right).$

Let $m$ be the minimum value attained by the continuous function $h$ on the compact interval $[a,b]$ and let $M$ be the maximum value attained by $h$ on $[a,b].$ If $m=M,$ then $h$ is a constant on $[a,b]$ and $h^{\text{'}}\left(c\right)=0$ for all $c\in (a,b).$ Hence, the theorem is true if $M=m,$ and we could use any $c\in (a,b).$ If $m\ne M,$ then at least one of these two extreme values is not equal to $h\left(a\right).$ Suppose $m\ne h\left(a\right).$ Of course, $m$ is also not equal to $h\left(b\right).$ Let $c\in [a,b]$ be such that $h\left(c\right)=m.$ Then, in fact, $c\in (a,b).$ By [link] , $h^{\text{'}}\left(c\right)=0.$

We have then that in every case there exists a point $c\in (a,b)$ for which $h^{\text{'}}\left(c\right)=0.$ This completes the proof.

REMARK The Mean Value Theorem is a theorem about real-valued functions of a real variable, and we will see later that it fails forcomplex-valued functions of a complex variable. (See part (f) of [link] .) In fact, it can fail for a complex-valued function of a real variable.Indeed, if $f\left(x\right)=u\left(x\right)+iv\left(x\right)$ is a continuous complex-valued function on the interval $[a,b],$ and differentiable on the open interval $(a,b),$ then the Mean Value Theorem certainly holds for the two real-valued functions $u$ and $v,$ so that we would have

which is not $f^{\text{'}}\left(c\right)(b-a)$ unless we can be sure that the two points $c_1$ and $c_2$ can be chosen to be equal. This simply is not always possible.Look at the function $f\left(x\right)=x^2+i{x}^3$ on the interval $[0,1].$

On the other hand, if $f$ is a real-valued function of a complex variable (two real variables),then a generalized version of the Mean Value Theorem does hold. See part (c) of [link] .

One of the first applications of the Mean Value Theorem is to show that a function whose derivative is identically 0 is necessarily a constant function.This seemingly obvious fact is just not obvious. The next exercise shows that thisresult holds for complex-valued functions of a complex variable, even though the Mean Value Theorem does not.

The next exercise establishes, at last, two important identities.