0
$\begingroup$

In the exercise section of the integration chapter of Bartles's book it says that if we define the function

$f(x)=\begin{cases}0&x\in [0,1) \\ 1 & x\in [1,2]\;,\end{cases}$

this Riemann-integrable function is not the derivative of any function. But

$F(x)=\begin{cases}1 & x\in [0,1)\;\\x &x\in [1,2]\end{cases}$ has f as a derivative, right? Then am I wrong? This is in the context of verifying the hypothesis for the Fundamental Theorem of Calculus.

Thanks in advance for the answers.

  • 0
    I have [an old answer](http://math.stackexchange.com/questions/62904/differentiability-of-fx-and-continuity-of-fx-same-thing-or-different/62910#62910) that seems relevant.2011-09-16

1 Answers 1

2

The function $F$ defined by $1$ when $x\in(0,1)$ and $x$ when $x\in[1,2]$ does not have a derivative at the point $x=1$. To see this analytically (it should be obvious graphically!), consider the definition of the derivative with the limit $\lim_{h\to0}\frac{F(x+h)-F(x)}{h}.$ Note that by definition this limit doesn't exist if it isn't the same from both directions. At the point $x=1$, from the negative direction, i.e. $h<0$, we have $\Delta F=1-1=0$, but from the positive direction (h>0) we have $\Delta F = (x+h)-x=h$. Thus the limit from the left is $0/h\to0$ and from the right is $h/h\to1$. But $0\ne1$.

This problem will arise in whatever function you try to have a derivative of $f(x)$ for the simple reason that $F$ would be constant from the left of $x=1$ and a line segment with slope $m=1$ on the right, from which the same side discrepancy occurs in evaluating the limit definition of the derivative. In other words, the fundamental theorem of calculus gives \lim_{h\to0+}\frac{1}{h}\int_{1-h}^1 f(x)dx=F'(1^-)=0 but \lim_{h\to0+}\frac{1}{h}\int_1^{1+h}f(x)dx=F'(1^+)=1 hence F'(1) cannot exist.