Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics II / 6

A function $f:(a,b)\to \mathbb{R}$ is differentiable on $(a,b)$ when, at each $c\in (a,b)$, its difference quotient has a finite limit of a function:Here $h\ne 0$ and $c+h\in (a,b)$.

For sufficiently small nonzero $h$, the difference quotient has absolute value at most $|f^{\prime }(c)|+1$, because it tends to $f^{\prime }(c)$. ConsequentlyThis is continuity at $c$. Since $c$ was arbitrary, differentiability implies continuity on $(a,b)$.

On $(0,1)$, the product rule and chain rule give $f^{\prime }(x)=2x\sin (1/x)-\cos (1/x)$. On $(-1,0)$, the function is constant and has derivative zero. At zero,Its absolute value is at most $|h|$, so the squeeze theorem gives $f^{\prime }(0)=0$. ThereforeFor $x_n=1/(2\pi n)$, we have $x_n\to 0$ but $f^{\prime }(x_n)=-1$ for every $n$. This differs from $f^{\prime }(0)=0$, so $f^{\prime }$ is not continuous at zero. Indeed, $y_n=1/((2n+1)\pi )$ gives $f^{\prime }(y_n)=1$, so the right-hand limit of a function does not exist. Thus $f$ is differentiable everywhere on $(-1,1)$, while $f^{\prime }$ is discontinuous at zero.

Fix $d$ strictly between $f^{\prime }(a)$ and $f^{\prime }(b)$, and set $F(x)=f(x)-dx$. The function $F$ is continuous on $[a,b]$: interior differentiability gives interior continuity, and the finite one-sided derivatives give continuity at the endpoints. The extreme value theorem therefore gives a point $c$ where $F$ attains its minimum.

Since $F_{+}^{\prime }(a)=f^{\prime }(a)-d<0$, for sufficiently small $h>0$ we have $F(a+h)<F(a)$, so $a$ cannot be a minimizing point. Since $F_{-}^{\prime }(b)=f^{\prime }(b)-d>0$, the quotient $(F(b-h)-F(b))/(-h)$ is positive for sufficiently small $h>0$, giving $F(b-h)<F(b)$. Thus $b$ cannot be a minimizing point either, and $c\in (a,b)$.

Fermat's theorem on stationary points states that a differentiable function at an interior local minimum or local maximum has derivative zero there. Applying it to $F$ givesThis proves the relevant case of Darboux's theorem; the argument uses continuity of $f$, with no assumption of continuity of $f^{\prime }$.

Choose $h_n=\min \{1/n,(b-c)/2\}>0$. Then $h_n\to 0$ and $[c,c+h_n]\subset (a,b)$. The mean value theorem states that a continuous function on a closed interval, differentiable in its interior, has an interior derivative equal to its endpoint difference quotient. It gives $x_n\in (c,c+h_n)$ withHere $x_n\ne c$, $x_n\to c$, and the right-hand side tends to $f^{\prime }(c)$ by differentiability at $c$. If $L={\lim }_{x\to c}{f}^{\prime }(x)$ exists, the same values $f^{\prime }(x_n)$ must tend to $L$. Uniqueness of the limit of a sequence givesThus a derivative cannot have a removable discontinuity at a point where the original function is differentiable. The proof also excludes an infinite limit of a function there, since the displayed difference quotients have a finite limit of a sequence.