Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics II / 4

A function $f:(a,b)\to \mathbb{R}$ is differentiable on $(a,b)$ if, for every $x\in (a,b)$, the finite limit of a functionexists. Here $h\ne 0$ and is sufficiently small that $x+h\in (a,b)$. This limit of a function is the derivative of $f$ at $x$.

The mean value theorem states: if a real-valued function $f$ is continuous on a closed interval $[u,v]$, where $u<v$, and differentiable on $(u,v)$, then there is $c\in (u,v)$ with

Take any $a<u<v<b$. By differentiability implies continuity, $f$ is continuous on $[u,v]$, and it is differentiable on $(u,v)$. The mean value theorem gives $c\in (u,v)$ for whichTherefore $\boxed{f(u)\le f(v)\text{ whenever }u<v}$: $f$ is a nondecreasing function, which is the non-strict meaning of increasing here.

For any $u<v$ in the interval, the mean value theorem, applicable by differentiability implies continuity, givesAll pairs of values are equal. Thus $\boxed{f\text{ is a constant function}}$, the constancy theorem.

Write the symmetric difference quotient as the average of two ordinary difference quotients:If $f$ is differentiable at $x$, both terms tend to the derivative $f^{\prime }(x)$. Hence the symmetric derivative satisfies $\boxed{f^s(x)=f^{\prime }(x)}$.

The absolute value function $f(x)=|x|$ is continuous, and $f^s(0)=0$ because $|h|=|-h|$. Its right one-sided derivative at zero is $1$ and its left one-sided derivative is $-1$, so it is not differentiable at zero. Away from zero it is differentiable, hence has a symmetric derivative there as well. This supplies an example meeting the hypotheses on, for example, $(-1,1)$.

We prove the symmetric derivative criterion for monotonicity using an antiderivative. Fix $a<u<v<b$ and defineThe fundamental theorem of calculus gives $F^{\prime }=f$ because $f$ is continuous. For each interior $x$,Since $f^s(x)>0$, the definition of symmetric derivative gives $\delta >0$ such that $f(x+t)-f(x-t)>t{f}^s(x)$ for $0<t<\delta$. Therefore, for $0<h<\delta$,

We next show that $F$ is a convex function. On any closed interval $[r,s]\subset (a,b)$ let $\ell$ be the affine function joining $(r,F(r))$ to $(s,F(s))$. If $F-\ell$ were positive anywhere, the extreme value theorem would give a positive maximum at an interior point $x$. For every sufficiently small $h>0$, that maximum would implyThe affine function contributes zero to this symmetric difference, contradicting the strict positive inequality above. Thus $F\le \ell$ on every such interval, exactly convexity.

For a differentiable convex function, secant slopes bound endpoint derivatives. Explicitly, the chord inequality at $u+h$ and at $v-h$, followed by $h\downarrow 0$, givesSo $f$ is nondecreasing. If $f(u)=f(v)$ for some $u<v$, monotonicity would make $f$ constant on $[u,v]$, forcing $f^s(x)=0$ for $u<x<v$, a contradiction. Consequently $\boxed{f(u)<f(v)\text{ for every }u<v}$: $f$ is strictly increasing. This proof requires no continuity of $f^s$ and never applies the ordinary mean value theorem to a merely symmetrically differentiable function.

If there were no interior $x_2$ with $f^s(x_2)\le 0$, the symmetric derivative would be positive everywhere in $(a,b)$. Part ii would make $f$ strictly increasing there. Choose $a<u<v<b$. Continuity at the endpoints and monotonicity on the interior implycontradicting $f(a)=f(b)$. Hence such an $x_2$ exists.

If there were no interior $x_1$ with $f^s(x_1)\ge 0$, then $(-f{)}^s=-f^s>0$ everywhere. Apply the same argument to the continuous function $-f$, whose endpoint values are equal. This is again impossible. ThusThe two points need not be distinct.