Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 1 / 10E (Analysis)

The mean value theorem states that a real-valued function continuous on $[a,b]$ and differentiable on $(a,b)$, with $a<b$, has a point $\xi \in (a,b)$ such thatFor the proof, subtract the secant line and put $h(t)=f(t)-f(a)-(t-a)[f(b)-f(a)]/(b-a)$. Then $h(a)=h(b)=0$. If $h$ is identically zero, every interior point works. Otherwise it has a positive maximum or a negative minimum. The extreme value theorem supplies that extremum, and its nonzero value makes its location interior. At an interior differentiable extremum, the difference quotients on the two sides have opposite weak signs, so their common limit is zero. Hence $h^{\prime }(\xi )=0$, proving the formula. This also gives the Rolle theorem argument rather than merely citing it.

Because $f^{\prime \prime }$ exists everywhere, $f^{\prime }$ is differentiable everywhere and therefore continuous everywhere; continuity of $f^{\prime \prime }$ is not required. For $x<y$, apply the mean value theorem to $f^{\prime }$ on $[x,y]$. It gives $\xi \in (x,y)$ withFor $x=y$ equality is immediate. Hence $f^{\prime }$ is nondecreasing: $x\le y$ implies $f^{\prime }(x)\le f^{\prime }(y)$.

Put $q=\lambda a+(1-\lambda )b$. Since $0<\lambda <1$ and $a<b$, $a<q<b$. Apply the mean value theorem separately to $[a,q]$ and $[q,b]$. There exist $x\in (a,q)$ and $y\in (q,b)$ such thatThese are the requested points, and $x<y$. Part (i) makes $f^{\prime }(x)\le f^{\prime }(y)$, so the corresponding secant slopes satisfyMultiply by the positive denominators and collect $f(q)$ to obtainThis is precisely the defining chord inequality for a convex function. The proof uses only existence and nonnegativity of $f^{\prime \prime }$, with no additional regularity hypothesis on that derivative.