Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics II / 7

Let $f:[a,b]\to \mathbb{R}$ be a bounded function, with $a<b$. A partition of an interval is $\mathcal{P}=(a=x_0<x_1<\cdots <x_N=b)$; its mesh of a partition is $\Vert \mathcal{P}\Vert ={\max }_i(x_i-x_{i-1})$. Choose arbitrary tags ${\xi }_i\in [x_{i-1},x_i]$. The corresponding Riemann sum isThe Riemann sum convergence theorem says that $f$ is Riemann integrable with Riemann integral $I$ if and only if, for every $\epsilon >0$, some $\delta >0$ satisfiesfor every partition of an interval with $\Vert \mathcal{P}\Vert <\delta$ and every permitted choice of tags. Equivalently, along every sequence of such tagged partitions of an interval with mesh of a partition tending to zero, the Riemann sums tend to $I$. Both the shrinking mesh of a partition and freedom of tags are essential.

SetThe points strictly increase from zero to $x_{n,n}=n/(n+1)$. With $f(t)=\sin t$, the given expression is exactly the midpoint Riemann sumfor the partition of an interval ${\mathcal{P}}_n=(x_{0,n},\dots ,x_{n,n})$ of $[0,n/(n+1)]$. The sine argument is precisely the midpoint tag, since it is half the sum of the consecutive endpoints.

For a fixed integration interval $[0,1]$, append $x_{n+1,n}=1$ and choose the final tag ${\xi }_{n,n}=1$. The resulting Riemann sum $T_n$ satisfiesThis makes explicit the small final interval that the original sum omits.

We verify that the augmented partition of an interval in part i has vanishing mesh of a partition. Define ${\varphi }_n(t)=t^2/(t^2+n)$ for $t\ge 0$. Its derivative isSince $2u\le 1+u^2$ for $u\ge 0$, we have $0\le {\varphi }_n^{\prime }(t)\le 1/\sqrt{n}$. The mean value theorem therefore givesThe appended interval has length $1/(n+1)$, so $\Vert {\mathcal{P}}_n\Vert \le \max \{n^{-1/2},(n+1{)}^{-1}\}\to 0$ for the augmented partition of an interval.

The sine is continuous, hence Riemann integrable on $[0,1]$. By the Riemann sum convergence theorem, $T_n\to {\int }_0^1\sin tdt=1-\cos 1$, using the fundamental theorem of calculus and the cosine antiderivative. As $T_n-S_n=\sin 1/(n+1)\to 0$, the answer is

The stated weighted inequality is the Jensen inequality for a concave function. Assume $a<b$; if $a=b$, both sides of the required inequality are zero. Write $x=\lambda a+(1-\lambda )b$ with $\lambda =(b-x)/(b-a)$. Applying the weighted inequality at the two endpoints givesThus the affine function $\ell$ joining the endpoint values lies below $f$. By monotonicity of the Riemann integral,This is the lower bound in the Hermite-Hadamard inequality for a concave function.

For $a<b$, choose the uniform partition of an interval into $n$ pieces and its midpoint tagsTheir arithmetic mean is exactly $(a+b)/2$. Apply the given Jensen inequality with equal weights ${\lambda }_i=1/n$:Multiplying by $b-a$ gives a midpoint Riemann sum on the left. Its mesh of a partition is $(b-a)/n\to 0$, so the Riemann sum convergence theorem givesFor $a=b$ the equality is immediate. Together with part i, this is the Hermite-Hadamard inequality with the directions appropriate to a concave function.