TITLE: Dependence of error on mesh for Riemann sums
QUESTION [6 upvotes]: Suppose $f$ is continuous on $[a,b]$ with $I = \int_a^b f(x)\: dx$,
and for every $\epsilon > 0$ let $\delta(\epsilon)$ be the largest
$\delta > 0$ such that every Riemann sum arising from a partition
of $[a,b]$ with mesh less than $\delta$ differs from $I$ by less
than $\epsilon$.
Is it true that (leaving aside the case where $f$ is constant)
$\delta(\epsilon)$ goes to zero like $\epsilon^2$, in the sense
that $\delta(\epsilon)/\epsilon^2$ is bounded above and below
by constants as $\epsilon$ goes to zero?

REPLY [4 votes]: It is possible to choose continuous functions $f$ with really bad approximations, i.e., so that $\delta(\epsilon)$ drops to zero arbitrarily quickly as a function of $\epsilon$.  For example, let $f_0$ denote the periodic function of period 1 that linearly interpolates between $(0,0)$, $(1/2,2)$, and the horizontal translates of these points by integers.  For each $n>0$, let $f_n(x) = f_0(2^n x)$, and let $f(x) = \sum_{n\geq 0} a_n f_n(x)$, for a summable sequence of positive reals $a_n$.  $f$ is integrable on $[0,1]$, with integral $\sum a_n$, and meshes with spacing $2^{-N}$ yield the first N summands.  However, if the partial sums converge very slowly, we can have, e.g., $\delta(1/N) \leq 2^{-N!}$.