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Is there a "natural" class of integrable functions $f: {\mathbb R} \rightarrow {\mathbb R}$ for which it is true (and, preferably, not too hard to prove!) that $\sup_{0 \leq a < h} |h S(a,h) - I|$ is $O(h^2)$, where $S(a,h)$ denotes the sum $\sum_{n \in {\bf Z}} f(a+nh)$ and $I$ denotes the integral $\int_{-\infty}^{\infty} f(x) \: dx$? (Note that for fixed $h$, $S(a,h)$ is periodic in $a$ with period $h$, so the supremum over $0 \leq a < h$ is equivalently the supremum over all $a$ in ${\mathbb R}$.)

The class of functions $f(x)$ should include the functions $e^{-x^2}$, $e^{-|x|}$, $\max(1-x^2,0)$, and $\max(1-|x|,0)$.

I can prove this estimate for functions $f$ that (like the four functions listed above) have bounded first derivative and have bounded second derivative on the complement of a finite set of "bad" points, but this class of functions seems rather unnatural.

I would settle for a bound weaker than $O(h^2)$ as long as it's $o(h)$. (Note that the integrability of $f$ gives us $o(1)$ automatically.)

I'm including the fourier-analysis and harmonic-analysis tags because $S(a,h)$ looks like something that could be approached using harmonic analysis.

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2 Answers 2

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One class of functions you could consider is those $f$ for which $|f(x)|+|{\hat f}(x)| \le C(1+|x|)^{-1-\delta}$ for some constant $C$ and some $\delta>0$. Here ${\hat f}(x) = \int_{-\infty}^{\infty} f(t) e^{-2\pi i tx} dt$ is the Fourier transform. For this class of functions, one can use the Poisson summation formula, which gives $$ h\sum_{n\in {\Bbb Z}} f(a+nh) = \sum_{k\in {\Bbb Z}} {\hat f}(k/h) e^{2\pi i ak/h}. $$ Since $I={\hat f}(0)$, $$ |hS(a,h) - I| \le \sum_{k\neq 0} |{\hat f}(k/h)| \le 2C \zeta(1+\delta) h^{1+\delta}, $$ using our assumption on $f$. See e.g. Stein & Weiss, or the wikipedia article on Poisson summation. The functions you listed would fall into this category.

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How can one see that the four listed functions fall into this category? Does one need to first derive explicit formulas for $\hat{f}$ in each case, or are more qualitative arguments available (possibly outsourcing all the hard work to results in the literature)? –  James Propp Feb 21 at 20:24
    
Well three of your functions are very well known, and the fourth is easy enough to compute with. The Gaussian which is its own Fourier transform (up to scaling). The Fourier transform of $e^{-|t|}$ is $1/(1+x^2)$ (up to scaling again). The function $\max(1-|x|,0)$ is the Fejer kernel; Fourier transform $(\sin x/x)^2$ (up to scaling). The fourth function $\max (1- x^2, 0)$ needs a small computation using integration by parts (along the lines of calculating the Fejer kernel). I'm sure all would be found in a table; e.g. Gradshteyn & Ryzhik. –  Lucia Feb 21 at 21:14
    
@James Propp: Also, if your function $f$ has reasonable decay, and you can carry out integration by parts twice, then you would get bounds on the Fourier transform that decay like $1/(1+|x|)^2$. This is of course analogous to your twice differentiable type condition; but as you can see above the Fourier condition is a bit more general ... . I imagine you were using Euler Maclaurin in your argument; the point of this answer is simply that you could also use Poisson sum. –  Lucia Feb 21 at 21:21
    
That's very helpful. Thanks! –  James Propp Feb 21 at 22:12

One very cheap bound for the error is $$ c h^2 \sum_n \sup_{|x-n| \le 1} |f''(x)|\;, $$ for a suitable $c$. (I believe that $c=1/3$ works, but haven't checked very carefully.) This still works if your function satisfies that bound only piecewise, as long as there are finitely many pieces and the function is continuous. (You lose an additional term of order $h^2$ at each "junction". If the function jumps, this term becomes order $h$ and the bound breaks down.) This covers all the examples you gave.

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