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Suppose that $f$ is a continuous function of bounded variation from $R^2$ to $R$ that's negative outside of some bounded set, and let $F=\max(f,0)$. Let $S_n$ be the Riemann sum for the integral of $F$ over $R^2$ obtained by summing the values of $F$ at all points in the lattice $(Z/n)^2$ and dividing by $n^2$. What sort of bounds can be given for the difference between $S_n$ and the integral of $F$ over $R^2$? ($O(1/n)$ or $O(1/n^2)$ or what?)

Also how can this basic bound be improved if one knows more about $f$, e.g. that it is smooth or concave?

I'm restricting the question to functions on $R^2$ for definiteness, but I'd like to know the more general situation for $R^n$.

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I can imagine some cropped, two dimensional version of abs(sin(1/x)/x) having very bad estimates, especially if the local maxima were at points in the domain with irrational coordinates. I don't have the error estimates, but my gut suggests that worse than O(1/n) is acheivable, and that the situation does not get better as the dimension grows. Gerhard "Ask Me About System Design" Paseman, 2011.08.12 –  Gerhard Paseman Aug 14 '11 at 23:36
    
Sorry my question wasn't clearer. What I'm after are multi-dimensional analogues of Euler-Maclaurin summation, and things like that. –  James Propp Aug 15 '11 at 4:13
    
Also, I'd like an answer that is applicable to specific cases like $f(x,y)=\sqrt{1-x^2-y^2}$, for which the error of the Riemann sums appears to fall like $1/n^2$. –  James Propp Aug 15 '11 at 11:34
    
If you want good estimates on the error for a specific f, rather than for all functions of bounded variation, and you are summing over the periodic lattice (and not just on the "worst" point in each cube, which is what a Riemann sum does) I would recommend using the Poisson summation formula and getting good bounds on the Fourier transform, possibly after first smoothing away any discontinuities (cf. the proof of the Landau (or Voronoi) error term in the Gauss circle problem, see e.g. ams.org/notices/200106/fea-iosevich.pdf ). –  Terry Tao Aug 15 '11 at 19:04
    
I only discussed $\sqrt{1-x^2-y^2}$ because I expect it to exhibit behavior typical for the case where $f$ is concave (note the divergence of the derivative of $\sqrt{1-x^2-y^2}$ at the boundary; in this respect the example is about as bad as it can be). I'm more interested right now in finding out about existing general-purpose off-the-shelf estimates than about any examples in particular. But let's focus on this example anyway. Can anyone suggest a (preferably online) derivation of the Fourier transform of the indicator function of the disc? I gather that Bessel functions play a role. –  James Propp Aug 16 '11 at 6:40

2 Answers 2

up vote 6 down vote accepted

With the hypotheses given, one can't do better than $O(1/n)$ decay. Consider for instance the function $\frac{1}{n} \cos^2(2\pi n x_1)$ smoothly localised to a ball for some large $n$. This has a total variation norm of $O(1)$, but for this specific value of $n$, the Riemann sum will be off by $O(1/n)$.

Of course, this function depends on $n$. For an $n$-independent example, one could then consider the Weierstrass type function $\sum_{n=1}^\infty \frac{1}{j^2 n_j} \cos^2(2\pi n_j x_1)$ smoothly localised to the unit ball, where $n_j$ goes rapidly to infinity. This is still continuous and of bounded variation, but now the Riemann sum will be off by about $O(1/j^2 n_j)$ at scale $1/n_j$.

In dimensions $d$ greater than 1, the situation is much worse; one can't do much better than $O(1)$, basically because of the failure of the Sobolev embedding $W^{1,1} \subset L^\infty$ in higher dimensions. For instance, one can consider a function $f$ that consists of a bump function of height 1 localised to a ball of radius $O( n^{-d/(d-1)} )$ at each lattice point on $\frac{1}{n} {\bf Z}^d \cap B(0,1)$. This has total variation norm $O(1)$ and is bounded by $O(1)$, but the Riemann sum is off by $O(1)$. By superimposing several such examples together as in the Weierstrass type example we can then construct an $n$-independent function of bounded variation and continuous of compact support whose Riemann sum error decays as slowly as one pleases.

Once one does have enough regularity (in, say, a Sobolev class) to control local $L^\infty$ oscillation, then one can estimate the error term in the Riemann sum by partitioning space into cubes, using some sort of local Sobolev inequality on each cube, and summing up. This for instance gives an $O(1/n)$ error term in the one-dimensional bounded variation case.

One can also analyse Riemann sums by Littlewood-Paley theory. Functions whose Fourier transform is supported on frequencies much smaller than $n$ have excellent agreement between the integrals and their Riemann sums (particularly if one uses quadrature to improve the accuracy of the latter), and functions whose Fourier transform are supported on frequencies much larger than $n$ have a negligible integral. So the error term is basically the same thing as the Riemann sum of the high-frequency component of the function $f$.

Concavity should be very helpful, ruling out the oscillatory counterexamples mentioned above and giving some new bounds on first and second derivatives of $f$ that can be plugged into the local Sobolev inequality method, but I don't immediately see what the best bounds would be with this hypothesis.

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By mollification one can see that a function is convex if and only if its second (distribution-theoretic) derivative is non-negative -- the point being that these two classes of functions are both invariant under translation and are closed under the limiting process. Non-negative distributions are all locally finite, non-negative measures. In frequency space, this observation gives a convex function a decay of $\frac{C}{|\xi|^2}$. We should be able to continue from here? (This is only using the fact that the function is subharmonic -- maybe there's a better way to use true convexity.) –  Phil Isett Aug 16 '11 at 0:13
    
I'll follow up on this branch of the question in a new thread: mathoverflow.net/questions/73142/… . –  James Propp Aug 18 '11 at 12:38

If $f$ is of bounded variation, then there are bounds given by a (multi-dimensional generalization of a) theorem of Koksma. A reference is Kuipers and Niederreiter, Uniform Distribution Of Sequences.

EDIT: Here are a couple of results from that book.

Theorem 5.5: Koksma-Hlawka Inequality. Let $f(x)$ be of bounded variation on $[0,1]^k$ in the sense of Hardy and Krause. Let $\omega$ be the finite sequence of points $${\bf x}_1,\dots,{\bf x}_N$$

in $[0,1]^k$, and let $\omega_{j_1,m\dots,j_p}$ denote the projection of the sequence $\omega$ on the $k-p$-dimensional face of $[0,1]^k$ defined by $x^{(j_1)}=\cdots=x^{(j_p)}=1$. Then we have

$$ \left|{1\over N}\sum_{n=1}^Nf({\bf x_{\it n}})-\int_{[0,1]^k}f({\bf x})d{\bf x}\right|\le\sum_{p=1}^k\sum_{1,\dots,k;p}^*D_N^*(\omega_{p+1,\dots,k})V^{(p)}(f(\dots,1,\dots,1)) $$

where $V^{(p)}(f(\dots,1,\dots,1))$ denotes the $p$-dimensional variation of $f(x^{(1)},\dots,x^{(p)},1,\dots,1)$ on $[0,1]^p$ in the sense of Vitali and where the term of the sum corresponding to $p=k$ is understood to be $D_N^*(\omega)V^{(k)}(f)$.

Here $D$ is a discrepancy, probably very simple to calculate for the situation at hand, but I'm not up to typing it out. I'm going to bail on typing out Theorem 5.6, too; it applies when $f$ has certain continuous partial derivatives, and replaces the variation $V$ with an integral of the absolute value of said derivatives.

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I'd be (pleasantly) surprised if Koksma-Hlawka gave a good bound for my problem, since Koksma-Hlawka deals with very general sets of sample points, and my question concerns the square grid, which has quite special properties (e.g., harmonic analysis becomes applicable, as Terry Tao points out above). But I agree that Koksma-Hlawka will yield some sort of answer to my question. –  James Propp Aug 16 '11 at 5:54

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