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The following question came up when thinking about equidistribution of Satake parameters of elliptic curves. Let $G$ be a compact Lie group with Haar measure $\mathrm{d} x$. Recall that a sequence $\{x_n\}$ of points in $G$ is equidistributed if for all $f\in C(G)$, we have equality: $$ \lim_{N\to\infty} \frac{1}{N} \sum_{n=1}^N f(x_n) = \int_G f(x)\, \mathrm{d}x . $$ Here, we fix the sequence $\{x_n\}$ and vary $f\in C(G)$. My question is based on what happens when we fix $f$ and vary the $\{x_n\}$. More precisely:

Let $f\in L^1(G)$. Call $f$ equidistribution-good if the equality above holds for all equidistributed $\{x_n\}$, even though $f$ is not necessarily continuous. Is there a good analytic condition (weaker than continuity) such that if $f$ satisfies this condition, then $f$ is equidistribution-good?

If $G=S^1$, then bounded variation works, though it's not clear to me that $f$ is equidistribution-good if and only if it has bounded variation (I think bounded + a.e.-continuity is iff here).

Any partial answers or pointers to references would be super helpful!

Even better, it would be awesome if there was a good analytic criterion for which functions $f$ aren't just equidistribution-good, but satisfy $$ \left|\lim_{N\to\infty} \frac{1}{N} \sum_{n=1}^N f(x_n) - \int_G f(x)\, \mathrm{d}x \right| = O_{\{x_n\},f}(\text{some kind of ``discrepancy''}) . $$

[EDIT: I've modified the question to reflect Christian's comments.]

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  • $\begingroup$ I think you mean "weaker than continuity" (not stronger). Also, continuity a.e. (on $S^1$) is certainly not enough, an e-g $f\in L^1$ must be bounded. $\endgroup$ – Christian Remling Jun 16 '16 at 16:50
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    $\begingroup$ $f$ bounded and a.e. continuous sounds like it could be the right condition. $\endgroup$ – Christian Remling Jun 16 '16 at 16:53
  • $\begingroup$ Thanks Christian - I'll see if I can prove it for bounded + a.e. continuous. $\endgroup$ – Daniel Miller Jun 16 '16 at 17:11
  • $\begingroup$ Christian—your answer is correct. If you make it into an official answer I'll accept it. $\endgroup$ – Daniel Miller Jun 16 '16 at 17:38
  • $\begingroup$ Oh, thanks, but I was just guessing, didn't have any profound thoughts. You could just write it up yourself if you have a proof. $\endgroup$ – Christian Remling Jun 16 '16 at 17:42
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The paper "λ-equidistributed sequences of partitions and a theorem of the De Bruijn–Post type", by Chersi and Volčič, proves that if $(X,d,\lambda)$ is a separable metric space with probability measure whose support is $X$, then "$\frac{1}{N}\sum_{n\leqslant N} f(x_n)\to \lambda(f)$ for all $\lambda$-equidistributed $\{x_n\}$" if and only if $f$ is continuous almost everywhere.

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