Short version for people who don't like reading:

Let $f\colon\mathbb{R}\to\mathbb{R}$ be $1$-periodic, measurable and bounded. Is it true that, for almost all $x$, the average of $f(x)$, $f(x+\frac{1}{n})$, $f(x+\frac{2}{n})$, …, $f(x-\frac{1}{n})$ tends to $\int_0^1 f$ when $n\to+\infty$?

And now for a more detailed version of the question:

Let $\mathbb{T} := \mathbb{R}/\mathbb{Z}$ so that functions $\mathbb{T}\to\mathbb{R}$ can be identified with $1$-periodic functions $\mathbb{R}\to\mathbb{R}$.

If $f\colon\mathbb{T}\to\mathbb{R}$, we define $\mathscr{M}_n(f)\colon\mathbb{T}\to\mathbb{R}$ by $$(\mathscr{M}_n(f))(x) := \frac{1}{n}\sum_{k=0}^{n-1} f\Big(\!x+\frac{k}{n}\Big)$$ the $n$-th “Riemann sum” of $f$, i.e., the average of the $n$ translates of $f$ by $n$-th periods. If also $f \in L^1(\mathbb{T})$ we define $\mathscr{E}(f)\colon\mathbb{T}\to\mathbb{R}$ by $$(\mathscr{E}(f))(x) := \int_{\mathbb{T}} f(t)\,dt$$ (constant function!) the integral, i.e., overall average of $f$.

The general problem is in what ways and under what assumptions we can say that $\mathscr{M}_n(f) \to \mathscr{E}(f)$.

Precise questions are below (at end), but first let me first state a few simple known facts relevant to this situation, that might help provide some background:

  • If $f$ is a step function (where "step function" means a linear combination of characteristic functions of intervals) then $|\mathscr{M}_n(f) - \mathscr{E}(f)| \leq \frac{\|f\|_\infty}{n}$ everywhere. (Sketch of proof: if $f = \mathbf{1}_{[0,r/n)}$ with $r\in\mathbb{N}$ then in fact $\mathscr{M}_n(f) = \mathscr{E}(f)$, and if $f_c = \mathbf{1}_{[0,c)}$ with $\frac{r}{n}\leq c<\frac{r+1}{n}$ then $f_{r/n} \leq f_c \leq f_{(r+1)/n}$ everywhere so that the same inequality holds after applying $\mathscr{M}_n$, i.e., $\frac{r}{n} \leq \mathscr{M}_n(f_c) \leq \frac{r+1}{n}$, whence the conclusion for $f_c$, and then for a general step function by translating and taking linear combinations.)

  • If $f \in L^p(\mathbb{T})$ with $1\leq p<\infty$ then $\mathscr{M}_n(f) \to \mathscr{E}(f)$ in $L^p(\mathbb{T})$. (Follows from the above by density of step functions in $L^p$ and the fact that $\mathscr{M}_n$ and $\mathscr{E}$ have norm $1$.)

  • If $f$ is Riemann-integrable then $\mathscr{M}_n(f) \to \mathscr{E}(f)$ uniformly on $\mathbb{T}$. (Fairly obvious using the first point and the following definition of R-integrable functions: for every $\varepsilon>0$ there exist $h$ and $\varphi$ step functions such that $|f-h|\leq\varphi$ everywhere and $\int\varphi \leq \varepsilon$.)

  • If $u_n = n\mathbf{1}_{[0,1/n)}$ then $\mathscr{M}_n(f) - \mathscr{E}(f) = \mathscr{M}_n(f - (f*u_n))$ (writing $*$ for convolution), and when $f$ is measurable we have $f*u_n \to f$ almost everywhere (by the existence of right Lebesgue points).

  • The Fourier coefficients of $\mathscr{M}_n(f)$ are those of $f$ at indices multiple of $n$, the other being $0$; so they converge punctually (i.e., for a given idnex) to those of $\mathscr{E}(f)$. Also, if the Fourier coefficients of $f$ are $\ell^q$ then the convergence of Fourier coefficients of $\mathscr{M}_n(f)$ to those of $\mathscr{E}(f)$ holds in $\ell^q$.

  • Update 2016-02-10: If $f$ is $L^1(\mathbb{T})$, it is not necessarily the case that $\mathscr{M}_n(f) \to \mathscr{E}(f)$ almost everywhere, or indeed, anywhere: this is a theorem of Marcinkiewicz and Zygmund ("Mean values of trigonometrical polynomials", Fund. Math. 28 (1937), chapter II, theorem 3 on p. 157). Their counterexample (which is $(-\log|x|)/\sqrt{|x|}$ on $[0,\frac{1}{2}]$) is certainly not bounded, however.

  • Birkhoff's ergodic theorem is probably also worth mentioning here: if $\xi$ is irrational, then for all $f\in L^1(\mathbb{T})$, for almost all $x$ we have $\frac{1}{n}\sum_{k=0}^{n-1} f(x+k\xi) \to \mathscr{E}(f)$.

Now at last here are my questions, motivated by the gaps left in the above facts:

If $f \in L^\infty(\mathbb{T})$, do we have $\mathscr{M}_n(f) \to \mathscr{E}(f)$ in $L^\infty(\mathbb{T})$? If not, do we have $\mathscr{M}_n(f) \to \mathscr{E}(f)$ in $L^\infty(\mathbb{T})$ almost everywhere? [This is the "short version" at the start of this post.]


1 Answer 1


The answer is "no", even for $f$ measurable and bounded (or indeed, a characteristic function), and even just for convergence almost everywhere (or indeed, anywhere). This is the main result in: Walter Rudin, "An Arithmetic Property of Riemann Sums", Proc. Amer. Math. Soc. 15 (1964), 321–324.


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