I can try to define an averaging operator for functions, namely let $$A: D \subset L^\infty([0,\infty]) \to \mathbb{R}$$ by $$Af = \lim_{N\to\infty} \frac{1}{N}\int_0^N f(x)dx$$ whenever the limit converges. My question is: is there any intuitive description of the domain $D$ (other than the above limit converging)? A few observations:

  • Clearly if $f$ is compactly supported then $Af = 0$, in particular $f \in D$.
  • There exist non-compactly supported $f \in D$. E.g. constant functions, $sin$, $cos$, and many others.
  • There are bounded functions which are not in $D$. Maybe there is a more concise example but here is an intuitive one: $f$ will take either the value 0 or 3. It should start say as 0 on [0,1], then it will be 3 for long enough to bring the average up to 2, then it will return to 0 for long enough to bring the average down to 1, and so on. Thus the average will fluctuate between $1$ and $2$ and never converge.


  • $\begingroup$ A simple thought: what you're looking at is a simple analogue for integrals of Cesaro summation, although not, apparently, the same as what is referred to as Cesaro summability of an integral: en.wikipedia.org/wiki/Ces%C3%A0ro_summation $\endgroup$ May 12, 2012 at 18:16
  • $\begingroup$ Perhaps what you really want is the closure of $D$? $\endgroup$ May 13, 2012 at 14:39

1 Answer 1


I would say that you can define $A$ in all $L^\infty$. Indeed, let $V$ be the set of essentially bounded functions such that your limit exists. $V$ is a linear subspace of $L^\infty$ and the operator $A$ is linear and continuous and then it admits an extension to all of $L^\infty$ by the Hahn-Banach theorem. This extension is neither constructive nor unique and this may create some problems. However, I think that this is the natural way to proceed, having a perfect counterpart on the natural numbers, where the average is the natural density that can be extended to a (non-unique) invariant measure. For instance, your example of a function not belonging to $D$ is the analogue of a set of natural numbers without density.

As I told you the extension is not unique and this means that some functions $f\in L^\infty$, as the one you explicitly considered, may have different averages, depending on which extension you consider. In other cases you may be lucky and obtain a function having a unique average. Maybe, in $D$ you want to put all functions having a unique average. I do not how to characterize them, but the same problem on the integers was studied in a paper by Rosenblatt called Functions with a unique mean value that is available on the web. I think that this can be a good starting point if you want to characterize $D$ as the set of functions having a unique mean value.

  • $\begingroup$ Thanks. The operator is indeed very close to density of subsets of the natural numbers. Do you have a reference that discusses the Hahn-Banach extension of the density? If I recall Hahn-Banach correctly, the extension is both non-constructive and non-unique, which doesn't sound very useful to me. How do people actually work with the extension of the density? $\endgroup$
    – Eric
    May 12, 2012 at 19:05
  • $\begingroup$ You're welcome. I've just included a few details and also some bibliography that may be of interest for you. $\endgroup$ May 13, 2012 at 14:40

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