Let G be a countable group. We say that a sequence \((A_ n)\) of finite sets in G is a right averaging sequence for a function \(\phi\) iff there is a number \(M(\phi)\) (right-side mean value) such that \(\lim_{n}| A_ n|^{-1}\sum_{g\in A_ n}\phi(fg)= M(\phi)\) uniformly in \(f\in G\). Obviously “right” can be replaced by “left”. The sequence \((A_ n)\) is universally averaging for a linear translation invariant space of functions on G if it is right and left averaging for every function in that space. A typical case is that of almost periodic (a.p.) functions. It is not known if there exists a universal averaging sequence for a.p. functions in any countable group. Of course is does exist for amenable groups (owing to the Følner condition). In the free group \(F_ 2\) with two generators (a and b) Greenleaf has found universal averaging sequences for a.p. functions. He also has shown that some finite sets, which might be hoped to have this property do not have it (such is for example the sequence of sets \(E_ n=\{x_ 1...x_ n:\) \(x_ i=a\) or \(b\}\)).
The authors solve the problem in the affirmative for all free groups \(F_ k\) with finitely many generators \(g_ 1,...,g_ k\). More exactly, they prove the following theorem, which implies such solution: Let \((a_ n)\) and \((b_ n)\) be sequences of natural numbers with \(a_ n<b_ n\), \(b_ n-a_ n\to \infty\) and let \(s^{(n)}_{a_ n}\), \(s^{(n)}_{a_ n+1},...,s_{b_ n}^{(n)}\) and \(t_{a_ n}^{(n)},t^{(n)}_{a_ n+1},...,t_{b_ n}^{(n)}\) be integers such that \(t_ i^{(n)}\geq s_ i^{(n)}\). Finally we put \[ A_ n=\{g\in F_ k:\;g=\prod^{b_ n}_{i=a_ n}g_{\bar i}^{j_ i},\quad s_ i^{(n)}\leq j_ i\leq t_ i^{(n)}-1\} \] where \(\bar {\i}\) means the residue of i mod k. If \[ \lim_{n}(b_ n-a_ n)^{- 1}\min \{t_ i^{(n)}-s_ i^{(n)}:\;a_ n\leq i\leq b_ n\}=\infty \] then, for every finite-dimensional unitary representation U of \(F_ k\), there exists \(\lim_{n}| A_ n|^{-1} \sum_{g\in A_ n}U(g)=P\) where P is the orthogonal projection onto the space \(\{x: U(g)x=x\) for every \(g\in F_ k\}\).