## Alternative proofs of two classical lemmas

Two of the most fundamental tools in ergodic Ramsey theory are the mean ergodic theorem and the van der Corput trick. Both have a classical and fairly simple proof, which I have presented before in the blog. Recently I came across alternative proofs for both results which seem to “not use anything” other than very well known general facts (such as Banach-Alaoglu theorem). Although both facts are related to Cesàro limits along Følner sequences in arbitrary amenable groups, the proofs presented in this post are not related. I don’t think either of the proofs is new, but I don’t remember ever seing them written out; so I decided to post them here.

After writing down the alternative proof of the van der Corput trick, I realized one actually needs somewhat advanced results to obtain the full generality… (However if we care only about ${{\mathbb Z}}$, or even abelian groups (or the weak version with only one average), then the proof is softer.)

— 1. Mean ergodic theorem —

I wrote a post about both the mean and the pointwise ergodic theorem before in this blog and there I presented the usual textbook proof. The typical proof uses the fact that the “ergodic” subspace (the orthogonal complement to the space of invariant functions) is generated by vectors of the form ${f-Uf}$. The proof below is even softer and proceeds in two stages: first we establish weak convergence — this portion is valid for any amenable group — and then we use a trick (which goes back at least to Furstenberg) to deduce strong convergence. Unfortunately this second step does not seem to go through in non commutative groups (the theorem is still true, just this proof doesn’t work).

Theorem 1 (Weak ergodic theorem) Let ${G}$ be an amenable group and let ${(U_g)_{g\in G}}$ be a unitary representation of ${G}$ on a Hilbert space ${H}$. Let ${(F_N)_{N\in{\mathbb N}}}$ be a Følner sequence in ${G}$. Then for every ${x\in H}$ we have

$\displaystyle \lim_{N\rightarrow\infty}\frac1{|F_N|}\sum_{g\in F_N}U_gx=Px\qquad\text{ weakly} \ \ \ \ \ (1)$

where ${P:H\rightarrow H}$ is the orthogonal projection onto the space of vectors in ${H}$ fixed by every ${U_g}$.

Proof: Let ${H_0=\{x\in H:U_gx=x~\forall g\in G\}}$; observe that ${H_0}$ is a closed subspace of ${H}$ invariant under all ${U_g}$. Let ${x\in H}$ be arbitrary and decompose it as ${x=x_0+x^\perp}$ where ${x_0\in H_0}$ and ${x^\perp\perp H_0}$. Since (1) depends linearly on ${x}$, and is trivially true when ${x\in H_0}$, it remains to show that it holds for ${x\perp H_0}$, in which case the right hand side is ${0}$.

Assume next that ${x\perp H_0}$ . By the Banach-Alaoglu theorem there exists a subsequence of ${(F_N)_{N\in{\mathbb N}}}$ (which will be a Følner sequence itself) along which the left hand side in (1) equals some ${y}$, in the weak topology. For any ${h\in G}$ and ${z\in H}$, using the Følner property we have

$\displaystyle \langle U_hy,z\rangle=\left\langle\lim_{N\rightarrow\infty}\frac1{|F_N|}\sum_{g\in F_N}U_hU_gx,z\right\rangle=\left\langle\lim_{N\rightarrow\infty}\frac1{|F_N|}\sum_{g\in F_N}U_gx,z\right\rangle=\langle y,z\rangle$

It follows from the Riesz representation theorem that ${U_hy=y}$ and hence ${y\in H_0}$. Since ${x\perp H_0}$ and ${H_0}$ is invariant we have that each ${U_gx\perp H_0}$ and hence also ${y\perp H_0}$. Therefore ${y=0}$. We showed that every weakly convergent subsequence of ${(1/|F_N|)\sum_{g\in F_N}U_gx}$ converges to ${0}$ which finishes the proof. $\Box$

We can now use Theorem 1 for the group ${G\times G}$ to deduce the strong convergence, at least when ${G}$ is abelian:

Theorem 2 (Mean ergodic theorem) Let ${G}$ be an abelian group and let ${(U_g)_{g\in G}}$ be a unitary representation of ${G}$ on a Hilbert space ${H}$. Let ${(F_N)_{N\in{\mathbb N}}}$ be a Følner sequence in ${G}$. Then for every ${x\in H}$ we have

$\displaystyle \lim_{N\rightarrow\infty}\frac1{|F_N|}\sum_{g\in F_N}U_gx=Px\qquad\text{ in norm} \ \ \ \ \ (2)$

where ${P:H\rightarrow H}$ is the orthogonal projection onto the space of vectors in ${H}$ fixed by every ${U_g}$.

Proof: As in the proof of the weak ergodic theorem, the result is trivially true when ${x\in H_0}$, so it suffices to prove it for ${x\perp H_0}$, in which case the right hand side of (2) is ${0}$. Observe that

$\displaystyle \left\|\frac1{|F_N|}\sum_{g\in F_N}U_gx\right\|^2=\sum_{g,h\in F_N}\langle U_gx,U_hx\rangle=\sum_{g,h\in F_N}\langle U_{g-h}x,x\rangle$

Define the unitary representation ${V}$ of ${G\times G}$ by ${V_{g,h}=U_{g-h}}$. Notice that ${(F_N\times F_N)_{N\in{\mathbb N}}}$ is a Følner sequence in ${G\times G}$. Using Theorem 1 we now conclude that indeed

$\displaystyle \left\|\frac1{|F_N|}\sum_{g\in F_N}U_gx\right\|^2=0.$

$\Box$

— 2. The van der Corput trick —

The original van der Corput trick deals with equidistribution of sequences in the ${[0,1]}$ interval. Bergelson observed that the same mechanism could be extended to sequences in Hilbert spaces and that it could be used as the engine behind several recurrence and convergence results.

Most proofs of the van der Corput trick are based on the Cauchy-Schwartz inequality; the proof below was inspired by results of Kamae and Mendès France on van der Corput sets and does not seem to need it. As with the mean ergodic theorem, I will first prove a weaker version of the van der Corput trick and then the stronger version. The reason for this is that in order to deal with the stronger version I actually need a (corollary of a) heavier result from functional analysis, called Naimark’s dilation theorem (which I only learned when trying to push the proof of the weak van der Corput trick to the strong version).

Theorem 3 (Weak van der Corput trick) Let ${G}$ be a countable amenable group and let ${u:G\rightarrow H}$ be a bounded map into a Hilbert space ${H}$. Let ${(F_N)_{N\in{\mathbb N}}}$ be a Følner sequence in ${G}$. Assume that for every ${h\in G}$, ${h\neq1_G}$ we have

$\displaystyle \lim_{N\rightarrow\infty}\frac1{|F_N|}\sum_{n\in F_N}\langle u(nd),u(n)\rangle=0$

Then

$\displaystyle \lim_{N\rightarrow\infty}\left\|\frac1{|F_N|}\sum_{n\in F_N}u(n)\right\|=0$

Proof: Let

$\displaystyle a_N=\frac1{|F_N|}\sum_{n\in F_N}u(n)$

Assume for the sake of a contradiction that the theorem is false. After passing to a subsequence if needed we can assume that ${a:=\lim_{N\rightarrow\infty}\|a_N\|}$ exists and is not ${0}$. Define, for each ${d\in G}$

$\displaystyle \gamma(d):=\lim_{N\rightarrow\infty}\frac1{|F_N|}\sum_{n\in F_N}\langle u(nd)-a_N,u(n)-a_N\rangle \ \ \ \ \ (3)$

Passing to a further subsequence we can assume that all the (countably many) limits exist. Using the Følner property we can rewrite ${\gamma(d)}$ for ${d\neq1_G}$ as

$\displaystyle \begin{array}{rcl} \displaystyle\gamma(d)&=&\displaystyle\lim_{N\rightarrow\infty}\frac1{|F_N|}\sum_{n\in F_N}\langle u(nd),u(n)\rangle-\langle a_N,u(n)\rangle-\langle u(n),a_N\rangle+\|a_N\|^2\\&=&\displaystyle\lim_{N\rightarrow\infty}-\|a_N\|^2+\frac1{|F_N|}\sum_{n\in F_N}\langle u(nd),u(n)\rangle\\&=&-a\end{array}$

Next, let ${F\subset G}$ be a finite set with cardinality ${|F|>1+\gamma(1_G)/a}$, so that ${(|F|^2-|F|)(-a)+|F|\gamma(1_G)<0}$. We have

$\displaystyle \begin{array}{rcl} 0&>&\displaystyle(|F|^2-|F|)(-a)+|F|\gamma(1_G)\\ &=&\displaystyle\sum_{g,h\in F}\gamma(h^{-1}g)\\&=&\displaystyle\sum_{g,h\in G}\lim_{N\rightarrow\infty}\frac1{|F_N|}\sum_{n\in F_N}\langle u(nh^{-1}g)-a_N,u(n)-a_N\rangle\\&=& \displaystyle\sum_{g,h\in G}\lim_{N\rightarrow\infty}\frac1{|F_N|}\sum_{m\in F_N}\langle u(mg)-a_N,u(mh)-a_N\rangle\\&=& \displaystyle\lim_{N\rightarrow\infty}\frac1{|F_N|}\sum_{m\in F_N}\left\|\sum_{g\in G}\big(u(mg)-a_N\big)\right\|^2\geq0 \end{array}$

which is the desired contradiction.

$\Box$

The same proof essentially gives the strong version, but there is a technical step which, though it seems rather natural, requires the Naimark’s dilation theorem to prove. This technical step is spelled out in the next lemma.

Definition 4 Let ${G}$ be a group and ${\gamma:G\rightarrow{\mathbb C}}$ a function. We say that ${\gamma}$ is positive definite if for every ${f:G\rightarrow{\mathbb C}}$ with finite support (i.e. ${f}$ vanishes outside a finite set) we have

$\displaystyle \sum_{n,m\in G}f(n)\overline{f(m)}\gamma(m^{-1}n)\geq0$

Lemma 5 Let ${G}$ be an amenable group, let ${(F_N)_{N\in{\mathbb N}}}$ be a Følner sequence in ${G}$ and let ${\gamma:G\rightarrow{\mathbb C}}$ be a positive definite function. If

$\displaystyle L:=\lim_{N\rightarrow\infty}\frac1{|F_N|}\sum_{g\in F_N}\gamma(g)$

exists, then ${L\geq0}$.

Proof: Since ${\gamma}$ is positive definite, it follows from Naimark’s dilation theorem (Theorem 5.20 in these notes of V. Paulsen) that there exists a unitary representation ${(U_g)_{g\in G}}$ of ${G}$ on a Hilbert space ${H}$ and a vector ${x\in H}$ such that ${\langle U_gx,x\rangle=\gamma(g)}$. Applying the weak ergodic theorem (Theorem 1 above) one deduces that

$\displaystyle L=\lim_{N\rightarrow\infty}\frac1{|F_N|}\sum_{g\in F_N}\gamma(g)=\lim_{N\rightarrow\infty}\frac1{|F_N|}\sum_{g\in F_N}\langle U_gx,x\rangle=\langle Px,x\rangle=\|Px\|^2\geq0$

where ${P:H\rightarrow H}$ is the orthogonal projection onto the space of vectors fixed by every ${U_g}$. $\Box$

We can now prove the strong version of the van der Corput trick:

$\displaystyle \lim_{D\rightarrow\infty}\frac1{|F_D|}\sum_{d\in F_D}\limsup_{N\rightarrow\infty}\left|\frac1{|F_N|}\sum_{n\in F_N}\langle u(nd),u(n)\rangle\right|=0$

Then we have

$\displaystyle \lim_{N\rightarrow\infty}\left\|\frac1{|F_N|}\sum_{n\in F_N}u(n)\right\|=0$

Proof: Assume for the sake of a contradiction that the theorem is false. Let ${a_N}$, ${a>0}$ and ${\gamma}$ be as in the proof of Theorem 3. Using the Følner property we can rewrite ${\gamma(d)}$ as

$\displaystyle \gamma(d)=-a+\lim_{N\rightarrow\infty}\frac1{|F_N|}\sum_{n\in F_N}\langle u(nd),u(n)\rangle \ \ \ \ \ (4)$

Therefore

$\displaystyle \lim_{D\rightarrow\infty}\frac1{|F_D|}\sum_{d\in F_D}\gamma(d)=-a \ \ \ \ \ (5)$

On the other hand, it follows directly from (3) that ${\gamma}$ is a positive definite function. Indeed, let ${f:G\rightarrow{\mathbb C}}$ be a function with finite support. Then

$\displaystyle \begin{array}{rcl} \displaystyle\sum_{g,h\in G}f(g)\overline{f(h)}\gamma(h^{-1}g)&=&\displaystyle\sum_{g,h\in G}f(g)\overline{f(h)}\lim_{N\rightarrow\infty}\frac1{|F_N|}\sum_{n\in F_N}\langle u(nh^{-1}g)-a_N,u(n)-a_N\rangle\\&=& \displaystyle\sum_{g,h\in G}f(g)\overline{f(h)}\lim_{N\rightarrow\infty}\frac1{|F_N|}\sum_{m\in F_N}\langle u(mg)-a_N,u(mh)-a_N\rangle\\&=& \displaystyle\lim_{N\rightarrow\infty}\frac1{|F_N|}\sum_{m\in F_N}\left\|\sum_{g\in G}f(g)\big(u(mg)-a_N\big)\right\|^2\geq0 \end{array}$

Lemma 5 now implies that the limit ${-a}$ in (5) is non-negative, contradicting the fact assumption that ${a>0}$. This contradiction finishes the proof. $\Box$