Posted by: yglima | April 8, 2012

## CF4: law of large numbers for certain cylinder flows

The present post will focus on the paper Law of large numbers for certain cylinder flows in collaboration with Patricia Cirilo and Enrique Pujals, in which we construct a class of cylinder flows that are rationally ergodic along a subsequence of iterates (and thus have explicit law of large numbers). See CF3 for the definitions.

Firstly, we remind what is known: let ${T:\mathbb T\rightarrow\mathbb Z}$ be the Haar function, defined as

$\displaystyle T(x)=\left\{ \begin{array}{rl} 1\ ,&\text{ if }x\in\left[0,\dfrac{1}{2}\right)\\ &\\ -1\ ,&\text{ if }x\in\left[\dfrac{1}{2},1\right)\,. \end{array} \right.$

As we proved in CF2, the associated cylinder flow ${F}$ is ergodic, for any irrational ${\alpha}$. With respect to rational ergodicity, the only result is from J. Aaronson and M. Keane. They proved in The visits to zero of some deterministic random walks that if ${\alpha}$ is quadratic surd, then ${F}$ is rationally ergodic. An irrational number ${\alpha}$ is quadratic surd if it satisfies a quadratic equation with integer coefficients or, equivalently, if its continued fraction expansion is pre-periodic. Specifically, they proved that ${F}$ is rationally ergodic along the sequence of denominators ${(q_n)}$ of ${\alpha}$. When ${\alpha}$ is quadratic surd, the jumps between two consecutive ${q_n}$s are not dramatic and then one can interpolate the rational ergodicity along ${(q_n)}$ to every positive integer.

It is not known to what extent their result can be extended to every irrational basis. Indeed, for a generic ${\alpha}$, the jumps between the ${q_n}$s are large and so it is natural first to ask about rational ergodicity along a subsequence of iterates, for some roof function (not necessarily ${T}$). Our goal is to prove that such phenomenon happens for almost every ${\alpha}$.

Theorem 1 (Cirilo-L.-Pujals) For almost every ${\alpha\in\mathbb R}$, there exists a function ${\phi:\mathbb T\rightarrow\mathbb Z}$ belonging to ${L^p(\mathbb T)}$, for every ${p\ge 1}$, such that the associated cylinder flow is rationally ergodic along a subsequence of iterates.

Our class of ${\alpha}$ will satisfy the following properties: there exists a sequence of (non consecutive) denominators ${(q_n)}$ of ${\alpha}$ such that

1. ${q_n\|q_n\alpha\|}$ converges to zero as ${n}$ goes to infinity, and
2. ${2q_n}$ divides ${q_{n+1}}$.

Appendix B of the paper is devoted to prove that the above properties define a set of full Lebesgue measure.

Throughout this post we will eventually change the basis rotation and for this reason we will denote the ${n}$th Birkhoff sum of a function ${\psi:\mathbb T\rightarrow\mathbb Z}$ with respect to the irrational rotation ${x\mapsto x+\alpha}$ by ${S_n(\alpha,\psi)}$. We remind that ${\nu}$ denotes the Lebesgue measure on ${\mathbb T}$.

1. The roof function ${\phi}$

The roof function we construct is different in nature from the others used in this context. Consider a sequence ${(q_n)}$ satisfying properties 1 and 2 above and let

$\displaystyle \phi(x)=\dfrac{1}{2}\sum_{j\ge 1}\Big[T(q_j(x+\alpha))-T(q_jx)\Big].$

One can see ${\phi}$ as the limit of worser and worser coboundaries

$\displaystyle \phi_n(x)=\dfrac{1}{2}\sum_{j=1}^n\Big[T(q_j(x+\alpha))-T(q_jx)\Big].$

On one hand, the telescoping character of ${\phi_n}$ allows to easily calculate its Birkhoff sums. On the other hand, the increasing bad feature of ${\phi_n}$ is what will guarantee that ${\phi}$ has the required properties. In some sense, our construction resembles Anosov-Katok method of fast approximations developed in New examples in smooth ergodic theory, in which the authors construct differentiable maps sufficiently close to fibered maps of the torus (and, more generally, of any manifold that admits a ${\mathbb T}$-action) with exotic dynamical properties. Indeed, the referred maps are obtained as limits of periodic maps and here we will also use this perspective to prove Theorem 1.

Observe that the bigger ${n}$ is, the closer the points ${q_n(x+\alpha)}$ and ${q_nx}$ are. In particular, the (at least) exponential growth of ${(q_n)}$ guarantees that ${\phi\in L^p(\mathbb T)}$ for every ${p\ge 1}$. Furthermore, the sequence ${(\phi_n)}$ converges pointwise to ${\phi}$.

More than this, condition 2 implies that the Birkhoff sums of ${\phi}$ and ${\phi_n}$ agree up to iterate ${q_{n+1}}$ in a large subset of ${\mathbb T}$. To this purpose, we need the

Lemma 2 Let ${q}$ be a positive integer and ${\beta,\gamma\in\mathbb T}$. Then the set

$\displaystyle \{x\in\mathbb T\,;\,T(qx+\beta)\not=T(qx+\gamma)\}$

has Lebesgue measure equal to ${2\|\beta-\gamma\|}$.

Proof: First, observe that changing ${x}$ by ${x-\beta/q}$, we can assume that ${\beta=0}$. The function ${x\mapsto T(qx)}$ is ${1/q}$-periodic, with

$\displaystyle T(qx)=\left\{ \begin{array}{rl} 1\ ,&\text{ if }x\in\left[0,\dfrac{1}{2q}\right)\bigcup\left[\dfrac{2}{2q},\dfrac{3}{2q}\right)\bigcup \cdots\bigcup\left[\dfrac{2q-2}{2q},\dfrac{2q-1}{2q}\right)\\ &\\ -1\ ,&\text{ if }x\in\left[\dfrac{1}{2q},\dfrac{2}{2q}\right)\bigcup\left[\dfrac{3}{2q},\dfrac{4}{2q} \right)\bigcup\cdots\bigcup\left[\dfrac{2q-1}{2q},1\right). \end{array} \right.$

For each interval ${\left[\frac{i}{2q},\frac{i+1}{2q}\right)}$, ${T(qx)}$ is different from ${T(qx+\gamma)}$ if and only one of the discontinuities ${0,1/2}$ belong to the interval in ${\mathbb T}$ defined by the points ${qx}$ and ${qx+\gamma}$. This happens for an interval of length ${\|\gamma\|/q}$ and so, multiplying by the number ${2q}$ of these intervals, the desired assertion is proved. $\Box$

Let ${\mathcal D=\{0,1/2\}}$ be the set of discontinuities of ${T}$ and define

$\displaystyle \Lambda_n=\left\{x\in\mathbb T\,;\,d(q_jx,\mathcal D)>q_j\|q_j\alpha\|\text{ for }j>n\right\}. \ \ \ \ \ (1)$

Note that

$\displaystyle d(q_j(x+k\alpha),q_jx)=\|kq_j\alpha\|=k\|q_j\alpha\|\le q_j\|q_j\alpha\|$

whenever ${j>n}$ and ${k=1,\ldots,q_{n+1}}$. This implies that

$\displaystyle S_k(\alpha,\phi)(x)=S_k(\alpha,\phi_n)(x)\ \ ,\ x\in\Lambda_n,\ k=1,\ldots,q_{n+1}.$

The ${\Lambda_n}$s form an ascending chain of subsets of ${\mathbb T}$ and ${\mathbb T\backslash\Lambda_n}$ has, by Lemma 2, Lebesgue measure at most ${\sum_{j>n}q_j\|q_j\alpha\|}$, which by property 1 can be assumed to be small.

2. Ergodicity

In this section we prove that, with the above definition, the cylinder flow ${F}$ is ergodic. We proceed in two steps.

Step 1. For any ${A\subset\mathbb T\times\{0\}}$ of positive measure, the union ${\bigcup_{n\ge 1}F^nA}$ contains ${\mathbb T\times\{0\}}$.

Step 2. ${F(\mathbb T\times\{0\})\cap(\mathbb T\times\{1\})}$ and ${F(\mathbb T\times\{0\})\cap(\mathbb T\times\{-1\})}$ have positive measure.

Step 2 is easy: by Lemma 2, for ${s\in\{-1,1\}}$ the set of points ${x\in\mathbb T}$ such that

• ${T(q_1(x+\alpha))=T(q_1x)+2s}$, and
• ${T(q_j(x+\alpha))=T(q_jx)}$ for ${j>1}$

has Lebesgue measure at least ${\|q_1\alpha\|-2\sum_{j>1}\|q_j\alpha\|}$, which can be assumed to be positive if we pass to a subsequence of ${(q_n)}$.

We now give the idea to prove Step 1. The formal proof requires care with some technicalities which I think are not worth in a first reading. The interested reader might check the details in the paper. The main observation is the following: assuming the divisibility condition on the ${q_n}$s, the sequence ${(m_n)}$ of partial sums given by

$\displaystyle \begin{array}{rcrcl} m_n&:&\mathbb T&\longrightarrow &\mathbb Z\\ &&&&\\ & & x &\longmapsto &T(q_1x)+\cdots+T(q_nx) \end{array}$

defines a simple random walk in ${{\mathbb Z}}$. This is easy to see: each map ${x\mapsto T(q_jx)}$ is equal to 1 in ${q_j}$ intervals of length ${1/2q_j}$ each, and ${-1}$ in other ${q_j}$ intervals of length ${1/2q_j}$ each. We call each of these ${2q_j}$ intervals a plateau of order ${j}$ and denote by ${I_j(x)}$ the one that contains ${x}$. By the divisibility property, each ${I_j(x)}$ divides itself into ${q_{j+1}/2q_j}$ plateaux of order ${j+1}$, half of them on which ${T(q_{j+1}x)=1}$ and half on which ${T(q_{j+1}x)=-1}$. This proves the random walk character of ${(m_n)}$. In particular, it has the level-crossing property: almost every two random walks intersect infinitely often. Before going into the proof of ergodicity, we make a further remark: by the very definition of a plateau,

$\displaystyle y\in I_n(x)\ \Longrightarrow\ m_n(x)=m_n(y).$

Now let ${B\subset\mathbb T\times\{0\}}$ be a subset of positive measure, and let ${x}$ and ${y}$ be points of density of ${A}$ and ${B}$, respectively. The goal is to find ${k}$ such that the ${k}$th iterate of a neighborhood ${I}$ of ${x}$ is mapped into a neighborhood of ${y}$. This is equivalent to having, for every ${x'\in I}$,

• ${x'+k\alpha}$ is close to ${y}$, and
• ${S_k(\alpha,\phi)(x')=0}$.

As the ${\Lambda_n}$s converges to ${\mathbb T}$, we might assume that ${x}$ is a point of density of ${B\cap\Lambda_n}$ for large ${n}$ and, whenever ${k\le q_{n+1}}$, the second condition is equivalent to the simpler one

• ${S_k(\alpha,\phi_n)(x')=0}$.

This, in turn, means that ${m_n(x'+k\alpha)=m_n(x')}$. But, because of the first condition, very likely we will have ${x'+k\alpha\in I_n(y)}$ and thus ${m_n(x'+k\alpha)=m_n(y)}$. This hints us to do the following: let ${n}$ large such that

$\displaystyle m_n(x)=m_n(y).$

Then let ${k\le q_{n+1}}$ such that ${x+k\alpha}$ is close enough to ${y}$ to guarantee that ${I}$ is mapped inside ${I_n(y)}$. In this way, for any ${x'\in I}$,

$\displaystyle m_n(x'+k\alpha)=m_n(y)=m_n(x)=m_n(x')$

and so ${I\times\{0\}}$ is mapped, under ${F}$, to a neighborhood of ${(y,0)}$. This concludes the proof of Step 1 and thus of ergodicity.

Observe that, because we can always restrict ${(q_n)}$ to a subsequence, it is no loss of generality to assume that each orbit ${x,x+\alpha,\ldots,x+q_{n+1}\alpha}$ is ${\varepsilon_n}$-dense in ${\mathbb T}$, where ${\varepsilon_n>0}$ depends only on ${q_n}$, and so the term “close enough” in the previous paragraph makes sense.

3. Counting the number of returns to zero

The goal of this section is to calculate the number of returns of an arbitrary point ${(x,0)\in\mathbb T\times\{0\}}$ to ${\mathbb T\times\{0\}}$, under successive iterations of ${F}$. More specifically, we want to calculate the distribution of the map ${R_{n+1}:\mathbb T\rightarrow\mathbb N}$ given by

$\displaystyle R_{n+1}(x)=\#\{0\le k

Once we have such information, we will be able to prove in the next section that ${F}$ is rationally ergodic along the sequence of iterates ${(q_n)}$, by showing that ${\mathbb T\times\{0\}}$ is a sweep-out set for which the Renyi inequality holds (see Definition 6 in CF3).

Firstly, to the purpose of iterates up to order ${q_{n+1}}$, we can work with ${\phi_n}$ instead of ${\phi}$. Better than this, we consider the “rational” truncated versions of ${\phi}$ defined by

$\displaystyle \tilde\phi_n(x)=\dfrac{1}{2}\sum_{j=1}^n\Big[T_j(x+\alpha_{n+1})-T_j(x)\Big],$

where ${\alpha_{n+1}=p_{n+1}/q_{n+1}}$ is the good rational approximation associated to ${q_{n+1}}$. This reduction is, again, similar in spirit to the Anosov-Katok method of fast approximations, in which the authors define transformations as the limit of coboundaries, not from the proper irrational rotation, but from good rational approximations of it. We claim that ${S_k(\alpha,\phi_n)}$ and ${S_k(\alpha_{n+1},\tilde\phi_n)}$ coincide in a large set for ${k\le q_{n+1}}$. Just define, similarly to (1), the set ${\tilde\Lambda_n}$ by the relations

• ${T(q_j(x+k\alpha))=T(q_j(x+k\alpha_{n+1}))}$ for ${j=1,\ldots,n}$ and ${k=1,\ldots,q_{n+1}}$, and
• ${d(q_jx,\mathcal D)>q_j\|q_j\alpha\|}$ for ${j>n}$.

By Lemma 2, the complement of ${\tilde\Lambda_n}$ has ${\nu}$-measure at most

$\displaystyle \sum_{1\le k\le q_{n+1}\atop{1\le j\le n}}\|kq_j(\alpha-\alpha_{n+1})\|+\sum_{j>n}q_j\|q_j\alpha\|< |\alpha-\alpha_{n+1}| q_{n+1}^2\sum_{j=1}^n q_j+\sum_{j>n}q_j\|q_j\alpha\|,$

which goes to zero as ${n}$ goes to infinity (by again, if necessary, passing to a subsequence of ${(q_n)}$). So the proof boils down to understanding the map ${\tilde R_{n+1}:\mathbb T\rightarrow\mathbb N}$ given by

$\displaystyle \begin{array}{rcl} \tilde R_{n+1}(x)&=&\#\{0\le k

For each ${k}$, the value of ${m_n(x+k\alpha_{n+1})}$ is defined by the vector

$\displaystyle v_k(x)=(T(q_1(x+k\alpha_{n+1})),\ldots,T(q_n(x+k\alpha_{n+1})))\in\{-1,1\}^n.$

We claim that the possible combinatorics of this vector, as ${k}$ runs over the set ${\{0,1,\ldots,q_{n+1}-1\}}$, are equally distributed, for almost every ${x\in\mathbb T}$.

Proposition 3 For almost every ${x\in\mathbb T}$, the following holds: for each ${v\in\{-1,1\}^n}$,

$\displaystyle \#\{0\le k

The reader should interpret the above result as a binary tree: if we let, for ${s_1,\ldots,s_j\in\{-1,1\}}$, the set

$\displaystyle B_{(s_1,\ldots,s_j)}=\{0\le k

then

$\displaystyle B_{(s_1,\ldots,s_j)}=B_{(s_1,\ldots,s_j,1)}\sqcup B_{(s_1,\ldots,s_j,-1)},$

and each of the sets ${B_{(s_1,\ldots,s_j,1)},B_{(s_1,\ldots,s_j,-1)}}$ has half of the cardinality of ${B_{(s_1,\ldots,s_j)}}$.

Once the lemma is established, we know exactly what is ${\tilde R_{n+1}}$.

Corollary 4 For each ${m\in\{-n,\ldots,n\}}$ with the same parity of ${n}$,

$\displaystyle \tilde R_{n+1}(x)=\dfrac{q_{n+1}}{2^n}\cdot{n\choose\frac{n+m}{2}}$

for a set of measure equal to ${\frac{1}{2^n}{n\choose\frac{n+m}{2}}}$.

We leave the proof of the corollary as an exercise: just use the random walk character of ${(m_n)}$. We now proceed to prove Proposition 3: the idea is to interpret, for each ${j=1,\ldots,n}$, the map ${k\mapsto T(q_j(x+k\alpha_{n+1}))}$ as a function in ${\mathbb R}$. Define

$\displaystyle \begin{array}{rcrcl} \psi_j&:&{\mathbb R}&\longrightarrow &\{-1,1\}\\ &&&&\\ & &z &\longmapsto &T(q_jx + zq_j\alpha_{n+1})\\ \end{array}$

Each ${\psi_j}$ is a periodic function with period

$\displaystyle \dfrac{u_j}{v}=\dfrac{q_{n+1}/q_j}{p_{n+1}}\,\cdot$

Dilating ${\psi_j}$ to ${\tilde\psi_j:\mathbb R\rightarrow\{-1,1\}}$ given by

$\displaystyle \tilde\psi_j(z)=\psi_j(z/v),$

then each ${\tilde\psi_j}$ is periodic with period ${u_j}$ and to analyse ${\psi_1,\ldots,\psi_n}$ along ${\{0,1,\ldots,q_{n+1}-1\}}$ is the same as analysing ${\tilde\psi_1,\ldots,\tilde\psi_n}$ along ${\mathcal R=\{0,v,\ldots,(q_{n+1}-1)v\}}$. Observing that ${\mathcal R}$ is a complete residue system modulo ${q_{n+1}}$, Proposition 3 is thus a consequence of the following

Lemma 5 Let ${\tilde\psi_j:{\mathbb R}\rightarrow{\mathbb R}}$ be a periodic function with period ${u_j\in{\mathbb Z}}$, ${j=1,\ldots,n}$. Assume that

• ${u_n}$ is even and ${u_j}$ is a multiple of ${2u_{j+1}}$ for ${j=1,\ldots,n-1}$, and
• there are ${z_1,\ldots,z_n\in{\mathbb R}\backslash{\mathbb Q}}$ such that

$\displaystyle \tilde\psi_j|_{\left[z_j,z_j+\frac{u_j}{2}\right)}\equiv 1\ \text{ and }\ \tilde\psi_j|_{\left[z_j+\frac{u_j}{2},z_j+u_j\right)}\equiv -1$

for ${j=1,\ldots,n}$.

Let ${\mathcal R}$ be a complete residue system modulo ${r_1}$. Then, for any sequence ${(s_1,\ldots,s_n)\in\{-1,1\}^n}$,

$\displaystyle \#\{k\in\mathcal R\,;\,\tilde\psi_j(k)=s_j\text{ for }j=1,\ldots,n\}=\dfrac{u_1}{2^n}\,\cdot$

The proof of the above lemma is by induction on ${n}$. Here we prove the case ${n=1}$ and refer the reader to the paper for the full proof. Firstly, let us give an idea of why this must be true. If, instead of being interested in the behavior of ${\tilde\psi_1,\ldots,\tilde\psi_n}$ along integers, we want to calcule the Lebesgue measure of a set with a specific combinatorics and ${x=0}$, then the result is clear. In our case, once we understand how the residue classes of ${\mathcal R}$ modulo ${u_1,\ldots,u_n}$ are related, the induction argument holds without further problems, whenever the values ${q_jx+k/u_j}$, ${k\in\mathcal R}$, are not discontinuities of ${\tilde\psi_j}$ (this is guaranteed by the assumption that ${z_j\not\in\mathbb Q}$, and it holds for almost every ${x\in\mathbb T}$).

We now prove the case ${n=1}$: let ${\psi:{\mathbb R}\rightarrow{\mathbb R}}$ be a function with period ${u\in{\mathbb Z}}$ such that

• ${u}$ is even and
• there is ${z\in{\mathbb R}\backslash{\mathbb Q}}$ such that

$\displaystyle \psi|_{\left[z,z+\frac{u}{2}\right)}\equiv 1\ \text{ and }\ \psi|_{\left[z+\frac{u}{2},z+u\right)}\equiv -1,$

and let ${\mathcal R}$ be a complete residue system modulo ${u}$. We claim

$\displaystyle \#\{k\in\mathcal R\,;\,\psi(k)=1\}=\#\{k\in\mathcal R\,;\,\psi(k)=-1\}=\dfrac{u}{2}\,\cdot \ \ \ \ \ (2)$

To this matter, consider the sets

$\displaystyle \begin{array}{rcl} \Psi_{+}&=&\left\{i\in\mathbb Z\,;\,i\in\left[z,z+\dfrac{u}{2}\right)\right\}\hspace{.6cm}\pmod u \ \ \text{and}\\ &&\\ \Psi_{-}&=&\left\{i\in\mathbb Z\,;\,i\in\left[z+\dfrac{u}{2},z+u\right)\right\}\pmod u\,. \end{array}$

It is clear that ${\Psi_{+}\cup\Psi_{-}=\mathbb Z/u\mathbb Z}$ and that ${\#\Psi_{+}=\#\Psi_{-}=u/2}$. Also, ${\psi(k)=1}$ if and only if ${k\equiv i\pmod u}$ for some ${i\in\Psi_{+}}$. Because ${\mathcal R}$ is a complete residue system module ${u}$, (2) is proved.

4. Renyi inequality

It is a simple task to check that ${(\tilde R_{n+1})}$ satisfies a Renyi inequality. Here, we denote the asymptotic relation ${f(n)=O(g(n))}$ by ${f\lesssim g}$. On one hand,

$\displaystyle \begin{array}{rcl} \left\|\tilde R_{n+1}\right\|_1&=&\displaystyle\int_{\mathbb T}\tilde R_{n+1}d\nu\\ &&\\ &=&\displaystyle\sum_{-n\le m\le n\atop{m\equiv n(\text{mod }2)}}\left[\dfrac{q_{n+1}}{2^n}{n\choose\frac{n+m}{2}}\right]\cdot \left[\dfrac{1}{2^n}{n\choose\frac{n+m}{2}}\right]\\ &&\\ &=&\dfrac{q_{n+1}}{2^{2n}}\displaystyle\sum_{i=0}^n{n\choose i}^2\\ &&\\ &=&\dfrac{q_{n+1}}{2^{2n}}\displaystyle {2n\choose n}\\ &&\\ &\sim&\displaystyle \frac{q_{n+1}}{2^{2n}}\cdot\displaystyle \frac{2^{2n}}{\sqrt{\pi n}}\\ &&\\ &=&\dfrac{q_{n+1}}{\sqrt{\pi n}}\ , \end{array}$

where in the fifth passage we used Stirlings approximation formula. On the other hand,

$\displaystyle \begin{array}{rcl} \left\|\tilde R_{n+1}\right\|_2^2 &=&\displaystyle\sum_{-n\le m\le n\atop{m\equiv n(\text{mod }2)}}\left[\dfrac{q_{n+1}}{2^n}{n\choose\frac{n+m}{2}}\right]^2 \cdot\left[\dfrac{1}{2^n}{n\choose\frac{n+m}{2}}\right]\\ &&\\ &=&\dfrac{q_{n+1}^2}{2^{3n}}\displaystyle \sum_{i=0}^n {n\choose i}^3\\ &&\\ &\le&\dfrac{q_{n+1}^2}{2^{3n}}\displaystyle {n\choose\frac{n}{2}}\displaystyle\sum_{i=0}^n{n\choose i}^2\\ &&\\ &=&\dfrac{q_{n+1}^2}{2^{3n}}\displaystyle {n\choose\frac{n}{2}}\displaystyle {2n\choose n}\\ &&\\ &\sim&\sqrt{2}\cdot\dfrac{q_{n+1}^2}{\pi n} \end{array}$

and therefore

$\displaystyle \dfrac{\left\|\tilde R_{n+1}\right\|_2}{\left\|\tilde R_{n+1}\right\|_1}\lesssim \dfrac{\sqrt[4]{2}\cdot\dfrac{q_{n+1}}{\sqrt{\pi n}}}{\dfrac{q_{n+1}}{\sqrt{\pi n}}}\lesssim 1\,.$

Finally, we prove the Renyi inequality for ${(R_{n+1})}$. Note that

$\displaystyle \left|\|R_{n+1}\|_1-\|\tilde R_{n+1}\|_1\right| \le\int_{\mathbb T\backslash\tilde\Lambda_n}|R_{n+1}-\tilde R_{n+1}|d\nu \le q_{n+1}\nu(\mathbb T\backslash\tilde\Lambda_n)\lesssim \|\tilde R_{n+1}\|_1$

and thus ${\|R_{n+1}\|_1\lesssim\|\tilde R_{n+1}\|_1}$. Similarly, ${\|\tilde R_{n+1}\|_2\lesssim\|R_{n+1}\|_2}$ and so

$\displaystyle \|R_{n+1}\|_1\lesssim\|\tilde R_{n+1}\|_1\lesssim\|\tilde R_{n+1}\|_2\lesssim\|R_{n+1}\|_2\,,$

which concludes the proof.

Previous posts: CF0CF1, CF2, CF3.