Finding $f$ s.t. the sequence of functions $f_n(x)=f (x − a_n )$ is not a.e. convergent to $f$

Question

The following is an exercise from Bruckner's Real Analysis:

Let $a_n$ be any sequence of positive numbers converging to zero. If $f$ is continuous, then certainly $f (x − a_n)$ converges to $f (x)$. Find a bounded measurable function on $[0, 1]$ such that the sequence of functions $f_n(x)=f (x − a_n )$ is not a.e. convergent to f [Hint: Take the characteristic function of a Cantor set of positive measure.]

I don't understand how "$f_n(x)=f (x − a_n )$ is not a.e. convergent to f" can happen at all : $f (x − a_n )$ are 'moving' to become $f(x)$ as $n \to \infty$ and we only have a solution for not a.e. convergent to f when we consider any different $f$ that is not a.e. limit of $f_n$? So what is the use of fat Cantor sets here?

Added - There should be an easy solution within the scope of Bruckner's book because when an exercise is hard and even still is at the book's level, it is always mentioned in a parenthesis "This is hard", but no notice for this exercise is.

Answer 1

Attention Till now no one has succeed to find an answer to the OP at the level of Brucker's book as requested by the OP. My answer is the closest one to the OP at the level of Brucker's book.

This answer presents two "partial solutions". The first one is:

A bounded measurable function on $[0, 1]$ and a sequence $a_n$ of positive numbers converging to zero, such that the sequence of functions $f_n(x)=f (x − a_n )$ is not a.e. convergent to f.

Attention: Here we use a fat Cantor set $F$ and the sequence $a_n$ does not depend on the point $x \in F$. However, it is not a complete solution to the OP, since the sequence is not arbitrary.

Let us go step by step.

1. Using representation in base 3, Cantor set $C$ can be described as the set of elements of $[0,1]$ that have at least one representation not using the digit $1$ (remember that rational numbers have two representations, while irrational number have only one). It is easy to see that the measure of the excluded set is $\sum_{n=1}^\infty \frac{2^{n-1}}{3^n} =1$. So $\lambda(C)=0$.

2. A simple way to produce a fat Cantor set is to define $F$ the set of elements of $[0,1]$ that have at least one representation not containing the sequence $11$. In this case, the measure of the excluded set is $\sum_{n=1}^\infty \frac{2^{n-1}}{3^{n+1}} =\frac{1}{3}$. So $\lambda(F)=\frac{2}{3}$.

Note that if one representation of a rational number contains the sequence $111$ , then both representations contain the sequence $11$. For instance, consider $r= 0.2111$ then the other representation is $r= 0.21102222 \dots$. Both representations contain the sequence $11$. So, if $y \in [0,1]$ and one representation of $y$ contains the sequence $111$, then $y \notin F$.

Now let us define the sequence $(a_n)_n$.

The fist 26 elements are $a_1=0.001$ till $a_{26}=0.222$.

The next 26 elements are $a_{27}=0.0001$ till $a_{52}=0.0222$.

The next 26 elements are $a_{53}=0.00001$ till $a_{78}=0.00222$.

And so on.

It is clear that $a_n$ converges to $0$.

Let $x \in F$, consider the sequence $x-a_n$. It is easy to see that given any $N \in \Bbb N$, there is $n>N$ such that one representation of $x-a_n$ contains $111$. So, $x-a_n \notin F$.

Let $f$ be the indicator function of $F$. Then, for all $x \in F$, $f(x)=1$. But, from the previous paragraph, we have that given any $N \in \Bbb N$, there is $n>N$ such that $x-a_n \notin F$, that means $f(x-a_n)=0$.

So, for all $x \in F$, $f(x-a_n)$ does not converge to $f$.

Since $\lambda(F) > 0$, the functions $f_n(x)= f(x-a_n)$ do not converge a.e to $f$.

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Here is a second "partial solution". It presents:

For any sequence $a_n$ of positive numbers converging to zero, a bounded function on $[0, 1]$ such that the sequence of functions $f_n(x)=f (x − a_n )$ is not a.e. convergent to f.

Attention: Although the sequence $a_n$ is arbitrary, this is not a complete solution to the OP, because $f$ will, in general, not be measurable.

The idea is, given any sequence of positive numbers converging to zero, to build a Vitalli set $G$ adapted to the sequence.

Given any sequence $(a_n)_n$ of positive numbers converging to zero, let

$$S=\left\{ \sum_{k=0}^m p_k a_{n_k}: (a_{n_k})_k \text { is a finite set of elements in the sequence } (a_n)_n \text { and } \\ p_k \in \Bbb Z \text{, for all } k\in \Bbb N \right\}$$ Of course, $0 \in S$ (empty addition) and for each $n$, $a_n \in S$. It is easy to see that $S$ is countable.

Let us define $\sim$ in $[0,1]$ by $x \sim y$ if and only if $x-y \in S$. It is easy to see that $\sim$ is an equivalence relation.

Let us choose exactly one element of each equivalence class. Let $G$ be the such chosen elements.

Note that $[0,1] \subseteq \bigcup_{s \in S} (G+s)$ and that, for all $s \in S$, $\lambda^*(G+s) = \lambda^*(G)$. It follows that $\lambda^*(G)>0$.

Let $f$ be the indicator function of $G$. It follows that, for every $x\[0,1]$ if $f(x)=1$ then $x \in G$ and then, for all $n \in \Bbb N$, $x-a_n \notin G$, so $f(x-a_n) =0$. So $f_n(x)=f(x-a_n)$ does not converge pointwise to $f$ on $G$, and since $\lambda^*(G)>0$, we have that $f_n$ does not converge to $f$.

Note that $f$ is a bounded function, but it is not measurable.

Answer 2

I ignore whether a simple answer (as the hint might suggest) to this OP exists (I have tried the suggestion myself to no avail). I hope there is such simple a solution, and that someone comes up with one along the lines of the hint. Relaying on heavier machinery however, I present a solution to the OP.

1. First, I show that for any given sequence $a_n\searrow0$, there is $f\in L_1([0,1],m)$ such that $$\begin{align} m(\{x:f_n(x-a_n)\stackrel{n}{\nrightarrow}f(x)\})>0\tag{0}\label{zero} \end{align} $$
2. From such a function $f$, one can then construct a measurable bounded function $\phi$ in $[0,1]$ satisfying \eqref{zero}. Indeed, let $B=\{x:f_n\stackrel{n}{\nrightarrow}f\}$. As $f\in L_1$, there is $n_0\in\mathbb{N}$ such that $m(|f|>n_0)<\frac12 m(B)$. Then $m(B\cap\{|f|\leq n_0\})>0$ and the function $\phi:=|f|\mathbb{1}_{B\cap\{|f|\leq n_0\}}$ would do the job.

It is worth mentioning that there is a weaker (simple) version of the problem (see below) and that a related problem has been discussed in MSE before. In fact, @Ramiro's initial solution to this OP (still visible to some) solves that related problem.

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Weaker version of the problem: there is a bounded measurable function $f$ such that the $\tau_hf(x)=f(x-h)$ does not converge to $f(x)$ as $h\rightarrow0$ for any point $x$ in a set of positive Lebesgue measure.

To prove this, we follow the hint in the OP, and consider any fat Cantor set $F$ in $[0,1]$. Since $F$ is perfect and nowhere dense, for any $x\in F$ there are sequences $\{p_n:n\in\mathbb{N}\}\subset F$ and $\{q_n:n\in\mathbb{N}\}\subset [0,1]\setminus F$ such that $\lim_n p_n=x=\lim q_n$. Let $h_n=x-p_n$ and $k_n=x-q_n$. Then $\lim_nh_n=0=\lim_nk_n$ and $$\begin{align} \lim_n f(x-h_n)&=\lim_n f(p_n)=1=f(x)\\ \lim_n f(x-k_n)&=\lim_n f(q_n)=0\neq f(x) \end{align} $$ This complete the proof of the weak case.

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Observation: The weak version as such does not provide a solution to the OP since the sequence of translations $\tau_{h_n}f$ and $\tau_{k_n}f$ depend on the point $x\in F$.

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$L_1$-version of the problem: It will be convenient to consider the problem on $\mathbb{T}$ (the circle of radius one centered at $0$, or $\mathbb{R}\mod 1$) with sum $\mod 1$, and arc-length measure (Lebesgue's measure on $\mathbb{T}$). Following a suggestion by Anthony Quas in his answer to a question related to the Ergodic theorem, one may look instead for existence of integrable functions (or locally integrable functions) $f\in L_1(\mathbb{R},m)$ such that $\tau_{a_n}f$ does not converge to $f$ almost surely. In this case, one may use Maximal functions (similar to the classical techniques of Hardy-Littlewood's, Wiener's, Doobs, for example) to analyze point wise convergence of bounded linear positive operators $T_n :L_1(m)\rightarrow L_1(m)$.

• In general terms, the technique can be described as follows: For each $n\in\mathbb{N}$, define $M_nf=\max_{1\leq j\leq n}T_jf$ and $Mf:=\sup_n T_n f$.

Proposition I: If $M$ is of weak-type $(1,1)$, that is, there is $C>0$ such that $$\begin{align} \sup_{\lambda>0}\lambda m(M|f|>\lambda)\leq C\|f\|_1,\qquad f\in L_1(m),\tag{1}\label{type11} \end{align}$$ and $T_ng\xrightarrow{n\rightarrow\infty} g$ poitwise for all functions $g$ in some $\mathcal{G}\subset L_1(m)$, then $T_nf$ converges to $f$ a.s. for all $f$ in the closure $\overline{\mathcal{G}}$ of $\mathcal{G}$ in $L_1$.

The proof of this fact is standard in Analysis. I present a sketch in the addendum.

• The converse also holds for example when operators $T_n$ on $L_1(\mathbb{T})$ are translation invariant.

Proposition II: If $T_nf(x)\xrightarrow{n\rightarrow\infty} f(x)$ almost surely for any $f\in L_1(\mathbb{T})$, and $T_n$ commute with translations, then $M$ is of weak-type $(1,1)$.

Results of this kind are more delicate. Known results in this direction are Stein's maximal principle, and Swayer's theorem (hence our choice of $L_1(\mathbb{T})$). On the real line, there is a result for certain type of convolution operators (just like the ones we have $T_nf=f*\mu_n$ for compactly supported measures) for which a weak-type (1,1) inequality holds (See Harmonic Analysis, Stein, E., chapter X).

• With these tools at our disposal, we can show that the collection of $f\in L_1$ for which $f_n=\tau_{\alpha_n}f$ does not converge to $f$ point wise almost surely (that is, the $f\in L_1$ for which $m(f_n\stackrel{n}{\nrightarrow} f)>0$) is not empty. For this, it is enough to show that the maximal funciton$M:f\mapsto Mf=\sup_n|\tau_{a_n}f|$ does not satisfy a weak-type (1,1) maximal inequality. This is equivalent to showing that for any $m\in\mathbb{N}$, there is a function $f_m\in L_1(\mathbb{T})$ such that $\|f_m\|_1\leq1$ and $$ \sup_{t>0}t \,m(|Mf_m|>t)\geq m\|f\|_1$$ Without loss of generality, assume $1\geq a_n\searrow0$, i.e. $0<a_{n+1}<a_n\leq1$ for all $n\in\mathbb{N}$ and $a_n\xrightarrow{n\rightarrow\infty}0$

(Due to Anthony Quas:) For each $k$, let $\beta_k>0$ so that $$\beta_k<\min_{1\leq j\leq k}(a_{j-1}-a_j)$$ and define $f_k:=\mathbb{1}_{[0,\beta_k]}$ so that $\|f_k\|_1=\beta_k<1$. Then, the sets $[a_0,\beta_k+a_0],\ldots [a_k,\beta_k+a_k]$ are pairwise disjoint and so $M_kf_k(x)=\max_{1\leq j\leq k}\tau_{a_j}f_k(x)=1$ on $\bigcup^k_{j=0}[a_j,a_j+\beta_k]$; hence $$\sup_{t>0}t m(Mf_k>t)\geq k\|f_k\|_1$$ This means that $$\sup_{\|f\|_1=1}\sup_{t>0}t\, m(Mf>t)=\infty$$ Proposition II implies the existence of $f\in L_1(\mathbb{T})$ for which \eqref{zero} holds.

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Observation: It may be possible to use the functions $f_k$ defined above to construct an explicit function $f\in L_1(\mathbb{T})$ satisfying \eqref{zero} and from this, obtain a bounded measurable function $\phi$ as above.

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Addendum: (Sketch of proof of Prop. I) For $f\in \overline{\mathcal{G}}$, choose $\{\phi_n:n\in\mathbb{N}\}\subset \mathcal{G}$ such that $\|f-\phi_n\|_1\leq \frac1n$. Then $$ |T_kf-f|=|T_k(f-\phi_n)-(f-\phi_n)|+T_k\phi_n-\phi_n|\leq M|f-\phi_n|+|f-\phi_n|+|T\phi_n-\phi_n| $$ Setting $T_*f=\limsup_k|T_kf-f|$, and noticing that $T_*\phi_n=$ for all $n$, we obtain that $$ T_*f\leq M|f-\phi_n|+|f-\phi_n| $$ for all $n$. Then, from \eqref{type11} and Chebyshev-Markov's inequality $$ m(T_*f>2\lambda)\leq m(M|f-\phi_n|>\lambda)+m(|f-\phi_n|>\lambda)\leq\frac{(C+1)}{n\lambda}\xrightarrow{n\rightarrow\infty}0 $$ As a consequence $\lim_nT_nf(x)=f(x)$ for $m$-almost all $x$. In particular, of $\mathcal{G}$ is dense in $L_1$, then for any $f\in L_1(m)$, $T_nf(x)\xrightarrow{n\rightarrow\infty} f(x)$ almost surely.

Answer 3

Say, that a set $A\subset [0,1]$ is uniformly "porous" if to every $\Delta>0$ and every interval $I$ of length $\Delta$ there is $\delta=\delta(\Delta)>0$ so that $I\setminus A$ contains an interval of length $\delta$. In other words, $A$ has "holes" of length $\delta$ that are $\Delta$-dense.

A typical construction of a fat Cantor-set has this uniform porosity property, e.g the set described on the page wiki fat cantor set.

Let $N_\Delta=\lceil \Delta/\delta\rceil$ and let $\tau_a(x)=x+a$ be the translation by $a>0$. The porosity property implies that for every $\Delta>0$: $$ \bigcap_{k=0}^{N_\Delta-1} \tau^k_\delta(A) = \emptyset.$$ Thus, every point in $A$ will at some point go into a hole by one of the translations $\tau_\delta,...,\tau_{(N_\Delta-1)\delta}$. Now, choose (arbitrarily) a sequence $\Delta_n \to 0^+$ and let $\delta_n=\delta(\Delta_n)>0$, $N_n= N_{\Delta_n}$ be the corresponding data for the associated holes. Define $(a_j)_{j\geq 1}$ to be the sequence obtained by juxtaposing the "chunks" $(\delta_n,2\delta_n,...(N_n-1)\delta_n)$ starting from the left with $n=1$. By construction, $a_j\to 0$ as $j\to +\infty$.

Let $f={\bf 1}_A$ be the characteristic function of $A$. Then, for every point $x\in A$, the sequence $(f(x+a_j))_{j\geq 0}$ contains infinitely many zeros, i.e. there is no point $x\in A$ for which the sequence converges to $f(x)=1$. This follow from the fact that arbitrarily far out in the sequence you will find a chunk for which one of the translates will belong to a hole of $A$.