Convergence in norm but not almost everywhere of $f_{t_n}=f(\cdot-t_n)$ to $f$

Question

My question is : is there a $L^1(\mathbf R)$ function $f$ and a sequence $(t_n)_n$ of reals with limit $0$ such that $f_{t_n}=f(\cdot-t_n)$ does not converges to $f$ a.e. ?

There are several variations of this questions which are pretty much equivalent : one can drop the $L^1$ hypothesis and just assume mesurability of $f$, one could also assume that $f$ is the indicator function of a measurable subset of $[0;1]$.

What i already know :

The first thing to say is that $f_\tau$ converges to $f$ in $L^1$ norm when $\tau$ goes to zero. One way to prove that is to say that this is true for compactly suported continuous functions (because of absolute continuity) and conclude by density.

Since the $f_{t_n}$ are converging in $L^1$ norm there exists a subsequence $t_{\sigma(n)}$ such that $f_{t_{\sigma(n)}}$ converges almost everywhere to $f$. However one cannot directly use this result to conclude with the convergence a.e. of $f_{t_n}$, indeed there are many examples of sequences of functions that are converging in $L^1$ norm but not convergent a.e. In fact this show that the a.e. convergence does not corespond to any topological convergence, since the caracterisation of convergence with sub-sub-sequences doesn't apply here.

Another thing to note is that if $f$ is continuous at $x$ then $f_{t_n}(x)\to f(x)$ automatically, so if we search some $f$ such that $f_{t_n}$ does not converges a.e. to $f$ it has to be a function discontinuous on a set with positive measure. In fact for every function $g=0$ a.e. one must have that $f+g$ is discontinuous on a set with positive measure.

I believe the answer to my question is yes, but it's mostly an intuition, i don't even have euristic arguments for that. So i tried to find an example, the two kind of $f$ i have thought of are :

$f=\chi_{C}$ where $C$ is a fat cantor set or something like this kind of set and

$$f(x)=\chi_{[0;1]}\sum_{k=0}^\infty \frac{2^{-k}}{|x-q_k|^{\alpha}}$$ where $q_n$ is an enumeration of the rationals of $[0;1]$.

The problem is that it's hard to find the behavior of $|f_{t_n}(x)-f(x)|$ for general $x$ and such patological functions...

Answer 1

Your idea of a fat Cantor set works too. I'll repeat a construction I used in this answer.

Let $C$ be a compact nowhere dense set of positive measure. We can construct a sequence $t_n \to 0$ such that for every $x \in C$, there are infinitely many $n$ for which $x - t_n \notin C$. In particular we have $\liminf_{n \to \infty} \chi_C(x-t_n) = 0$ and therefore $\chi_C(x-t_n) \not\to 1 = \chi_C(x)$. Since $C$ has positive measure, we therefore do not have $\chi_C(\cdot-t_n) \to \chi_C$ a.e.

To construct the sequence, let $U = C^c$ and note that $U$ is open and dense. So for each integer $m > 0$ and each $x \in C$, there exists $u \in U$ such that $|x-u| < 1/m$, which is to say that $x \in U+(x-u)$. In other words, the collection of open sets $\{U+s : |s| < 1/m\}$ cover $C$ (indeed they cover $\mathbb{R}$). Since $C$ is compact, there is a finite subcover $\{U+s_{m,1},\dots, U+s_{m, k_m}\}$. That is to say, for every $x \in C$ there is some $1 \le i \le k_m$ for which $x - s_{m, i} \in U = C^c$.

Now let $(t_n)$ be the sequence $$(s_{1,1}, \dots, s_{1, k_1}, s_{2,1}, \dots, s_{2, k_2}, \dots).$$ Since $|s_{i,m}| \le 1/m$ for every $m$, we have $t_n \to 0$, and there are infinitely many $t_n$ such that $x-t_n \notin C$ (namely, at least one of the $s_{m,i}$ for each $m$).

Answer 2

Yes. Consider $f(x)=\sum 2^{-n}g(x-q_n)$, where $q_n$ is an enumeration of $\mathbb Q$, and $g\in L^1$ is an unbounded non-negative bump function such as $g(x)=\chi_{(-1,1)}(x)|x|^{-1/2}$.

I'm now going to produce a sequence $t_n\to 0$ for which $f(x-t_n)$ is unbounded for every $x\in [-1,1]$. My first bunch of $t_n$'s will satisfy $|t_n|\le 1$. Start out by locating a $q_n$ close to $-1$ ($q_n=-1$ would work nicely). The corresponding bump $2^{-n}g(x-q_n)$ will be $\ge 1$, say, on an interval centered at $q_n$. By shifting this bump around, I can make sure that $f(x-t_n)\ge 1$ for every $x\in [-1,0]$ for some $n$. Cover the remainder of $[-1,1]$ in the same way.

Let's summarize: I have chosen finitely many elements $t_1, \ldots , t_N$ of my sequence, in such a way that $f(x-t_n)\ge 1$ for every $x\in [-1,1]$ for some $1\le n\le N$. Moreover, these satisfy $|t_n|\le 1$.

Now just continue in this way: In the second step, find the next batch of $t$'s, such that $f(x-t_n)\ge 2$ for all $x\in [-1,1]$ for some $n$, and $|t_n|\le 1/2$ for these $t_n$ etc.

Answer 3

Here is a different solution to a slightly general problem that says that for any sequence $t_n\searrow 0$, and any $p\geq1$, there is a function $f\in L_p([0,1],m)$ ($m$ is Lebesgue measure) such that $$m\big(\{x\in I:f(x-t_n)\stackrel{n}{\nrightarrow}f(x)\})>0$$

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The solution is based on a nice result by Stein that considers linear operators on $L_p$ of the form $$ T_nf=f*\mu_n$$ where $\mu_n$ is a sequence of finite measure supported on a compact set $K$ ($[0,1]$ for example). It is easy to check that $\|T_nf\|_p\leq\mu_n(K)\|f\|_p$ for any $f\in L_p(\mathbb{R})$. Define the maximal function $$ Mf(x):=\sup_n|T_nf(x)|$$ The result states the following:

Theorem (Stein). If $m\big(x:Mf(x)<\infty\big)>0 $ for all $f\in L_p(\mathbb{R},m)$, then there exists a constant $A>0$ such that $$ \begin{align}m\big(\{x:Mf(x)>\lambda\}\big)\leq\frac{A}{\lambda^p}\|f\|^p_p\tag{1}\label{one}\end{align}$$ for all $f\in L_p(\mathbb{R})$

Observation: We stated Stein's result as real line result here for convenience, but as we will see in the explanation of the result, one can restrict to $L_p([0,1],m)$; moreover, the result is valid for $L(\mathbb{R}^d)$.

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Solution to the OP: Without loss of generality, assume that $0<t_n<1$. Then $$ T_nf(x)=f(x-t_n)=f*\delta_{t_n}(x)$$ where $\delta_{x}$ is Dirac's measure concentrated at $x$, i.e. $\delta_x(A)=\mathbb{1}_A(x)$ for any Borel set $A$. Notice that these are of the type of operators in Stein's result. If it were the case that for any $f\in L_1([0,1])$, $\lim_nf(x-t_n)=f(x)$ a.s., then $Mf(x)=\sup_n|f(x-t_n)|<\infty$ for almost surely all $x\in [0,1]$ and so, $Mf$ would satisfy an inequality of the form \eqref{one}. We now show that this is actually not the case. For each $k\in\mathbb{N}$ let $$\delta_k<\min_{1\leq j\leq k}(t_j-t_{j-1})$$ Let $f_k=\mathbb{1}_{[0,\delta_k]}$. The choice of $\delta_k$ implies that the intervals $[0,\delta_k]+t_j$, $1\leq j\leq k$ are disjoint. Also, notice that $ \max_{1\leq j\leq k}f_k(x-t_j)=1$ for $x\in \bigcup^k_{j=1}(\big[0,\delta_k]+t_j\big)$. Thus, $$ m\big(x\in[0,1]: Mf_k(x)>\tfrac12\big)\geq k\delta_n=\frac{k}{2}\frac{\|f_k\|_1}{\tfrac12} $$ Hence $\sup_{\lambda>0}\frac{\lambda}{\|f_k\|_1}\,m\big(x\in[0,1]: Mf_k(x)>\lambda)\geq \frac{k}{2}\xrightarrow{k\rightarrow\infty}\infty$. This shows that $M$ does not satisfy \eqref{one}. Consequently, there exists $f\in L_1[0,1]$ for which $m\big(x\in[0,1]:Mf(x)=\infty\big)>0$. For any $u\in\{x\in[0,1]:Mf(x)=\infty\}$, $f(u-t_n)\stackrel{n}{\nrightarrow}f(u)$.

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The rest of this answer is dedicated to sketch a proof of Stein's theorem. First we state a result similar to Borel Cantelli's lemma:

Lemma 1: Suppose $(E_n:n\in\mathbb{N})$ is a sequence of measurable sets in $Q=[0,1]^d$ such that $\sum_nm(E_n)=\infty$. Then, there exists translations $F_n=E_n+x_n$ such that $$\limsup_nF_n=\bigcap_n\bigcup_{m\geq n}F_m =\mathbb{R^d}\qquad \text{a.s.}$$

Idea of proof of Lemma: This is based in the following geometric. If $A_1,\,A_2\subset Q$, then there is $h\in\mathbb{R}^d$ such that $m\big(A_1\cap(A_2-h)\big)\geq 2^{-d}m(A_1)m(A_2)$. This follows from the continuity of $g(x):=\big(\mathbb{1}_{A_1}*\mathbb{1}_{A_2}\big)(-x)$ and the fact that $\operatorname{supp}(g)\subset[-2,2]^d=Q'$. For $g$ attains in maximum in $Q'$ as some point, say $h$, and so $$m(A_1\cap(A_2-h)=g(h)\geq\frac{1}{m(Q')}\int_{Q'}g(x)\,dx=2^{-d}m(A_1)(A_2)$$ We show that there is a sequence $F_n=E_n+x_n$ of translates such that $$\bigcup_nF_n=Q\qquad\text{a.s.}$$ The construction is done induction: $F_1=E_1$. If $F_1,\ldots,F_{j-1}$ have been constructed, then let $A_1=Q\cap(F_1\cap\ldots\cap F_{j-1})$ and $A_2=E_j$. Let $h\in\mathbb{R}^d$ be such that $m(A_1\cap(A_2-h))\geq 2^{-d}m(A_1)m(A_2)$. Set $F_j=A_1-h=E_j-h$. Analysis of the proportions $p_j=m(Q\cap(F_1\cap\ldots\cap F_j))$ shows that $p_j-p_{j-1}\geq 2^{-d}(1-p_{j-1})m(E_j)$ from where to follows that $\lim_jp_j=1$.

By decomposing the sequence $E_n$ into countably infinite sub collections, each of which members have measures adding to infinity, and applying the argument above to appropriate translates of the cube $Q$, the conclusion of the Lemma follows.

Sketch of proof of Theorem: Assume that $\operatorname{supp}(\mu_n)\subset K$. and let $B$ a ball around the origin the contains the support of $\mathbb{1}_Q*\mathbb{1}_K$. Then for any $f\in L_p(\mathbb{R}^d)$ with support in $Q$, $\operatorname{supp}(f*\mu_n)\subset B$. and the same holds for $Mf$. The conclusion of the theorem is first proved for $L_p$-functions with support $Q$ (this is enough for the OP).

Suppose \eqref{one} does not hold. Then, for each $k\in\mathbb{N}$ there exist $g_k\in L_p(Q)$ and $\alpha_k>0$ such that $$ m\big(x\in B: Mg_k(x)>\alpha_k\big)\geq 2^k\frac{\|g_k\|^p_p}{\alpha^p_k}$$ Setting $g'_k=\frac{k}{\alpha_k}g_k$, we have that $$m\big(x\in B: Mg'_k(x)>k\big)\geq \frac{2^k}{k^p}\|g'_k\|^p_p$$ For each $k$, let $n_k$ the smallest integer such that $$n_k\frac{2^k}{k^p}\|g'_k\|^p_p\geq1$$ Then $(n_k-1)\|g'_k\|^p_p <\frac{k^p}{2^k}$ and so $$\sum_kn_k\|g'_k\|^p_p<\infty$$ This allows us to obtain a sequence $\{f_k\}\subset L_p(\mathbb{R}^d)$ with support in $Q$ (repeating $g'_k$ $n_k$ times) and a sequence $R_k\xrightarrow{k\rightarrow\infty}\infty$ such that with $E_k=\{x\in B: Mf_k(x)>R_k\}$,

1. $\sum_km(E_k)=\infty$
2. $\sum_k\|f_k\|^p_p <\infty$.

By Lemma 1, there are translates $F_j=E_j+x_j$ such that $\limsup_j F_j=\mathbb{R}^d$ $m$-a.s. Let $\widetilde{f}_j(x)=f(x-x_j)$ and $F(x)=\sup_j\widetilde{f}_k(x)$. Since the operators $T_n$ are positive ($T_nf\geq0$ whenever $f_n\geq0$), $MF(x)\geq\sup_k M(\widetilde{f}_k)$. Notice that $ M(\widetilde{f}_k)(x)\geq R_k$ for all $F_k$; hence $MF(x)=\infty$ $m$-a.s. On the other hand, $F^p\leq\sum_k\widetilde{f}^p_k$ and so $$\|F\|^p_p\leq\sum_p\|f_k\|^p_p<\infty$$ This contradicts the assumption that $m(Mf<\infty)>0$ for all $f\in L_p(\mathbb{R})$. Therefore \eqref{one} holds for all $f\in L_p(\mathbb{R}^d)$ supported in $Q$.

To extend the result to any $f\in L_p(\mathbb{R}^d)$ consider translates $Q_j=Q+j$, $j\in\mathbb{Z}^d$ and decompose $$f=\sum_{j\in\mathbb{Z}^d}f\mathbb{1}_{Q_j}$$ Then $Mf\leq \sum_jMf_j$, and $Mf_j$ is supported in $B_j=B+j$. Any point in $\mathbb{R}^d$ is in at most $N_d$ translates of the form $B+j$ ($N_d$ is a constant depending on $d$ and on the radius of $B$). Applying the result for functions supported on $Q$ we have $$ m\big(x\in B_j: Mf_j(x)>\lambda\big)\leq\frac{A}{\lambda^p}\|f_j\|^p_p$$ Putting things together, $$\begin{align} m\big(x: Mf(x)>\lambda\big)&\leq\frac{A N_d}{(\lambda/N_d)^p}\sum_j\|f_j\|^p_p=\frac{AN^{p+1}_d}{\lambda^p}\|f\|^p_p\end{align}$$

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More details and generalizations of results of these type can be count in Stein's paper

1. E. Stein. On limits of sequence of operators, Annals of Mathematics, 1961. Vol. 74, pp. 140-170.
2. S. A. Sawyer. Maximal inequalities of weak type, Annals of Mathematics, 1966, Vol 84, pp. 157-174