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...

  • $\begingroup$ something does not look right between $L^1$ and $\mathcal L^1$ $\endgroup$ – marmouset Sep 9 '16 at 13:47
  • $\begingroup$ It's $L^1$, not $\mathcal L^1$, that's why i use sequence of $t_n$ and not just $t\to 0$. By using sequences instead of $t\to 0$ the choice of representative for $f$ doesn't change the convergence ae property. Note that the only time i don't use a sequence and use $\tau\to 0$ this is not a problem since i'm talking about norm convergence. $\endgroup$ – Renart Sep 9 '16 at 14:01
  • $\begingroup$ Correct, and beautiful. $\endgroup$ – marmouset Sep 9 '16 at 14:22

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.

| cite | improve this answer | |
  • $\begingroup$ Ah you're right ! It's very close to the "wandering rectangle" sequences that converges in probability but not almost surely in fact. I'm surprised that your solution mimic this wandering rectangle patern, i didn't thought this was possible. Nice answer ! $\endgroup$ – Renart Sep 10 '16 at 17:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.