*Note: This is a refinement of a [previous problem](https://mathoverflow.net/questions/426759/convergence-of-integral-averages-in-l1).*


Let $f \in L^1 (\mathbb R)$. Suppose $g\_n \in L^1 (\mathbb R)$ are a sequence of positive functions.


Define, for each $n$, the function $f\_n$ by


$$f\_n (x) := \frac{1}{2g\_n (x)} \int\_{x - g\_n (x)}^{x + g\_n (x)} f(y) \, dy.$$


**Question:** Is it true that if $g\_n \to 0$ in $L^1$ strong, then $f\_n \to f$ in weak $L^1$ norm?


*Remarks:*


Yuval Peres has shown in the linked post that the implication $g\_n \to 0$ in $L^1$ $\Rightarrow$ $f\_n \to f$ in $L^1$ does not generally hold.


In fact it is possible that none of the $f\_n$ are even in $L^1$. As he points out, this is related to the fact that the Hardy-Littlewood maximal function of $f$ is not in $L^1$, and one can capture the behaviour of the maximal function adequately.



Weak convergence of integral averages
This is true, because we can approximate $f$ in $L^1$ by continuous functions. For them the statement clearly holds, and for the difference we can apply the classical
Hardy-Littlewood maximal inequality. Next, I will add some details.


Define the operator $T\_n$ on functions in $L^1(\mathbb R)$ by $$(T\_nf)(x) := \frac{1}{2g\_n (x)} \int\_{x - g\_n (x)}^{x + g\_n (x)} f(y) \, dy.$$


By [1], $T\_n$ maps $L^1$ to functions with finite weak L^1 norm, see [2] for the definition.


Given $f \in L^1(\mathbb R)$ and $\epsilon>0$, there exists a continuous function $h$ with compact support, $h \in C\_c(\mathbb R)$, such that $\|f-h\|\_1<\epsilon.$ Let $\mu$ be Lebesgue measure. By [1], for all $\lambda>0$,
$$\mu\{x \in \mathbb R: |T\_n(f-h)(x)|>\lambda/2 \} \le \frac{2C\_1 \epsilon}{\lambda} \,, \tag{1}$$ where $C\_1$ is an absolute constant.


Since $h$ is uniformly continuous and $g\_n \to 0$ in probability, the bounded convergence theorem gives that
$\|T\_n h-h\|\_1 \to 0$ as $n \to \infty$. Thus for all $n>n\_\epsilon$, we have
$\|T\_n h-h\|\_1<\epsilon$, so
$$\mu\{x \in \mathbb R: |T\_n(h)(x)|>\lambda/2 \} \le \frac{2\epsilon} {\lambda}\,. \tag{2}$$


By (1) and (2), for all $n>n\_\epsilon$,
$$\mu\{x \in \mathbb R: |T\_n(f)(x)|>\lambda \} \le \frac{(2+2C\_1)\epsilon} {\lambda}\,,\tag{2}$$
as needed.


[1] <https://en.wikipedia.org/wiki/Hardy%E2%80%93Littlewood_maximal_function>


[2] <https://en.wikipedia.org/wiki/Lp_space#Weak_Lp>


