Is the truncated Hardy-Littlewood maximal function of an $L^1$ function also in $L^1$? It is well known that the Hardy-Littlewood maximal function of a non-null $L^1(\mathbb{R})$ function doesn't belong to $L^1(\mathbb{R})$. What about the truncated Hardy-Littlewood maximal function? I.e., if $f\in L^1_{loc}(\mathbb{R})$ and $r>0$, define the $r$-truncated Hardy-Littlewood maximal function of $f$ as: $$M^*_r(f):\mathbb{R}\rightarrow[0,+\infty], x\mapsto \sup_{a\in(0,r)} \frac{1}{2a}\int_{[x-a,x+a]} |f(t)|\operatorname{d}t$$
Question 1: is it true that $M^*_r(f)\in L^1 _{loc}(\mathbb{R})?$
Question 2: if $f\in L^1(\mathbb{R})$, is it true that $M^*_r(f)\in L^1 (\mathbb{R})?$
My guess is that the answer to the first question is not true in general (and then also the answer to the second question is no in general). Anyway I have a little trouble in working out a counter-example.
 A: If $\delta$ is a delta "function" at the origin then $M_1\delta(x)$ is like $1/|x|$ near the origin, not locally integrable. Hence $L^1$ functions that approximate $\delta$ show that there does not exist $c$ such that $$\int_0^1 M_1f\le c||f||_1.$$Not that I've worked out the details, but "surely" you can concoct an infinite linear combination of translates of functions showing that inequality fails to show that $M_1f$ need not be integrable for $f\in L^1$.
(If that's what you've been trying to do my apologies - all you told us was you were having trouble with a counterexample...)
Edit: The OP says he got it. For the benefit of anyone who hasn't got it yet: If $M_1$ were linear I'd claim were done, but it's not. Took me a minute...
For $0<\delta<1$ let $$f_\delta=\frac1\delta\chi_{(-\delta,0)}$$and $$I_\delta=(\delta,\delta^{1/2}).$$If I did the calculations correctly then $$\lim_{\delta\to0}\int_{I_\delta}M_1f_\delta=\infty.$$ One should really verify that. Assuming that's right:
Choose $\delta_k\to0$ so that the $I_{\delta_k}$ are disjoint. Let $$\alpha_k=\int_{I_{\delta_k}}M_1f_{\delta_k}.$$Since $\alpha_k\to\infty$ there exists $c_k>0$ with $\sum c_k<\infty$ and $\sum c_k\alpha_k=\infty$. Let $f=\sum c_kf_{\delta_k}$. Then $$\int_0^1M_1f\ge\sum_k c_k\int_{I_{\delta_k}}M_1f_{\delta_k}=\infty.$$
Edit${}^2$: I hate the way the above depends on the explicit calculation of $\int_{I_\delta}M_1f_\delta$. Here's a softer version, that can't possibly be wrong:
Choose a sequence of disjoint closed intervals $I_k\subset(0,1)$ with $I_k\to0$ such that $$\lim_k\int_{I_k}\frac{dx}x=\infty.$$It's clear that there exists $f_k\ge0$, supported on $(-1,0)$, with $\int f_k=1$ and $$M_1f_k(x)\ge\frac1{2x}\quad(x\in I_k).$$Now let $f=\sum c_kf_k$ as above...
