Let $f\in L^{\infty}(\mathbb{R}^{n})$ be supported in the unit ball and have mean zero. Let $\phi\in L^{1}(\mathbb{R}^{n})\cap C^{\alpha}(\mathbb{R}^{n})$ be a Holder continuous function with exponent $\alpha>0$. Define an approximation to the identity $\left\{\phi_{t}\right\}_{t>0}$ by $\phi_{t}:=t^{-n}\phi(\cdot/t)$, and consider the maximal convolution operator $M_{\phi}$ defined by $$M_{\phi}(f)(x)=\sup_{t>0}\left|(f\ast\phi_{t})(x)\right|, \qquad\forall x\in\mathbb{R}^{n} \tag{1}$$
If $\phi$ is compactly supported in a ball about the origin $B_{A}(0)$ of radius $A>0$, then it is not hard to show that $M_{\phi}f\in L^{1}(\mathbb{R}^{n})$. Indeed, for $\left|x\right|\leq 2$, we have from Young's inequality
$$M_{\phi}(f)(x)\leq\sup_{t>0}\left\|f\right\|_{L^{\infty}}\left\|\phi_{t}\right\|_{L^{1}}=\left\|f\right\|_{L^{\infty}}\left\|\phi\right\|_{L^{1}} \tag{2}$$
Now consider case $\left|x\right|>2$. Using $\int f=0$, we may write \begin{align*} (f\ast\phi_{t})(x)=t^{-n}\int_{\mathbb{R}^{n}}f(y)\left[\phi\left(\dfrac{x-y}{t}\right)-\phi\left(\dfrac{x}{t}\right)\right]\mathrm{d}y \tag{3} \end{align*} for any $t>0$. Since $\left|x\right|>2$ and $\left|y\right|\leq 1$ (by the support hypothesis for $f$), $\left|x-y\right|\geq\left|x\right|/2$ by the reverse triangle inequality. Since $\phi((x-y)/t)=0$, unless $\left|x-y\right|/t\leq A$, we must have $t\geq\left|x\right|/(2A)$. Whence by Holder continuity, $$t^{-n}\left|\phi\left(\dfrac{x-y}{t}\right)-\phi\left(\dfrac{x}{t}\right)\right|\leq\left\|\phi\right\|_{C^{\alpha}}t^{-n-\alpha}\leq\dfrac{\left\|\phi\right\|_{C^{\alpha}}}{(2A)^{\alpha}}\left|x\right|^{-n-\alpha}\cdot\left|y\right|^{\alpha}$$
Substituting this result into (3), we obtain that \begin{align*} \left|(f\ast\phi_{t})(x)\right|\leq \left|x\right|^{-n-\alpha}\dfrac{\left\|\phi\right\|_{C^{\alpha}}}{(2A)^{\alpha}}\int_{\mathbb{R}^{n}}\left|f(y)\right|\left|y\right|^{\alpha}\mathrm{d}y\lesssim_{\phi}\left\|f\right\|_{L^{1}}\left|x\right|^{-n-\alpha}, \qquad\forall t>0 \tag{4} \end{align*} which is integrable on $\mathbb{R}^{n}$ away from the origin.
I am interested in how much we can relax the decay hypothesis on $\phi$ (i.e. compact support) and still obtain $M_{\phi}f\in L^{1}(\mathbb{R}^{n})$. Motivated by this related question here, where a commenter suggests such an $f$ above will work for an arbitrary Holder continuous, integrable function $\phi$, I ask:
Question: If $\phi\in C^{\alpha}(\mathbb{R}^{n})\cap L^{1}(\mathbb{R}^{n})$, then for $f$ as above, does it hold that $M_{\phi}(f)\in L^{1}(\mathbb{R}^{n})$? If not, does there exist a function $g\in L^{1}(\mathbb{R}^{n})$ such that $M_{\phi}(g)\in L^{1}(\mathbb{R}^{n})$?
For example, if we have that $\phi\in C^{1}(\mathbb{R}^{n})\cap L^{1}(\mathbb{R}^{n})$ and $\left|\nabla\phi\right|$ satisfies the decay conditions
$$\left|\nabla\phi(x)\right|\leq C(1+\left|x\right|)^{-n-\delta},\qquad\forall x\in\mathbb{R}^{n} \tag{5}$$
for some $\delta>0$ fixed, then I believe the answer is affirmative.
Indeed, for notational convenience, assume $\delta=1$. Fix $\left|x\right|>2$, and partition the set $\left\{t>0, \left|y\right|<1\right\}$ into disjoint annuli $$\mathcal{A}_{0}:=\left\{0<\left|\dfrac{x-y}{t}\right|\leq A\right\}\cup\mathcal{A}_{j}:=\left\{2^{j-1}A<\left|\dfrac{x-y}{t}\right|\leq 2^{j}A\right\}, \qquad j\geq 1$$ We have already dealt with the case of the first annulus. If $(y,t)\in\mathcal{A}_{j}$, then $$\dfrac{\left|x\right|}{2t}\leq 2^{j}A, \quad 2^{j-1}A<\dfrac{2\left|x\right|}{t}$$ By the mean value theorem, we have \begin{align*} t^{-n}\left|\phi\left(\dfrac{x-y}{t}\right)-\phi\left(\dfrac{x}{t}\right)\right|\leq t^{-n}\left|\nabla\phi(\bar{x})\right|\dfrac{\left|y\right|}{t}\leq \left|\nabla\phi(\bar{x})\right|\dfrac{(2^{j+1}A)^{n+1}}{\left|x\right|^{n+1}}, \tag{6} \end{align*} where $\bar{x}=\theta\frac{x-y}{t}+(1-\theta)\frac{x}{t}$ for some $\theta\in (0,1)$. Since $\left|\bar{x}\right|\geq\left|x\right|/2t>2^{j-2}A$, the decay condition for $\left|\nabla\phi\right|$ implies $$\left|\nabla\phi(\bar{x})\right|\leq C\dfrac{1}{(1+(\left|x\right|/2t))^{n+1}}\leq \dfrac{C}{(2^{j-2}A)^{n+1}} \tag{7}$$
Putting these results together, we see that for $(y,t)\in\mathcal{A}_{j}$
\begin{align*} \left|(f\ast\phi_{t})(x)\right|\leq\int_{\mathbb{R}^{n}}\left|f(y)\right|\dfrac{C(2^{j+1}A)^{n+1}}{(2^{j-2}A)^{n+1}\left|x\right|^{n+1}}\mathrm{d}y=\dfrac{C(8A)^{n+1}\left\|f\right\|_{L^{1}(\mathbb{R}^{n})}}{\left|x\right|^{n+1}} \tag{8} \end{align*}
The constants are independent of $x$ and $\mathcal{A}_{j}$ and only depend on $\phi$. Since $\left|x\right|^{-n-1}$ is integrable on $\mathbb{R}^{n}$, we conclude that $M_{\phi}(f)\in L^{1}(\mathbb{R}^{n})$.
Without some integrability condition for $\nabla\phi$, I don't know how to control cases where $\left|x\right|/t$ is very large, so the above argument doesn't seem to suffice to answer my question, as it is written, affirmatively.
Three examples.
Example 1
No, $M_\phi f$ need not be in $L^1$. In fact if $n=1$ it can die arbitrarily slowly at infinity; it need not be in weak $L^1$ either. Even if $\phi$ is infinitely differentiable with all derivatives vanishing at infinity. (We can throw in $\int\phi=0$ too if you want.)
I'm going to take $n=1$ and $$f=\chi_{[0,1]}-\chi_{[-1,0)}.$$
True Fact: If $h(x)\to0$ as $x\to\infty$ there exists $\phi\in C^\infty(\mathbb R)$ with all derivatives vanishing at infinity and $\int\phi=0$ such that $M_\phi f(x)\ge h(x)$ ($x\ge 10$).
Proof, leaving details to you: Let $\psi$ be infinitely differentiable with support in $[-1,1]$. Say $\psi$ is odd, and assume $\psi\ge0$ in $[-1,0]$.
Let $$\phi(x)=\sum_{n=1}^\infty c_n\psi(x-r_n),$$where $c_n\ge0$, $\sum c_n<\infty$, and $(r_n)$ is a sequence of positive numbers increasing to infinity. We'll be specifying the $r_n$ at the end; for now note that one constraint is $$r_{n+1}\ge 2r_n.$$
Now suppose $$10\le x\le r_n$$and set $$\delta_n=x/r_n.$$It follows that $$f*\phi_{\delta_n}(x)=c_n||\psi||_1.$$You can work that out as well as I can, unless it's not quite true, in which case there's an off by one error or a dropped minus sign in my scribblings; the picture is clear enough to me that I know any errors can be fixed. Two things that will be relevant in the calculation: First, since $\delta_n\le1$, $$(-\delta_n,\delta_n)\subset[-1,1].$$ Second, the fact that $r_{k+1}\ge 2r_k$ shows that $$[-1,1]\cap\left( x\left(1-\frac{r_m}{r_n}\right)-\delta_n,\,x\left(1-\frac{r_m}{r_n}\right)+\delta_n\right)=\emptyset\quad(m\ne n).$$
So we have $$M_\phi f\ge||\psi||_1\sup_n c_n\chi_{[10,r_n]}.$$ Given $(c_n)$ it's easy to see that there exist $r_n$ such that this is larger than $h$ on $[10,\infty)$.
Or, if $\alpha>0$ and we set $r_n=2^n$, $c_n=r_n^{-\alpha}$ we obtain $|\phi(x)|\le c(1+|x|)^{-\alpha}$ and $M_\phi f(x)\ge c(1+|x|)^{-\alpha}$ for $x>10$. (Modifying $\phi$ to be even would give the last inequality for all $x$.)
Now what about higher dimensions? I'm not going to give an example in $\mathbb R^n$; the letter $n$ is too useful to waste. Instead I'll give examples in $\mathbb R^d$. (It's not hard to show that $\mathbb R^n$ and $\mathbb R^d$ are isomorphic if $d=n$.)
Example 2
A construction like $$\phi(x)=\sum_{j=1}^\infty c_j\psi(|x|-r_j)$$ does everything you want in $\mathbb R^d$. Things come out particularly nice if you allow $\mathbb R^d$-valued functions, taking $f(t)=t/|t|$ for $|t|<1$.
I'm not going to prove this. Instead I'm going to give another example, which is just as simple to work out as Example 1 was, but which I claim has the additional property that if you "understand" how Example 3 works you will then "see" that Example 2 also works. (Comments on the connection at the very end.)
Example 3
Points in $\mathbb R^d$ will often be written $x=(x_1,x')$, where $x_1\in\mathbb R$ and $x'\in\mathbb R^{d-1}$. Similarly $B'(x',r)$ denotes a ball in $\mathbb R^{d-1}$, etc.
Let $K$ be the set $|x_1|\le 1$, $|x'|\le 1$: $$K=[-1,1]\times B'(0,1).$$ Take $f$ vanishing outside $K$; for $x\in K$ let $f(x)=1$ for $x_1>0$, $f(x)=-1$ for $x_1<0$.
Let $\psi\ge0$ be a smooth function supported in $[-1,0]$.
Note this is not quite the same as in Example 1.
(Comment that will make no sense to readers who haven't worked out the details for Example 1: In the first example we took $f$ and $\psi$ both odd, and then the estimate on the integral used the facts that $(-1)^2>0$ and $1^2>0$. That was cute but a little silly. We could set Example 3 up that way, but then my claim that Example 2 works the same as Example 3 would be problematic. Instead in the relevant estimate in Example 3 we will use $(-1)(0)\ge0$ and $1^2>0$.)
Let $\chi$ be a smooth function with $$\chi_{[-1,1]}\le\chi\le\chi_{[-2,2]}.$$
Given sequences $c_j$ and $r_j>0$ with $r_{j+1}\ge2r_j$, set $$\phi(x)=\sum_{j=1}^\infty c_j\psi(x_1-r_j)\chi\left(\frac{|x'|}{r_j}\right).$$
(For your immediate purposes there's no reason not to take $r_j=2^j$; I'm leaving the more general version for future applications.)
Define a truncated sector $S$ by $$S=\{x\in\mathbb R^d\,:\,x_1>10,\,\,|x'|\le x_1/2\}.$$
Now suppose $x\in S$ and $x_1\le r_n$. Let $\delta=x_1/r_n\le1$. As in Example 1 the condition $r_{j+1}\ge2r_j$ shows that only one term in the definition of $\phi$ matters, and we get $$f*\phi_\delta(x)=\delta^{-d}c_n\int_Kf(y)\,\psi\left(\frac{-y_1}\delta\right)\chi\left(\frac{|x'-y'|}{r_n\delta}\right)\,dy.$$Now the conditions we put on $\psi$ show that the integrand vanishes except at points where $f$ and $\chi$ equal $1$, so$$f*\phi_\delta(x)=\delta^{-d}c_n\int_K\psi\left(\frac{-y_1}\delta\right)\,dy=C\delta^{-d+1}c_n\ge C\,c_n.$$
And there you are.
Now about Example 2 being really the same. In Example 3 we had $\delta\le1$, and we essentially ended up with the integral of $c_n$ over the set $[0,\delta]\times B'(0,1)$. We could have set it up slightly differently, getting $0<\delta\le1/4$, and getting an integral over the set $$R=[1/2-\delta,1/2+\delta]\times B'(0,1).$$Imagine a picture of $R$ in the $x_1,|x'|$ plane: $R$ is a "vertical" rectangle. Take that rectangle and curve it just a little bit, so instead of a rectangle it becomes a part of an annulus with center at a point $(x_1,0)$ where $x_1$ is large.
The new $R$ looks the same as the old $R$ if you take your glasses off. Nothing essential has changed, we get the same sort of estimate. And this shows that Example 2 works. (Ok, there are other differences, in the definition of $\phi$. But none of the differences really matter. QED.)