Integrability of the maximum with respect to a parameter Let $f:\mathbb R^2\to \mathbb R$ be a smooth function that belongs to $L^1(\mathbb R^2)$.
I would like to have the following property:
$$ x\mapsto\max_{|y-y_0|\leq r} \Big|f(x,y)\Big| \ \in L^1(\mathbb R)$$
for every $y_0\in \mathbb R$ and some $r=r(y_0)>0$ arbitrary small.
How can I check it? Can it follow from the continuity and integrability of $f$?
Another strategy: is it possible to bound $f$ by a product
$$ |f(x,y)| \leq |f_1(x)|\, |f_2(y)|$$
with $f_1\in L^1(\mathbb R)$ and $f_2$ continuous? The desired property would follow easily.
 A: Consider the function
$$f(x,y)= \frac {\sin(x)}{x}e^{\lvert -y \rvert}$$
This function is smooth and lebesgue integrable over $\mathbb{R}^2$ but $|f(x,y)|$ is not lebesgue integrable. (All of this stems from $\frac {\sin(x)}{x}$). Clearly
$$ \max_{|y-y_0|\leq r} \Big|f(x,y)\Big| \geq \Big|f(x,y)\Big|$$ Therefore $x\mapsto\max_{|y-y_0|\leq r} \Big|f(x,y)\Big| \ \notin L^1(\mathbb R)$.
Edit: This answer is wrong i confuse my definitions. I thought $f$ is lebesgue integrable is $\int f(x)dx<\infty $ while $f\in L^1(\mathbb{R})$ was $\int |f(x)|dx<\infty $ but apparently they were the same.
New answer
Consider the function $f(r,\theta)$ (Which can be written as $f(x,y)$) given by,
$$\large{f(r,\theta)= \begin{cases} r\frac{\theta-\pi/2}{e^\theta} \ \normalsize\textrm{if} & 0 \leq r\leq 1 \ \normalsize\mathrm{and} \ 0\leq\theta\leq\pi/2 \\ 
{\frac{\theta-\pi/2}{e^{r\theta}r}} \ \normalsize\textrm{if} & r\geq 1 \ \normalsize\mathrm{and} \ 0\leq\theta\leq\pi/2 \\
r\frac{-\theta-\pi/2}{e^{-\theta}} \ \normalsize\textrm{if} & 0 \leq r\leq 1 \ \normalsize\mathrm{and} \ 0\geq\theta\geq-\pi/2\\
{\frac{-\theta-\pi/2}{e^{-r\theta}r}} \ \normalsize\textrm{if} & r\geq 1 \ \normalsize\mathrm{and} \ 0\geq\theta\geq-\pi/2 \\
0 \ \normalsize\textrm{if} & -\pi/2 \geq \theta\geq \pi/2 \\
\end{cases}} $$
Where $r\in [0,\infty)$ and $\theta \in (-\pi,\pi]$. So function is defined everywhere and it is smooth,integrable etc. For example when $x\leq 0$, our function is $0$.
As you can see this function is integrable.
$$ \int _0^\infty \int_0^{\pi/2} \frac{\theta-\pi/2}{e^{r\theta}r} d\theta dr<\infty $$(You can check this from wolfram if you want.) So we have
$$\int_{\mathbb{R}^2}f(r,\theta)drd\theta<\infty$$ Now if we choose $y_0=0$ you can see that $\max_{|y-0|\leq c}|f(x,y)|=\frac{1}{x}$ for any $c>0$ when $x>1$. (This $ c$ is your $r$.). But as we know
$$ \int_1^\infty \frac{1}{x}dx=\infty$$ and
$$ \int_1^\infty \frac{1}{x}dx \leq \int_{-\infty}^{\infty} \max_{|y-0|\leq c}|f(x,y)|dx$$
