Absolute convergence: multiplying $2$ functions How to multiply $f(x)$ and $g(x)$, so that the the integral $\int_{-\infty}^{+\infty} f(x) \cdot g(x) dx$ is absolutely convergent?
Other than zero-ing out numbers, like this:
$$f(x) = 
\begin{cases} 
1 & \mbox{ for } |x| \le \tfrac 12  \\
\\
 0 & \mbox{ for } |x| > \tfrac 12 
\end{cases}$$
$$g(x) = 
\begin{cases} 
1 & \mbox{ for } |x| \le \tfrac 12  \\
\\
 0 & \mbox{ for } |x| > \tfrac 12 
\end{cases}$$
 A: Short answer: convolution. Now let us give some details.
Let $L^1(\mathbb{R})$, or $L^1$ for short, denote the space of integrable functions over $\mathbb{R}$. That's what you call absolutely convergent. But I think it is slightly more common to reserve this terminology for series, and to talk about integrability for general functions. Note that $L^1$ is a particular case of the family of Banach spaces called $L^p$ spaces.
Pointwise product: if $f,g\in L^1$, the pointwise product $(fg)(x):=f(x)g(x)$ is not necessarily in $L^1$. For example, consider the infinite step function
$$
f=\sum_{n\geq 1}n\cdot 1_{[n,n+1/n^3]}\qquad \Rightarrow\qquad  f^2=\sum_{n\geq 1}n^2\cdot 1_{[n,n+1/n^3]}.
$$
Then
$$
\int_{-\infty}^{+\infty}f(x)dx=\sum_{n=1}^{+\infty}\int_n^{n+\frac{1}{n^3}}ndx=\sum_{n=1}^{+\infty}n\cdot\frac{1}{n^3}=\sum_{n=1}^{+\infty}\frac{1}{n^2}=\frac{\pi^2}{6}
$$
while
$$
\int_{-\infty}^{+\infty}f^2(x)dx=\sum_{n=1}^{+\infty}\int_n^{n+\frac{1}{n^3}}n^2dx=\sum_{n=1}^{+\infty}n^2\cdot\frac{1}{n^3}=\sum_{n=1}^{+\infty}\frac{1}{n}=+\infty.
$$
So $f$ is in $L^1$, but $f\cdot f=f^2$ is not in $L^1$.
Note that this fails also if the interval of integration is bounded. For example, the function $f(x):=\frac{1}{\sqrt{x}}$ is in $L^1((0,1))$, but $f^2(x)=\frac{1}{x}$ is not in $L^1((0,1))$.
Convolution product: this is the product you need to consider if you want $L^1$ stability. Is is defined, for $f,g$ in $L^1$, by
$$
(f*g)(x):=\int_{-\infty}^{+\infty}f(t)g(x-t)dt.
$$
It is not hard to see, by Fubini, that $f*g$ is also in $L^1$ with 
$$
\int_{-\infty}^{+\infty}|(f*g)(x)|dx\leq \int_{-\infty}^{+\infty}|f(x)|dx\cdot \int_{-\infty}^{+\infty}|g(x)|dx.
$$
