# 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:

$$f(x) = \begin{cases} 1 & \mbox{ for } |x| \le \tfrac 12 \\ \\ 0 & \mbox{ for } |x| > \tfrac 12 \end{cases}$$

$$g(x) = \begin{cases} 1 & \mbox{ for } |x| \le \tfrac 12 \\ \\ 0 & \mbox{ for } |x| > \tfrac 12 \end{cases}$$

-
I don't understand the question. What are the quantifiers on $f$ and $g$? – Qiaochu Yuan Mar 29 '13 at 0:39

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 $$\int_{-\infty}^{+\infty}|(f*g)(x)|dx\leq \int_{-\infty}^{+\infty}|f(x)|dx\cdot \int_{-\infty}^{+\infty}|g(x)|dx.$$

-
@user1095332 Then $f$ is in $L^1$= integrable over $\mathbb{R}$, but $ff=f^2$ is not in $L^1$= not integrable over $\mathbb{R}$. I hope it is english enough. Note: integrable=absolutely convergent. – 1015 Mar 28 '13 at 23:40
@user1095332 I've added more details. Let me know if it's enough or if you still need more. – 1015 Mar 29 '13 at 12:35
more, I love more mathicecreams :) So, much more that I can put it on a powerpoint presentation and show it to anyone and it will make absolutely sense. Take for example: Banach spaces, Pointwise product, {[n,n+1/n^3]}, 1/n2=π^2/6, Convolution product, L1 stability, Fubini. And is it possible to just define a f(x)=...x^2+...x+... and g(x)=...x^2+...x+...? – user1095332 Mar 30 '13 at 11:58
@user1095332 What do you mean? – 1015 Mar 30 '13 at 12:39
@user1095332 I don't know your background. This is very detailed for most people who know what an integrable (absolutely convergent if you prefer, but that's not really correct) function is. So please ask precise questions about what details you need and I will happily supply them. – 1015 Mar 30 '13 at 13:06