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In an attempt to find two integrable functions $f$ and $g$, say $\in L^1(\mathbb{R}$), whose product is not in $L^1(\mathbb{R})$, I thought about finding a more general example :

Let $f(x) = \frac{1}{x} $ and $g(x) = x$, we have that

\begin{equation} \int_{-\infty}^{\infty} f(x)g(x) dx = \int_{-\infty}^{\infty}1 dx = \infty \end{equation}

But is it correct to say the following (with a similar reasoning for $g(x)$) ?

\begin{equation} \int_{-\infty}^{\infty}\frac{1}{x} dx = \int_{0}^{\infty} \frac{1}{x}+\frac{-1}{x} dx = \int_{0}^{\infty} 0 dx = 0 + C \end{equation} We'd have found an example of two integrable functions whose product is not integrable, but I wonder if the reasoning is correct since $\int_{0}^{\infty} 1/x dx$ is not finite. This would be similar to say $\infty - \infty = 0$ , which is not always true.

Also, any idea on the initial problem, i.e finding $f,g\in L^1(\mathbb{R})$ s.t. $fg \not \in L^1(\mathbb{R})$ ?

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    $\begingroup$ check the definition of $L^1(X)$ function, you're missing an absolute value in the integrand $\endgroup$ Commented Jan 22 at 15:06
  • $\begingroup$ Yes I know, sorry if I have been unclear. That's why I put "more general example" $\endgroup$
    – Sileo
    Commented Jan 22 at 17:20
  • $\begingroup$ In general you say that a function $f : X \to \mathbb{C}$ is integrable if $\int_X |f| d \mu$ exists and is finite. This because one wants to avoid problems with things like $\infty - \infty$, so $L^1(X)$ is precisely the vector space of integrable functions. $\endgroup$ Commented Jan 23 at 8:42

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The Hölder inequality proves that $$ \|fg\|_1 \leq \|f\|_2 \|g\|_2 $$ so such a $f,g$ must be searched among non-$L^2(\mathbb{R})$ functions. We can consider $$ f(x)= \begin{cases} \frac{1}{\sqrt{|x|}} & \text{if } x \in [-1,1]\\ x^{-2} & \text{otherwise} \end{cases} $$ This is $L^1(\mathbb{R})$ as $\int_{-1}^1 f(x) dx = 4$ and $\int_1^{+\infty} f(x) dx = 1 $. But $$ f(x)^2= \begin{cases} \frac{1}{|x|} & \text{if } x \in [-1,1]\\ x^{-4} & \text{otherwise} \end{cases} $$ is not $L^1(\mathbb{R})$ as $|x|^{-1}$ is not integrable in any neighborhood of the origin. Therefore $\|f\|_1 < \infty$ but $\|f^2\|_1 = \infty$.

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