# How do $\int_0^{\infty} \left\lvert f(x) \right\rvert dx$ and $\int_0^{\infty}\int_0^{\infty} \left\lvert f(x+y) \right\rvert dx dy$ compare?

I would like to compare the integrals $\int_0^{\infty} \left\lvert f(x) \right\rvert dx$ and $\int_0^{\infty}\int_0^{\infty} \left\lvert f(x+y) \right\rvert dx dy$?

In particular, I would like to know whether there exists $\alpha \in (0,\infty)$ such that for all $f$

$$\int_0^{\infty} \left\lvert f(x) \right\rvert dx \le \alpha \int_0^{\infty}\int_0^{\infty} \left\lvert f(x+y) \right\rvert dx dy?$$

Does anybody know how to approach such a question?

• Nope, since for every $f$, $$\int_0^{\infty}\!\!\int_0^{\infty} \left\lvert f(x+y) \right\rvert dx dy=\int_0^{\infty}x \left\lvert f(x) \right\rvert dx$$ – Did Mar 13 '18 at 22:46
• OP, a justification for the above is the change of variables $x+y\mapsto x$ followed by Fubini's theorem – Operatorerror Mar 13 '18 at 23:17
• Interestingly, if we change the integration region to $\mathbb R$, then the claim is true for all $a>0$. In fact, the right-hand side is zero or $+\infty$. – daw Mar 14 '18 at 7:33

Nope. Considering the function $g(x, y) = f(x+y)$, then $g$ is constant on strips of the form $L_t = \{(x, y) \mid x+y = t\}$. So this is equal to $$\int_0^\infty \int_{L_t} |f(t)| \, dx \, dt = \int_0^\infty t |f(t)| \, dt.$$ So the integral $\int_0^\infty |f(t)| \, dt$ may be defined, while the above integral may not. For example, take $f$ to be the function $1/x^2$ for $x \geq 1$ and zero otherwise. (There is probably an easier counterexample; I am very rusty with these things)