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?

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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?

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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)