TITLE: $f(y-x)$ integrable implies $f=0$ a.e.
QUESTION [7 upvotes]: If $f(y-x)$ is in $L^p(\mathbb R^d\times\mathbb R^d)$, then I seem to conclude that $f=0$ a.e. (which seems wrong). My reasoning is that by Fubini and the integral's shift invariance (assume $p=1$ for convenience)
$$\infty>\iint|f(y-x)|dydx = \int\left( \int |f(y-x)|dy\right) dx =  \int\left( \int |f(y)|dy\right) dx = \int \text{const}\ dx $$ The const is non-negative but can't be positive else the right hand side will diverge, thus const $=0$. But this implies that $f=0$.
Where am I going wrong?

REPLY [2 votes]: As pointed out in the comments, your reasoning is correct and as ˈjuː.zɚ79365 said:

There is a difference between the function $x\mapsto K(x)$ being integrable on $\mathbb R^d$ and the function $(x,y)\mapsto K(x−y)$ being integrable on $\mathbb R^d\times \mathbb R^d$. The former is true for all $L^1$ functions; the latter is true only for functions that are zero a.e.