Let $U \in L^1(\mathbb{R}^d)$ and $\rho \in L^1(\mathbb{R}^d)$ such that $\rho \ge 0$ and the support of $\rho$ is included in $B(0,1)$ (the euclidean unit ball of $\mathbb{R}^d$). Is there a way to bound the integral, where $K$ is a compact of $\mathbb{R}^d$, $$\int_K \int_{B(0,1)} U(x-y) \rho(y) \text{d}y \text{d}x $$ by the $L^1$-norm of $U$ on a compact (may be on $K+B(0,1)$) ?
Thank you
With $K-B(0,1)$ it does work. Indeed,
\begin{eqnarray} \int_{K}|U(x-y)||\rho(y)|dx &=& |\rho(y)|\int_{K}|U(x-y)|dx \\ &=& |\rho(y)|\int_{K-\{y\}}|U(z)|dx \tag{1} \end{eqnarray}
where $K-\{y\}=\{x-y,\ x\in K\}$. We get from $(1)$ that $$\int_K |U(x-y)||\rho(y)|dx\leq |\rho(y)|\|U\|_{L^1(K-B(0,1))}\tag{2}$$
We conclude from $(2)$ that $$\int_{B(0,1)}\int_K|U(x-y)||\rho(y)|dx dy=\|\rho\|_{L^1(B(0,1))}\|U\|_{L^1(K-B(0,1))}\tag{3}$$
To finalize, apply Fubini's Theorem.
Remark: I just realized that $K- B(0,1)=K+B(0,1)$.