Let $\Omega\subset\Bbb R^d$ be an open set. Let $k:\Bbb R^d\to [0,\infty)$ be measurable such that $0\in \operatorname{supp}k$. This implies that $\Omega\subset \Omega_k=\Omega+\operatorname{supp}k$. Recall that see here \begin{split} \operatorname{supp} k &= \Bbb R^d\setminus\bigcup\big\{O\,:\, \text{$O$ is open and $k=0$ a.e. on $O$}\big\}\\ &= \bigcap\big\{ \Bbb R^d\setminus O\,:\, \text{$O$ is open and $k=0$ a.e. on $O$}\big\}. \end{split}

Let $u:\Bbb R^d\to \Bbb R$ be measurable. Assume that

• $k$ is symmetric, i.e. $k(x)=k(-x)$.
• $k(x)>0$ for all $x\in \operatorname{supp}k$.
• $u=0$ a.e. $\Bbb R^d\setminus\Omega_k$ and $u=0$ a.e. $\Bbb \Omega$.

$$I=\iint\limits_{\Omega \Omega_k}|u(y)|^2k(x-y)d y dx=0.$$

Question: Prove or disprove that $u=0$ a.e. on $\Bbb R^d$ that is, we do have $u=0$ a.e. on $\Omega_k\setminus \Omega$?

Note that when $supp k=\Bbb R^d$ we easily have $u=0$ a.e. on $\Bbb R^d$. Indeed, in this case we get $\Omega_k= \Omega+\Bbb R^d= \Bbb R^d$ and hence

$$I=\iint\limits_{\Omega \Bbb R^d}|u(y)|^2k(x-y)d y dx= \iint\limits_{\Omega \Bbb R^d}|u(x+h)|^2k(h)d h dx=0.$$ Then we pick $a\in \Omega$ such that $$\int_{\Bbb R^d}|u(a+h)|^2k(h)d h=0.$$ Since $k>0$ on $\Bbb R^d$ we $u(a+h)=0$ for almost every $h\in \Bbb R^d$ and thus $u=0$ a.e. on $\Bbb R^d$.

General observations Note that $x-y\not\in supp k$ for $x\in \Omega$ and $y\not\in \Omega_k$. Thus $k(x-y)=0$ for $x\in \Omega$ and $y\not\in \Omega_k$. This implies

$$I=\iint\limits_{\Omega \Omega_k}|u(y)|^2k(x-y)d y dx=\iint\limits_{\Omega \Bbb R^d}|u(y)|^2k(x-y)d y dx\\= \iint\limits_{\Omega \Bbb R^d}|u(x+h)|^2k(h)d h dx= \iint\limits_{\Omega \operatorname{supp} k}|u(x+h)|^2k(h)d h dx$$

PS: A good answer to this question may be awarded a bounty