Let $F$ be a $p$-adic field. Let $G=F^\times$ and $X=F^2-\{(0,0)\}$. Let $G$ acts on $X$ by $a.(x,y)=(ax,ay),a\in F^\times, (x,y)\in X$. Let $U=\{(x,y)\in X, xy\ne 0\}$.
$\textbf{Question 1}$: Is there a $G$-invariant distribution $T$ on $X$ such that the restriction of $T$ to $U$ is non-zero?
Here is one example in my mind that illustrates there might not exist such $T$ in Question 1. Consider the multiplicative action of $F^\times$ on $F$, and let $U=\{x\in F,x\ne 0\}$, which is an open subset of $U$. Then by Tate's thesis, there is no $F^\times$-invariant distribution $T$ on $F$ such that $T|_U$ is non-zero. In fact, the only $F^\times$-invariant distribution on $F$ is the Dirac measure.
If Question 1 DOES have an affirmative answer, then here is
$\textbf{Question 2}$: Let $\psi$ be a non-trivial additive character of $F$ and let $\mu=\psi(y/x)d^*xd^*y$, viewed as a measure on $U$. It is clear that $\mu$ is $G$-invariant. Is there a $G$-invariant distribution on $X$ such that $T|_U=\mu$?
Here is another similar question
$\textbf{Question 3}$: Let $G=F^\times \times F^\times,X=F^3-\{0\}$ and $U=(F^\times)^3$. Let $G$ act on $X$ by $(a,b).(x,y,z)=(ax,by,abz)$. Is there a $G$-invariant distribution $T$ on $X$ such that $T|_U$ is nontrivial? If such $T$ exists, what can we say about $Supp(T|_U)$?
Edits:
Many thanks to @paulgarrett to give an affirmative answer to $\textbf{Question 1}$. There does exist $G$-invariant distribution $T$ such that $T|_U$ is non-zero. For example, take $T$ as $$<T,\varphi>=\int_{F^\times}\varphi(\alpha,\alpha)d^*\alpha,\varphi\in S(X).$$
Here is a variant of Question 1.
$\textbf{Question 1'}$: Let $Y=F^2$ and $G=F^\times$ act on $Y$ by $a.(x,y)=(ax,ay).$ Is there a distribution $T\in S^*(Y)^G$ such that $Supp(T|_U)=U$? Here $U=(F^\times)^2\subset Y$.
I am sorry that I also asked the same question in mathoverflow https://mathoverflow.net/questions/279930/non-existence-of-certain-invariant-distributions-on-p-adic-spaces , because I thought it might draw more attention there. If it is not permitted, I will delete one.
Thanks.