Compactness of translation operator in weighted spaces

Question

Let $x,v\in\Bbb R^d$, $t\in \Bbb R$ and $m(x,v)$ be a smooth strictly positive function rapidly decaying on infinity - think $m(x,v) = \exp(-|x|^2-|v|^2)$.

Define Banach spaces $X$ and $Y$ by $$\|h(x,v)\|^2_X = \int_{\Bbb R^d\times\Bbb R^d}|h(x,v)|^2m(x,v)\,dx\,dv$$ $$ \|g(t,x,v)\|^2_Y = \sup_{t\in \Bbb R}\int_{\Bbb R^d\times\Bbb R^d}|g(t,x,v)|^2m(x-tv,v)\,dx\,dv.$$

Finally, we have a linear isometry $A:X\to Y$, $A[h](t,x,v) = h(x-tv,v)$.

I want to answer the question whether $A$ is compact or not. Most standard techniques apply only to operators in Hilbert spaces (in our case $Y$ is only a Banach space) and/or integral operators, so any hints/ideas/links will be greatly appreciated.

Answer 1

A linear isometry defined on an infinite-dimensional space is never compact.

To see this, note that (as shown in this (somehow misformulated) question (Is it true that the unit ball is compact in a normed linear space iff the space is finite-dimensional?)), a normed linear space is finite dimensional iff the closed unit ball is compact.

Thus, since your space $X$ is (obviously) not finite dimensional, there is a sequence $(x_n)_n$ in the unit ball of $X$ that has no convergent subsequence (and thus no Cauchy subsequence, since your space is complete).

If $A$ was a compact map, there would be a convergent (hence Cauchy) subsequence $(A x_{n_k})_k$. Since $A$ is a linear isometry, this easily implies that also $(x_{n_k})_k$ is Cauchy, in contradiction to the choice of $(x_n)_n$.

In particular, your map $A$ can not be compact.