
> 
> **Question**. Let $G$ be a non-compact, finite dimensional Lie group, and let $(X, \mu)$ be a [Radon measure space](https://en.wikipedia.org/wiki/Radon_measure). Let $$\rho\colon G\to U(L^2(X))$$
>  be a unitary, strongly continuous, representation. Is it true that, if $g\_n\to \infty$, then
>  $$
> \int\_X \overline{h(x)}\rho\_{g\_n} f(x)\, d\mu\to 0, \qquad \forall f, h\in L^2(X)?$$
>  Appropriate hypotheses on $X$ may be assumed.
> 
> 
> 


Here, $g\_n\to \infty$ means that, for any compact $K\subset G$, $g\_n\notin K$ for all sufficiently big $n\in\mathbb N$. 



> 
> **EDIT** (for reference). I learned in the comments that a fairly complete answer is given by the "Howe-Moore vanishing theorem". I found a reference in [this book](https://www.cambridge.org/core/books/ergodic-theory-and-topological-dynamics-of-group-actions-on-homogeneous-spaces/D44A57C1D660059C5C4AA7FA8E244659) (Bekka - Mayer, "Ergodic theory and topological dynamics of group actions on homogeneous spaces"); it is Theorem 1.1 at page 81. 
> 
> 
> 




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This property is true in the following cases. 


1. $G=(\mathbb R^n, +)$, $X=\mathbb R^n$ with Lebesgue measure and $\rho\_g f(x):=f(x-g)$.
2. $G=(\mathbb R\_{>0}, \cdot)$, $X=\mathbb R^n$ with measure $d\mu=\frac{dx}{\lvert x\rvert^n}$, and $\rho\_g f(x):=f(x/g)$.
3. $G=SU(1, 1)$, $X=\mathbb D$, the unit disk, with measure $d\mu=\frac{4dxdy}{(1-(x^2+y^2)^2)^2}$, and $$\rho\_g f(z):=f\left(\frac{az+b}{\overline b z + \overline a}\right), \qquad g=\begin{bmatrix} a & b \\ \overline b & \overline a\end{bmatrix}, $$ where $|a|^2-|b|^2=1$.
4. $G=SL(2, \mathbb R)$, $X=\mathbb H=\{z\in \mathbb C\ :\ \Im z>0\}$, with measure $d\mu=\frac{1}{y^2}dxdy$, and $$\rho\_g f(z):=f\left(\frac{az+b}{c z + d}\right), \qquad g=\begin{bmatrix} a & b \\ c & d\end{bmatrix}, $$ where $ad-bc=1$. This case is actually isomorphic to the previous one, via the Cayley transform $z\mapsto \frac{z-i}{z+i}$ that maps $\mathbb H$ onto $\mathbb D$.


[I learned the proof for the example 2 in this Math.SE post](https://math.stackexchange.com/q/805982/8157). The same idea works for the other examples, and it is even slightly simpler; indeed, in all three cases, for all $f\in L^2(X)$, and for all compact $A\subset X$, $$\lVert \rho\_{g\_n} f\rVert\_{L^2(A)}\to 0, $$ provided that $g\_n\to \infty$.$^{[1]}$ Thus,
$$
\left\lvert \int\_X \overline{h(x)}\rho\_{g\_n}f(x)\, d\mu\right\rvert \le \lVert h\rVert\_{L^2(A)}\lVert \rho\_{g\_n}f\rVert\_{L^2(A)}+ \lVert h\rVert\_{L^2(X\setminus A)}\lVert \rho\_{g\_n}f\rVert\_{L^2(X\setminus A)}. $$ 
The first summand tends to zero, while the second can be made arbitrarily small by choosing a sufficiently big $A$, because $h\in L^2(X)$. Here we use that $\rho$ is unitary.




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$^{[1]}$ As mentioned, this case is slightly simpler than the dilation one, because for the dilation group we must also consider the possibility that the $L^2$ norm concentrates at the origin.



Do all unitary representations weakly converge to zero at infinity?
No, there are simple counterexamples. E.g., take $G = \mathbb{R}$ and $X = \mathbb{C}$ with Lebesgue measure, and define $\rho\_t f(z) = f(e^{2\pi i t}z)$ for $t \in \mathbb{R}$ and $f \in L^2(\mathbb{C})$. Then $\rho\_t$ is the identity for any integer $t$, so $\rho\_n \to {\rm id}$ strongly, not to zero.


