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Question. Let $G$ be a non-compact, finite dimensional Lie group, and let $(X, \mu)$ be a Radon measure space. 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 $n\in\mathbb N$.

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. The same idea works for the other examples, and it is even slightly simpler; indeed, in both 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.

$^{[1]}$ 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.

Question. Let $G$ be a non-compact, finite dimensional Lie group, and let $(X, \mu)$ be a Radon measure space. 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 (Bekka - Mayer, "Ergodic theory and topological dynamics of group actions on homogeneous spaces"); it is Theorem 1.1 at page 81.

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. 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.

$^{[1]}$ 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.

Question. Let $G$ be a non-compact, finite dimensional Lie group, and let $(X, \mu)$ be a Radon measure space. 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 $n\in\mathbb N$.

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 $d\mu=\frac{dx}{\lvert x\rvert^d}$, 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 $$\rho_g f(x):=f\left(\frac{az+b}{\overline b z + \overline a}\right), \qquad g=\begin{bmatrix} a & b \\ \overline b & \overline a\end{bmatrix}, $$ where $|a|^2-|b|^2=1$.

I learned the proof for the example 2 is in this Math.SE. The same idea works for the same two examples, and it is even slightly simpler; indeed, in both 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 $g_n\to \infty$. Thus, we can argue that $$ \left\lvert \int_X \overline{h(x)}\rho_{g_n}f(x)\, d\mu\right\rvert \le \lVert h\rVert_{L^2(A)}\lVert \rho_{g_n}f\rVert_{L^2(A)}+ \lVert h\rVert_{L^2(X\setminus A)}\lVert \rho_{g_n}f\rVert_{L^2(X\setminus A)}. $$ The first summand tends to zero, while the second can be made arbitrarily small because $h\in L^2(X)$. Here we use that $\rho$ is unitary.

Question. Let $G$ be a non-compact, finite dimensional Lie group, and let $(X, \mu)$ be a Radon measure space. 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 $n\in\mathbb N$.

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. The same idea works for the other examples, and it is even slightly simpler; indeed, in both 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.

$^{[1]}$ 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.