For $(\theta,x_{0},\xi_{0},\lambda)\in (\mathbb{R}/Z)\times\mathbb{R}^{d}\times\mathbb{R}^{d}\times (0,\infty)$, let $g_{\theta,x_{0},\xi_{0},\lambda}: L^{2}(\mathbb{R}^{d})\rightarrow L^{2}(\mathbb{R}^{d})$ be the unitary operator defined by
$$g_{\theta,x_{0},\xi_{0},\lambda}f(x):=\lambda^{-\frac{d}{2}}e^{i\theta}e^{ix\cdot\xi_{0}}f(\frac{x-x_{0}}{\lambda}), \quad x\in\mathbb{R}^{d}$$
and let $G$ denote the collection of such operators. One can show that $G$ has a group structure with
- Identity $g_{0,0,0,1}$
- Inverse $g_{\theta,x_{0},\xi_{0},\lambda}^{-1}:=g_{-\theta-x_{0}\cdot\xi_{0},-\lambda\xi_{0},-x_{0}/\lambda,\lambda^{-1}}$
- Group Law $$g_{\theta,x_{0},\xi_{0},\lambda}g_{\theta',x_{0}',\xi_{0}',\lambda'}:=g_{\theta+\theta'-x_{0}\cdot\xi_{0}'/\lambda,\xi_{0}+\frac{\xi_{0}'}{\lambda},x_{0}+\lambda x_{0}',\lambda\lambda'}$$
but we will not need this this. The motivation for this group are the symmetries of certain nonlinear Schrodinger equations. Endow $G$ with the strong operator topology (SOT). I want to show that $G$ is a closed subset of $B(L^{2}(\mathbb{R}^{d})$, the space of bounded linear operators on $L^{2}(\mathbb{R}^{d})$. It's clear to me that the SOT limit of a sequence $g_{n}$ belongs to $B(L^{2}(\mathbb{R}^{d})$, but I am having difficulty show that the limit is also an element of $G$. Any suggestions?
I think the factorization $$g_{\theta,x_{0},\xi_{n},\lambda}=g_{\theta,0,0,1}g_{0,x_{0},0,1}g_{0,0,\xi_{0},1}g_{0,0,0,\lambda}$$ to which you allude is more useful than you think. Let $\{g_{n}=g_{\theta_{n},x_{n},\xi_{n},\lambda_{n}}\}_{n=1}^{\infty}$ be a sequence in $G$ which converges to some limit in $T\in B(H)$ in the strong operator topology (SOT). Since $g_{n}$ are unitary, we note that $T\neq 0$. First, I claim that it suffices to show that, passing to a subsequence and relabeling, there exists $(\theta,x_{0},\xi_{0},\lambda)\in (\mathbb{R}/(2\pi\mathbb{Z}))\times \mathbb{R}^{d}\times\mathbb{R}^{d}\times(0,\infty)$ such that $(\theta_{n},x_{n},\xi_{n},\lambda_{n})\rightarrow (\theta,x_{0},\xi_{0},\lambda)$, as $n\rightarrow\infty$. Indeed, for any $f,\varphi\in C_{c}(\mathbb{R}^{d})$, we can apply dominated convergence to conclude that \begin{align*} \lim_{n\rightarrow\infty}\langle{g_{\theta_{n},x_{n},\xi_{n},\lambda_{n}}f,\varphi}\rangle_{L^{2}}&=\lim_{n\rightarrow\infty}\lambda_{n}^{-d/2}\int_{\mathbb{R}^{d}}e^{i\theta_{n}}e^{ix\cdot\xi_{n}}f(\frac{x-x_{n}}{\lambda_{n}})\varphi(x)dx\\ &=\lambda^{-d/2}\int_{\mathbb{R}^{d}}e^{i\theta}e^{ix\cdot\xi_{0}}f(\frac{x-x_{0}}{\lambda})\varphi(x)\\ &=\langle{g_{\theta,x_{0},\xi_{0},\lambda_{0}}f,\varphi}\rangle_{L^{2}} \end{align*} By an approximation argument, it follows that $g_{n}\rightarrow g:=g_{\theta,x_{0},\xi_{0},\lambda}$ in the weak operator topology. By uniqueness, we conclude that $T=g\in G$.
Since $\mathbb{R}/(2\pi\mathbb{Z})$ is compact, passing to a subsequence, we may assume that $\theta_{n}\rightarrow\theta$. Let $f,\varphi$ be as above. Suppose $\sup_{n}\lambda_{n}=\infty$; the argument if $\inf_{n}\lambda_{n}=0$ is completely analogous. Passing to a subsequence, we may assume that $\lambda_{n}\rightarrow\infty$. Then \begin{align*} \lim_{n\rightarrow\infty}\lambda_{n}^{-d/2}|\int_{\mathbb{R}^{d}}e^{i\theta_{n}}e^{ix\cdot\xi_{n}}f(\frac{x-x_{n}}{\lambda_{n}})\varphi(x)dx|&\leq\lim_{n}\lambda_{n}^{-d/2}\|f\|_{\infty}\|\varphi\|_{L^{1}}\\ &=0 \end{align*} By an approximation argument, we conclude that $g_{n}\rightharpoonup0$, which implies that $T=0$, a contradiction. So passing to a subsequence, we may assume that $\lambda_{n}\rightarrow\lambda\in (0,\infty)$.
If $\sup_{n}|x_{n}|=\infty$, then passing to a subsequence, we may assume that $x_{n}\rightarrow\infty$. With $f,\varphi$ as above, we have that $$\langle{g_{0,x_{n},0,1}f,\varphi}\rangle_{L^{2}}=0$$ when $(x_{n}+\mathrm{supp}(f))\cap\mathrm{supp}(\varphi)=\emptyset$. As above, this implies $g_{0,x_{n},0,1}\rightharpoonup 0$. By the factorization of $G$, triangle inequality and unitarity of the operators, this implies that $g_{n}\rightharpoonup 0$, a contradiction. Passing to a subsequence, we have that $x_{n}\rightarrow x_{0}\in\mathbb{R}^{d}$.
If $\sup_{n}|\xi_{n}|=\infty$, then passing to a subsequence, we may assume that $\xi_{n}\rightarrow\infty$. With $f,\varphi$ as above, we have that \begin{align*} \lim_{n\rightarrow\infty}\langle{g_{0,0,\xi_{n},1}f,\varphi}\rangle_{L^{2}}=\lim_{n\rightarrow\infty}\int_{\mathbb{R}^{d}}e^{ix\cdot\xi_{n}}f(x)\overline{\varphi}(x)dx=0 \end{align*} Since $f\varphi\in L^{1}(\mathbb{R}^{d})$, we can apply Riemann-Lebesgue limit to conclude the RHS is zero. By a similar argument, one can show that $g_{0,0,\xi_{n},1}f$ is not Cauchy in $L^{2}$. By the same argument as in the previous paragraph, this implies that $g_{n}\rightharpoonup 0$, which is a contradiction. So passing to a subsequence, we have that $\xi_{n}\rightarrow\xi_{0}\in\mathbb{R}^{d}$. This completes the proof.