Let $H$ be the Hilbert space $L^2(\mathbb{R}^n)$. For $x, y \in \mathbb{R}^n$, define the unitary operators $U_x$ and $V_y$ on $H$ by
$U_x(f)(t)=f(t-x)$ and $V_y(f)(t)=e^{-i\lambda\langle y|t\rangle}f(t)$. I want to prove that both these unitary representations are strongly continuous. That is I need to prove that for any $f \in L^2(\mathbb{R}^n)$, the map $x \mapsto U_x(f)$ and the map $y \mapsto V_y(f)$ are continuous.
It suffices to show $U_x(f)$ is continuous at $x=0$, i.e. we want to show
\begin{align} \lim_{|x|\rightarrow 0}\|U_x(f)-f\|_{L^2}^2 = \lim_{|x|\rightarrow 0}\int dy\ |f(x-y)-f(y)|^2 =0. \end{align}
By density argument, it suffices to assume $f$ is smooth with compact support. Since \begin{align} \lim_{|x|\rightarrow 0}(f(x-y)-f(y)) = 0 \end{align} for all $y \in \mathbb{R}^n$ and $|f(x-y)-f(y)|\leq 3|f(y)|$ then by Lebesgue Dominated Convergence theorem you are done. The other operator is similar (after you apply Plancherel).
Additional: Observe \begin{align} \widehat{V_y(f)} =&\ \int dx\ e^{-ix\cdot\xi} V_y(f) = \int dx\ e^{-ix\cdot\xi}e^{-i\lambda x\cdot y}f(x)\\ =&\ \int dx\ e^{-ix\cdot(\xi+\lambda y)}f(x) = \hat f (\xi+\lambda y). \end{align}