Uniformly continuous unitary representations. Let $U$ be a unitary rep. of $\mathbb{R}^d$ on a separable Hilbert space $H$, and $H\cong\oplus L^2_{\mu_v}(\mathbb{R}^d)$ be the spectral decomposition (according to the spectral theorem for these representations). Then it seems that $t\mapsto U(t)$ is continuous (in the operator norm) iff there is a bounded set $K$ in $\mathbb{R}^d$ such that the suppport of all the spectral measures $\mu_v$ is contained in $K$. Is this true, and how could one prove it?
I would be happy if someone gives me a sketch of proof and hints, containing the main idea.
 A: I must be missing something - this seems trivial. The argument below maybe takes more than one line, but there's nothing clever about it, every single step is an "ok, just work it out" thing.
Say $K_n$ is the support of $\mu_n$. Then $$||U_x-I||=\sup_n\sup_{t\in K_n}
|e^{ix\cdot t}-1|.$$ So for $\delta>0$ we have $$\sup_{|x|<\delta}||U_x-I||=\sup_n\sup_{t\in K_n}
\sup_{|x|<\delta}|e^{ix\cdot t}-1|.$$
Lemma Suppose $\delta>0$ and $t\in\Bbb R^d$.


*

*$\sup_{|x|<\delta}|e^{it\cdot x}-1|\le\delta|t|$.

*If $\delta|t|>\pi$ then $\sup_{|x|<\delta}|e^{it\cdot x}-1|=2$.


Lemma Suppose $\delta>0$ and $K\subset \Bbb R^d$. Let $r=\sup_{t\in K}|t|$.


*

*$\sup_{t\in K}\sup_{|x|<\delta}|e^{it\cdot x}-1|\le\delta r$.

*If $\delta r>\pi$ then $\sup_{t\in K}\sup_{|x|<\delta}|e^{it\cdot x}-1|=2$.


Now let $r_n=\sup_{t\in K_n}|t|$.
Suppose $r_n$ is unbounded. Then for every $\delta>0$ there exists $n$ so $r_n\delta>\pi$, hence $$\sup_{|x|<\delta}||U_x-I||=2\quad(\delta>0)$$and so our representation is not continuous in norm.
Suppose on the other hand that $r_n\le r$ for all $n$. Then $$\sup_{|x|<\delta}||U_x-I||\le r\delta,$$and so we do have continuity in norm.
