Stone's theorem for bounded operators

Question

Let $H$ be a Hilbert space (assume separable if you like), and let $(U_t)_{t\in\mathbb{R}}$ be a unitary representation of $\mathbb{R}$ on $H$. Let us assume that $t\mapsto U_t$ is continuous, where we use the uniform operator topology on the bounded operators. Define $$D=\{v\in H \ | \ \lim_{t\rightarrow 0}\frac{U_t v-v}{t} \ exists\}.$$

I would like to show that under these assumptions we have $D=H$, and that the operator $A:H\rightarrow H$ given by $A(v)=\lim_{t\rightarrow 0}\frac{U_t v-v}{t}$ is bounded. Can someone give me some hints how to go about it?

Answer 1

EDIT: "uniform operator topology" was just "operator topology" in the original, which I assumed meant either the strong or weak operator topology, in which case it's false:

Say $(X,\mu)$ is a measure space, suppose $\alpha:X\to\Bbb R$ is measurable, and define $U_t:L^2\to L^2$ by $$U_tf(x)=e^{it\alpha(x)}f(x).$$ Then $(U_t)$ is strong-operator-continuous group of unitary operators. But ignoring convergence for a second, it looks like $$\lim_{t\to0}\frac{U_tf-f}{t}(x)=i\alpha(x)f(x).$$If $\alpha\notin L^\infty$ then $A$ is only densely defined and unbounded.

EDIT: Turns out that the OP was referring to the norm topology. In that case it's true. Using the Spectral Theorem makes it too easy:

First, we have a commuting family of normal operators, so my favorite version of the Spectral Theorem shows that after an isomorphism on $H$ we can assume that $H=L^2(\mu)$ for some measure $\mu$ on some set $X$, and that $$U_tf(x)=m_t(x)f(x)$$for some collection of measurable functions $m_t$. Now $U_t$ unitary implies that $|m_t|=1$ almost everywhere. And $m_{s+t}=m_sm_t$. Note that $$||m_t-1||_\infty=||U_t-I||\to0$$as $t\to0$. Some measure-theoretic argument that I haven't concocted shows now that for almost every $x$ the function $x\mapsto m_t(x)$ is continuous. Hence there exists $\alpha(x)$ so $$m_t(x)=e^{it\alpha(x)}.$$ And now $\lim_{t\to0}||m_t-1||_\infty=0$ shows that $\alpha\in L^\infty$. As before $A$ is multiplication by $i\alpha(x)$; this time this shows that $A$ is bounded and defined on all of $H$.

EDIT: That measure-theoretic argument doesn't exist; each $m_t$ is defined only almost everywhere, we might be able to modify each one on a different set of measure zero in such a way that there does not exist $t$ such that $x\mapsto m_t((x)$ is continuous. Things are slightly more subtle.

As above we have $m_t\in L^\infty$ such that $U_t$ is multiplication by $t$. Now for each rational $t$ choose an actual function $n_t$ (as opposed to almost-everywhere equivalence class) with $n_t=m_t$ almost everywhere. Since we're talking about only countably many functions, we can throw out a set of measure zero and obtain $$\sup_{x\in X}|n_t(x)-n_s(x)|=||U_{t-s}-I||\to0\quad(s,t\in\Bbb Q, s-t\to0).$$ Now for every $x$ the function $t\mapsto n_t(x)$ is uniformmly continuous on $\Bbb Q$. So there exists $\alpha(x)$ such that $$n_t(x)=e^{it\alpha(x)}\quad(t\in\Bbb Q).$$ Define $n_t(x)=e^{it\alpha(x)}$ for all $t\in\Bbb R$ and proceed. (Continuity of $U_t$ shows that $U_t$ is multiplication by $n_t$ for every $t$.)

FINAL EDIT Come to think of it, you get a cleaner proof, without all that messing around with sets of measure zero, by using the other version of the Spectral Theorem, or for that matter just the fact that a commutative $C^*$ algebra with identity is $C(K)$. I've typed enough...

Answer 2

If $t\in [0,\infty)\mapsto U(t)\in\mathcal{L}(H)$ is continuous in the operator norm, then, for $\delta$ small enough, $\frac{1}{\delta}\int_{0}^{\delta}U(t)dt$ is invertible because there exists $\delta > 0$ such that $$ \left\|\frac{1}{\delta}\int_{0}^{\delta}U(t)dt-I\right\|= \left\|\frac{1}{\delta}\int_{0}^{\delta}(U(t)-I)dt\right\| < 1. $$ So fix $\delta$ so that $V=\int_{0}^{\delta}U(t)dt$ is invertible in $\mathcal{L}(H)$. For $h \in (0,\delta)$, \begin{align} \frac{1}{h}(U(h)-I)V & =\frac{1}{h}\int_{0}^{\delta}U(t+h)dt-\frac{1}{h}\int_{0}^{\delta}U(t)dt \\ & = \frac{1}{h}\int_{h}^{\delta+h}U(t)dt-\frac{1}{h}\int_{0}^{\delta}U(t)dt \\ & = \frac{1}{h}\int_{\delta}^{\delta+h}U(t)dt-\frac{1}{h}\int_{0}^{h}U(t)dt. \end{align} Assuming $t \in [0,\infty) \mapsto U(t) \in \mathcal{L}(H)$ is continuous, then the right side converges to $U(\delta)-I$ in $\mathcal{L}(H)$ as $h\downarrow 0$. It follows that $$ \lim_{h\downarrow 0}\left\|\frac{1}{h}(U(h)-I) - (U(\delta)-I)V^{-1}\right\|_{\mathcal{L}(X)}=0 $$ Therefore, the generator of the semigroup is the bounded operator $$ A=(U(\delta)-I)V^{-1}. $$ Then you can show that $U(t)=e^{tA}$ by considering $e^{-tA}U(t)$.

Note: This proof works for $C_0$ semigroups, which includes your special case.

Answer 3

I was gonna shut up on this, but I can't stand it. Here's a much cleaner proof.

Let $X$ be the $C^*$ algebra generated by $(U_t)$. Basic $C^*$ algebra stuff shows that $X$ is isomorphic to $C(K)$ for some compact Hausdorff space $K$. The isomorphism is an isometry and takes adjoints to complex conjugates.

Say $f_t\in C(K)$ corresponds to $U_t$. The fact that $U_tU_t^*=I$ shows that $f_t\overline f_t=1$, so $|f_t|=1$. Now $$\sup_{x\in K}|f_t(x)-f_s(x)|=||U_s-U_t||\to0\quad(s\to t),$$so $f_t(x)$ depends continuously on $t$. So there exists $\alpha(x)$ with $$f_t(x)=e^{it\alpha(x)}.$$

Now since $f_t\to1$ uniformly as $t\to0$ it follows that $\alpha$ is bounded. And then given that $\alpha$ is bounded, the fact that $f_t(x)$ is continuous in $x$ for small $t$ shows that $\alpha$ is continuous. And so $$\frac{f_t-1}{t}\to i\alpha$$uniformly. Since $\alpha\in C(K)$ it corresponds to an operator $A\in X$, bounded and everywhere defined. And the uniform convergence of $(f_t-1)/t$ shows that in fact $(U_t-I)/t\to iA$ in norm.