I posted this on MSE a couple months ago and it got three upvotes but no answers or even comments so I decided to cross-post it here:


For every pair $ a,b $ of real numbers define the operator $ U\_{a,b} $ on $ L^2(\mathbb{R}) $ sending $ \psi \in L^2(\mathbb{R}) $ to $ U\_{a,b}\psi $ defined by the equation
$$
[U\_{a,b}\psi](x)=e^{ibx}\psi(x+a)
$$
Consider the set of operators
$$
\mathcal{B}:=\{ U\_{a,b}:a,b \in \mathbb{R} \}
$$
Let $ V $ be the closure in the operator norm topology of the span of the set $ \mathcal{B} $. Does anyone have a good idea for a nice characterization of what sort of operators are and are not in $ V $? Does $ V $ include all trace class operators? All compact operators? All unitary operators?


This is a follow up question to my question:
<https://math.stackexchange.com/questions/4303824/is-this-a-basis-for-the-bounded-operators-on-l2-mathbbr>



Characterize this subspace of the bounded operators on $ L^2(\mathbb{R}) $
Any linear combination $L$ of $U\_{a,b}$'s can be written $(L\psi)(x) = \sum\_{k=1}^n \alpha\_ke^{ib\_kx}\psi^{\to a\_k}(x)$, where $\psi^{\to a\_k}(x) = \psi(x + a\_k)$. Fix $L$.


Let $N \in \mathbb{N}$ be such that $Nb\_k$ is close to an integer multiple of $2\pi$, for all $k$. Then $$(L\psi^{\to N})(x) = \sum \alpha\_k e^{ib\_kx}\psi^{\to a\_k + N}(x) = (\sum \alpha\_k e^{ib\_k(x - N)}\psi^{\to a\_k}(x))^{\to N} = (\sum \alpha\_k'e^{ib\_kx}\psi^{\to a\_k}(x))^{\to N}$$ where each $\alpha\_k'$ is close to $\alpha\_k$. Okay, now find a sequence $N\_j \to \infty$ such that as $j \to \infty$ the multiples $N\_jb\_k$ get arbitrarily close to integer multiples of $2\pi$. Then $(L\psi^{\to N\_j})^{-N\_j} \to L\psi$ in $L^2(\mathbb{R})$, for every $\psi$.


But any compact operator $T$ satisfies $T\psi^{\to N\_j} \to 0$ in $L^2(\mathbb{R})$. Taking $\psi$ with $\|\psi\|\_2 = 1$ and $\|L\psi\|\_2$ close to $\|L\|$, we have $\|T\psi^{\to N\_j}\|\_2 \to 0$ but $\|L\psi^{\to N\_j}\|\_2 \to \|L\psi\|\_2 \cong \|L\|$. This shows you that $\|T - L\| \geq \|L\|$; that is, the distance from $L$ to the compact operators is $\|L\|$. Every element of $V$ will have the same property, so in particular $V$ contains no compact operators besides $0$.


Every operator is a linear combination of four unitaries, so if $V$ contained every unitary then it would be all of $B(L^2(\mathbb{R}))$, which we've just seen is not the case. On the other hand, the WOT closure of $V$ does equal $B(L^2(\mathbb{R}))$; this follows from the double commutant theorem, since any operator that commutes with $U\_{0,b}$ for all $b$ must be a multiplication operator and hence won't commute with $U\_{a,0}$ for all $a$ unless it is a scalar. That is, $V' = \mathbb{C}\cdot I$, so $V'' = B(L^2(\mathbb{R}))$.


