Let $ L^2=L^2(\mathbb{R}) $. For every pair $ a,b $ of real numbers define the operator $ U_{a,b} $ on $ L^2 $ sending $ \psi \in L^2 $ 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: Is this a basis for the bounded operators on $ L^2(\mathbb{R}) $?
It turns out that the only compact operator in $ V $ is the $ 0 $ operator. That means, for example, that $ V $ is missing all the nonzero compact self adjoint operators on $ L^2(\mathbb{R}) $ and all nonzero trace class operators and all the nonzero finite rank projections/finite rank operators in general.
Since unitary operators on infinite dimensional spaces are never compact this does not a priori rule out the possibility of $ V $ containing all unitary operators. But a moment of thought reveals that, since $ V $ is clearly not all of $ B(L^2(\mathbb{R})) $, then it cannot include all unitary operators either since every bounded operator is a linear combination of four unitary operators.
For more details see this answer:
https://mathoverflow.net/questions/414596/characterize-this-subspace-of-the-bounded-operators-on-l2-mathbbr/414613#414613