Is this a basis for the bounded operators on $ L^2(\mathbb{R}) $?

Question

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 $$ \mathcal{B}:=\{ U_{a,b}:a,b \in \mathbb{R} \} $$ Is $ \mathcal{B} $ a basis for the space of bounded operators on $ L^2 $? They seem linearly independent and it also seems like the span should be a subalgebra at least. But is the closure of the span of $ \mathcal{B} $ all of the bounded operators on $ L^2 $?

Answer 1

First of all, note that the most appropriate notion for bases in Banach spaces are probably Schauder bases. The concept that you are interested in (being linearly independent and spanning a dense subspace) is not enough for this.

Next, as @user940347 noted, it is easy to see that your operators (which are called time-frequency shifts) do not span the space of all bounded operators.

A more interesting question (that you are actually asking) is whether your operators span a dense subspace. Then, the next question is which topology or notion of convergence you are interested in.

First, if you are interested in the strong or weak operator topology, then the Stone-von Neumann theorem combined with a version of Schur's lemma implies that the commutator of (the closed linear span of) your set $\mathcal{B}$ is $\{\alpha \cdot \mathrm{id} \colon \alpha \in \mathbb{C} \}$, so that the double commutant theorem shows that the span of $\mathcal{B}$ is dense in the set of all bounded linear operators in the strong and weak operator topology.

Finally, I will show below that

The span of $\mathcal{B}$ is not dense in the set of all bounded linear operators on $L^2(\mathbb{R})$ with respect to the operator norm.

The proof uses the following notation:

• The set $\mathcal{B}(L^2(\Bbb{R}))$ of all bounded linear operators on $L^2(\Bbb{R})$.
• The shift operator $T_x$ defined by $(T_x f) (y) = f(y-x)$,
• The modulation operator $M_\omega$ defined by $M_\omega f(x) = e^{2\pi i \omega x} f(x)$,
• for any measurable set $\Omega \subset \Bbb{R}$ the multiplication operator $$ A_\Omega : \quad L^2(\mathbb{R}) \to L^2(\mathbb{R}), \quad f \mapsto \boldsymbol{1}_\Omega \cdot f. $$

Let $V := \overline{\mathrm{span}} \{ T_x M_\omega \colon x,\omega \in \mathbb{R} \} \subset \mathcal{B}(L^2(\mathbb{R}))$. In what follows, we show that $A_{[0,1]} \notin V$, showing that the time-frequency operators $T_x M_\omega$ do not span a dense subspace of $\mathcal{B}(L^2(\mathbb{R}))$ with respect to the operator norm. We divide the proof into five steps.

Step 1: Given $T \in \mathcal{B}(L^2(\mathbb{R}))$, define $$ f_T : \quad \mathbb{R} \to [0,\infty), \quad x \mapsto \| T \circ A_{[x,x+1]} \| . $$ Note for $S,T \in \mathcal{B}(L^2(\mathbb{R}))$ that $$ |f_T (x) - f_S(x)| \leq \| T \circ A_{[x,x+1]} - S \circ A_{[x,x+1]} \| \leq \| T - S \| . $$ In particular, if $\| S_n - T \| \to 0$, then $f_{S_n}\to f_T$, with uniform convergence.

Step 2: Note that \begin{align*} \bigl(T_x A_{[0,1]} T_{-x} f\bigr) (y) & = (A_{[0,1]} T_{-x} f) (y-x) \\ & = \boldsymbol{1}_{[0,1]} (y-x) \cdot (T_{-x} f) (y-x) \\ & = \boldsymbol{1}_{[x,x_1]} (y) \cdot f(y) \\ & = (A_{[x,x+1]} f) (y) . \end{align*} Hence, $ f_T (x) = \| T \circ T_x \circ A_{[0,1]} \circ T_{-x} \| = \| T \circ T_x \circ A_{[0,1]} \| . $

Step 3: Note for $\varphi : \mathcal{B}(L^2(\mathbb{R})) \to [0,\infty), T \mapsto \| T \circ A_{[0,1]} \|$ that $$ |\varphi(S) - \varphi(T)| \leq \| S \circ A_{[0,1]} - T \circ A_{[0,1]} \| \leq \| S - T \| , $$ so that $\varphi$ is continuous.

Step 4: Consider $T = \sum_{j=1}^n a_j T_{x_j} M_{\omega_j}$ with $a_j \in \mathbb{C}$ and $x_j, \omega_j \in \mathbb{R}$. Note that $$ (M_{\omega_j} T_x f)(y) = e^{2 \pi i \omega_j \cdot ((y - x) + x)} f(y-x) = e^{2 \pi i \omega_j x} (T_x M_{\omega_j} f) (y) . $$ Hence, using the identity from Step 2, we see \begin{align*} f_T (x) & = \| T \circ T_x \circ A_{[0,1]} \| \\ & = \bigg\| \bigg( \sum_{j=1}^n a_j \, T_{x_j} M_{\omega_j} T_x \bigg) \circ A_{[0,1]} \bigg\| \\ & = \bigg\| \bigg( \sum_{j=1}^n a_j e^{2 \pi i \omega_j x} T_{x_j} T_x M_{\omega_j} \bigg) \circ A_{[0,1]} \bigg\| \\ & \overset{(\ast)}{=} \bigg\| \bigg( \sum_{j=1}^n a_j e^{2 \pi i \omega_j x} T_{x_j} M_{\omega_j} \bigg) \circ A_{[0,1]} \bigg\| \\ & = \varphi \bigg( \sum_{j=1}^n a_j e^{2 \pi i \omega_j x} T_{x_j} M_{\omega_j} \bigg) , \end{align*} which shows that $f_T$ is continuous. Here, the step marked with $(\ast)$ used that $T_{x_j} T_x = T_x T_{x_j}$ and that $\| T_x S \| = \| S \|$.

Step 5: Overall, if we had $A_{[0,1]} \in V$, there would be a sequence $(T_m)_{m \in \Bbb{N}}$ of operators as in Step 4 satisfying $\| T_m - A_{[0,1]} \| \to 0$. By Step 1, this implies $f_{T_m} \to F_{A_{[0,1]}}$, with uniform convergence. Hence, since Step 4 shows that each $f_{T_m}$ is continuous, so is $f_{A_{[0,1]}}$. But we have \begin{align*} f_{A_{[0,1]}} (x) & = \| A_{[0,1]} \circ A_{[x,x+1]} \| = \| A_{[0,1] \cap [x,x+1]} \| \\ & = \begin{cases} 0, & \text{if } [0,1] \cap [x,x+1] \text{ is a null-set}, \\ 1, & \text{otherwise}, \end{cases} \end{align*} so that $f_{A_{[0,1]}} (\mathbb{R}) = \{ 0,1 \}$, in contradiction to the intermediate value theorem.

Answer 2

No. All operators in $\mathcal{B}$ and hence finite linear combinations of them map continuous functions on continuous functions. The operator $Tf := f 1_{[0,1]}$ is clearly a bounded linear operator on $L^2$ but it maps continuous functions not necessarily on continuous functions -- hene $T$ is not in the span of $\mathcal{B}$.

The next step would be to ask if your statement is true for the closed linear span. You would then have to clarify w.r.t. what topology (operator norm, weak/strong operator topology).

I think that is wrong as well...