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 $V'' = B(L^2(\mathbb{R}))$.

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 $V'' = B(L^2(\mathbb{R}))$.

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 $N \in \mathbb{R}$ 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 $V'' = B(L^2(\mathbb{R}))$.

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 $V'' = B(L^2(\mathbb{R}))$.