# Generator of an abelian von Neumann Algebra

I want to show that $L^{\infty}(S^1)$(where $S^1$ is equipped with its Haar Measure) as a von Neumann algebra is generated by the multiplication operators $M_{e^{in\theta}}$ where $n \in Z$ and $\theta\in R$. Note - 1. For any $f\in L^{\infty}(S^1)$,the multiplication operator $M_f : L^{2}(S^1) \longrightarrow L^{2}(S^1)$ is given by $M_f(g) = fg$ $\forall g\in L^{2}(S^1)$ and thus $L^{\infty}(S^1)$ can be identified with a subalgebra of $B(L^{2}(S^1))$. 2. A von Neumann algebra is a strongly closed(or weakly closed) self-adjoint subalgebra of $B(H)$ where $H$ is a Hilbert Space.

Continuous functions on a compact set are uniformly continuous. As such, they are the uniform norm limits of the partial sums of their fourier series. In the terminology of your problem, this means that if $\xi$ is the function $\xi:z\longmapsto z$, then $C^*(\xi)$ contains all of $C(S^1)$.

So now it is enough to argue that $C(S^1)$ generates $L^\infty(S^1)$ as a von Neumann algebra in $B(L^2(S^1))$. Given any $f\in L^\infty(S^1)$, by Lusin's Theorem there exist compact subsets $E_n\subset S^1$ with $m(S^1\setminus E_n)<1/n$ such that $f|_{E_n}$ is continuous. If we choose carefully, we can get $E_1\subset E_2\subset\cdots$; indeed, we first find $E_1$ with $m(S^1\setminus E_1)<1$ and $f|_{E_1}$ continuous; now we apply Lusin on $S^1\setminus E_1$ and we obtain $F_2\subset S^1\setminus F_1$ with $m(S^1\setminus E_1\setminus F_2)<1/2$ and $f|_{F_2}$ continuous. If $E_2=E_1\cup F_2$, then $F|_{E_2}$ is continuous (at worst we need to remove a point where they touch) and $m(S^1\setminus E_2)<1/2$; now we continue in the same way.

By Tietze's Extension we can extend each $f|_{E_n}$ to continuous functions $f_n:S^1\to\mathbb C$ with $\|f_n\|_\infty\leq\|f\|_\infty$.

Now, for any $g\in L^2(S^1)$ $$\|(M_f-M_{f_n})g\|_2^2=\|(f-f_n)g\|_2^2\leq4\|f\|_\infty^2\,\int_{S^1\setminus E_n}|g|^2 =4\|f\|_\infty^2\int 1_{S^1\setminus E_n}|g|^2\to0$$ by dominated convergence, since $1_{S^1\setminus E_n}\to0$ a.e. (this, because this is a decreasing sequence of sets and the intersection is a null-set). Indeed, let $E=\bigcap(S^1\setminus E_n)$; then $m(E)=\lim m(S^1\setminus E_n)=0$. If $x\not\in E$, then $x\in E_n$ for all $n$ sufficiently big; that says that $1_{S^1\setminus E_n}(x)=0$ for all $n$ sufficiently big.

The above shows that $M_f$ is the sot limit of $\{M_{f_n}\}$, which implies that $L^\infty(S^1)$ is the sot closure of $C(S^1)$. So $L^\infty(S^1)=W^*(\xi)$.

• Very nice proof.I just have one doubt, why does 1_(E_n) tends to 0 a.e?Please explain. – Ester Oct 8 '15 at 20:10
• I added some clarification and corrected some typos. Let me know if you still have questions. – Martin Argerami Oct 8 '15 at 22:13
• I guess that the second equality is not right, it should be an inequality <=4||f||^2 integral...,though the proof remains the same. – Ester Oct 8 '15 at 23:12
• Yes, you are right. – Martin Argerami Oct 9 '15 at 0:02