I'm trying to understand an assertion that appears in Luck's book on $L^2$-invariants. I believe that it's elementary functional analysis, but I can't seem to figure it out.

Here goes. Let $$T^n = \{\text{$(z_1,\ldots,z_n)$ $|$ $|z_i|=1$ for $1 \leq i \leq n$}\} \subset \mathbb{C}^n$$ be the $n$-torus and let $G$ be the set of bounded linear operators from $L^2(T^n)$ to itself. The group $\mathbb{Z}^n$ acts on $G$ in the following way. Consider $L=(\ell_1,\ldots,\ell_n) \in \mathbb{Z}^n$ and $\phi \in G$. Then $L \cdot \phi$ is the operator that takes $f \in L^2(T^n)$ to $z_1^{\ell_1} \cdots z_n^{\ell_n} \phi(f) \in L^2(T^n)$.

For any $h \in L^{\infty}(T^n)$, there is a multiplication operator $M_h \in G$. Clearly $M_h$ is invariant under $\mathbb{Z}^n$. Luck claims that in fact the set of $\mathbb{Z}^n$-invariants in $G$ is exactly $L^{\infty}(T^n)$. Can anyone help me prove this?

By the way, it is clear that if $\phi \in G$ is invariant under $\mathbb{Z}^n$, then $\phi$ should be equal to the multiplication operator $M_{\phi(1)}$. But I can't even prove that $\phi(1)$ is in $L^{\infty}(T^n)$, much less that $\phi = M_{\phi(1)}$.