If $f$ is in the span of eigenfunctions, then $|f|$ is also in the span of eigenfunctions.

Let $F$ be the $L^2$ closure of $L^{\infty}(X) \cap \mathscr{E}$.

To show that $F=\mathscr{E}$, it is enough to prove that every eigenfunction is in $F$.

So if $f$ is a $L^2$ eigenfunction with eigenvalue $\lambda$, then $|\lambda|=1$.

For each $R >0$, let $g_R(z)=z$ if $|z| \leq R$ and $g_R(z)=R\frac{z}{|z|}$. Denote $f_R=g_R \circ f$.

Since $g_R$ commutes with all rotations, $f_R$ is an eigenfunction $L^{\infty}$. Besides $f_R \rightarrow f$ in $L^2$.

Thus $L^{\infty} \cap \mathscr{E}$ is dense in $\mathscr{E}$. Actually, we proved a stronger result: $\mathscr{E}$ is the $L^2$-closure of the vector subspace generated by $L^{\infty}$ eigenfunctions (let’s call it $G$).

Note now that $L^{\infty}(X)$ is an algebra, and that $U_T$ is linear and multiplicative (from $L^{\infty}$ to itself), the set of $L^{\infty}$ eigenfunctions is thus multiplicative. Therefore, the subspace it spans in $L^{\infty}(X)$ (ie $G$) is a subalgebra. For a similar reason, it is also stable under complex conjugation.

Now we show that if $f \in G$, $f \geq 0$, then $\sqrt{f}$ is in the $L^{\infty}$ (hence $L^2$) closure of $G$.

Indeed, let $R >0$ be such that $f \leq R$ ae. There is a sequence of polynomials $P_n$ such that $P_n(x) \rightarrow \sqrt{x}$ uniformly in $0 \leq x \leq R$.

Thus, for each $n$, $P_n(f) \in G$, and $P_n(f)$ converges to $\sqrt{f}$ in $L^{\infty}$.

In particular, if $f \in G$, then $|f|=\sqrt{f\overline{f}}$ is in the $L^2$ closure of $G$ (aka $\mathscr{E}$)

Now, if $f \in \mathscr{E}$, there exists a sequence $f_n \in G$ converging to $f$ with the $L^2$ norm. Then $|f_n| \in \mathscr{E}$ for all $n$. Since $|f_n| \rightarrow |f|$, $|f|\in \mathscr{E}$.