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

Let $$(X,\mathscr{B},\mu,T)$$ be a measure preserving dynamical system. Then $$U_T:L^2(X,\mu)\rightarrow L^2(X,\mu)$$ defined by $$U_Tf=f\circ T$$ is an isometry. Let $$\mathscr{E}$$ be the eigenspace(closed) of $$U_T$$.

Problem: If $$f$$ is in $$\mathscr{E}$$, then show that $$|f|$$ is also in $$\mathscr{E}$$.

Maybe a possible hint. The above problem will be solved if we can prove that $$L^\infty(X)\cap \mathscr{E}$$ is dense in $$\mathscr{E}$$. See the discussion before Proposition 4.19.

This result is essentially used to prove Lemma 4.23 of Recurrence in Ergodic Theory and Combinatorial Number Theory by Furstenberg, but I am not able to decode the argument given there.

• What do you mean by "the eigenspace(closed) of $U_T$"? – levap May 11 at 22:01
• @levap closed span of the eigenfunctions of $U_T$ in $L^2(X,\mu)$. – Surajit May 11 at 22:36
• Not sure how what they mean with $\max(f,g)$ lie in $\mathscr{E}$. If there is an orthogonal basis of non-negative eigenfunctions then for $f \ge 0$ its projection $\bar{f}$ on $\mathscr{E}$ is non-negative. – reuns May 12 at 2:59
• Yeah this is tough. The only thing that I can offer, off the top of my head, is that the problem is trivial if $T$ is weak mixing (since in that case, the only eigenvalue of $U_T$ is 1, and the problem with my answer below is easy to solve). – pseudocydonia May 15 at 5:35
• I can prove even when $T$ is ergodic, as follows. First observe that if $f$ is an eigenfunction, then $|f|$ is $T$ invariant, hence by ergodicity of $T$, $|f|$ is constant, and so $f$ is in $L^\infty(X)$. So, each eigenfunction is in $L^\infty(X)$. So, any finite span of eigenfunctions is also in $L^\infty(X)$. Now, take the finite span $\mathscr{E}_0$ of eigenfunctions and take its $L^\infty$ closure. This form a $C^*$-algebra. So, by continuous functional calculus, if $f$ is in $\mathscr{E}_0$, then $|f|$ is in $\mathscr{E}_0$. Now observe that $\mathscr{E}_0$ is dense in $\mathscr{E}$. – Surajit May 15 at 6:18

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 $$|f_n| \rightarrow |f|$$, $$|f|\in \mathscr{E}$$.

Suppose that $$U_T f = \lambda f$$ for some eigenvalue $$\lambda$$. Since $$U_T$$ is an isometry, our only possible eigenvalues are $$\pm 1$$ (since otherwise, $$\Vert U_T f \Vert = | \lambda | \Vert f \Vert \neq \Vert f \Vert$$).

If $$\lambda = 1$$, we have that $$U_T |f| = |f \circ T|= |f|$$, so $$|f|$$ is in the eigenspace with eigenvalue 1. Likewise, if $$\lambda = -1$$, we have that $$U_T |f| = |f \circ T | = |-f| = |f|$$, so again $$|f|$$ is in the eigenspace with eigenvalue 1.

EDIT. This is not an answer, see comment below.

• I assume in the context of unitary operators that $f$ is complex-valued. It is still true, with the same argument you give that any eigenvalue $\lambda$ satisfies $|\lambda|=1$, and that if $f$ is an eigenfunction of $U_T$ for the eigenvalue $\lambda$, then $|f|$ is an eigenfunction for $|\lambda| = 1$. – Lukas Geyer May 14 at 20:58
• Indeed, the argument is identical. – pseudocydonia May 14 at 21:02
• @pseudocydonia You have shown that if $f$ is an eigenfunction, then $|f|$ is also an eigenfunction. But that's not enough. For example, how do you show $|cf+dg|$ is in $\mathscr{E}$ if $f$ and $g$ are eigenfunctions, and $c,d$ are arbitrarily fixed scalars? – Surajit May 15 at 3:23