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.

  • $\begingroup$ What do you mean by "the eigenspace(closed) of $U_T$"? $\endgroup$ – levap May 11 at 22:01
  • $\begingroup$ @levap closed span of the eigenfunctions of $U_T$ in $L^2(X,\mu)$. $\endgroup$ – Surajit May 11 at 22:36
  • $\begingroup$ 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. $\endgroup$ – reuns May 12 at 2:59
  • $\begingroup$ 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). $\endgroup$ – pseudocydonia May 15 at 5:35
  • $\begingroup$ 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 $\mathscr{E}$. $\endgroup$ – 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.

  • $\begingroup$ 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 $|\lambda| = 1$. $\endgroup$ – Lukas Geyer May 14 at 20:58
  • $\begingroup$ Indeed, the argument is identical. $\endgroup$ – pseudocydonia May 14 at 21:02
  • 1
    $\begingroup$ @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? $\endgroup$ – Surajit May 15 at 3:23
  • $\begingroup$ Ach! Good point. Let me think about this. $\endgroup$ – pseudocydonia May 15 at 4:49
  • $\begingroup$ I guess you have to go through something like in the proof of the Weierstrass approximation theorem, just as Furstenberg does in the quoted paper. $\endgroup$ – Lukas Geyer May 16 at 17:30

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