5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
3
+50
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.