Let $f\in L^2(\Bbb{T}),\ f\ne0$ and $M=\overline{f\wp_+}$. Prove that, $M$ is simply invariant for $\chi_1$ iff $f\notin \chi_N M$ for some $N>0$

Question

Here we denote $\chi_n$ to be the functions on $\Bbb{T}$ defined as $\chi_n(z)=z^n\ \forall z\in\Bbb{T}$ for $n\in\Bbb{Z}$. We define $\wp_+=\{\sum\limits_{n=0}^N a_n\chi_n|\ a_n\in\Bbb{C},N\ge0\}$ be the space of analytic trigonometric polynomials.

A non-trivial closed subspace $M$ of $L^2(\Bbb{T})$ is said to be simply invariant for $\chi_1$ if $\chi_1M\subset M $ and $\bigcap\limits_{n\ge0}\chi_n M=\{0\}$.

Here $f$ is a non-zero $L^2$ function. Then $M=\overline{f\wp_+}$ is non-trivial closed invariant subspace for $\chi_1$. If $M$ is imply invariant then $\bigcap\limits_{n\ge0}\chi_n M=\{0\}$. Then we have $N>0$ such that $f\notin \chi_N M$. So,one side is done.

But I am stuck with the converse part. Can anyone help me finish the proof? Thanks for your help in advance.

Answer 1

Currently, I am only be able to give a proof based on (the well-known) Wold–von Neumann decomposition theorem and the F. & M.Riesz theorem. Although there may exists some proofs without using any results about the Hardy space $H^2(\mathbb T)$.

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If there is a positive integer $N$, such that $f\notin \chi_N M$, then $\bigcap_{n\geq 0} \chi_n M=\{0\}$.

Proof. The assumption is exactly $\chi_N M\neq M$. Moreover, we can assume $N=1$, since $\chi_1M=M$ will implies $\chi_N M=M$.

Let $W=M\ominus \chi_1M$ (sometimes, such $W$ is called the wondering subspace). Note that the multiplication operator $M_{\chi_1}$ is an isometry on $M$. It follows from Wold–von Neumann decomposition theorem that $$M=\bigg(\bigoplus_{n\geq 0} \chi_n W\bigg)\bigoplus \bigg(\bigcap_{n\geq 0} \chi_n M \bigg).$$

The key is to observe that $\bigcap_{n\geq 0} \chi_n M$ is also an invariant subspace of $M_{\chi_1}^*$. Consequently, $$\bigcap_{n\geq 0} \chi_n M=\chi_E L^2,$$ where $\chi_E$ is the indicator function of some measurable subset $E$. Note that $W\neq \{0\}$, we chose a nonzero function $g\in W$.

Claim: $|g|$ is a nonzero constant almost everywhere. The proof of this claim is nothing else but copy the proof of Beurling's theorem. Indeed, because $W\perp \chi_n W$ for $n>0$, we see that $g\perp \chi_n g$ for $n>0$, or equivalently, $$\int_{\mathbb T}|g(z)|^2 z^n =0, \,\,\, n>0.$$ Note that $|g|^2$ is real, by considering the conjugation, we have $$\int_{\mathbb T}|g|^2 z^{n} =0,\,\,\, n<0.$$ The claim follows.

Note that $\chi_n W\perp \bigcap_{n\geq 0} \chi_n M$ and $\bigcap_{n\geq 0} \chi_n M=\chi_E L^2$. In particular, $$\chi_n g\perp \chi_E, \,\,\, n\geq0.$$ That is $$g\chi_E\perp \chi_{n},\,\,\, n\leq 0.$$ This shows $g\chi_E\in H^2(\mathbb T)$. Combining with the previous claim, it follows from F. & M. Riesz theorem that either $E$ is null or $E^c$ is null. We conclude that $\bigcap_{n\geq 0} \chi_n M=\chi_E L^2=\{0\}$ since $M$ is nontrivial. The proof is finished.