Factorization internal and also external features

I'm searching for a factorization for $f(z) = z - c$ in internal and also external features.

So, first I search for wherefore $c$, $f$ is cyclic for the independent change $M_z$. There is a theory that states that if $A$ is a bounded driver on a Hilbert room $H$ after that $x \in H$ is cyclic for $A$ iff absolutely no is the only component that is orthogonal to all $A^n x$ for $n = 0,1,2,\ldots$.

Allow $M_z$ be the reproduction driver by $z$ (" the independent change on $H^2$").

So, if $g_n(z) = M_z^n (z - c) = -cz^n + z^{n + 1}$ I locate that the power collection coefficients of $g_n$ (which is offered by $g_n(z) = \sum_m a_{n,m} z^m$) are $a_{n,n} = -c$ and also $a_{n,n+1} = 1$. If I insert this in the internal item (where $h(z) = \sum \overline{b_n} z^n$), $(g_n, h) = 0$ for all $n = 0, 1, 2, \ldots$. I get that $b_{n + 1} - c b_n = 0$ for $n \geq 0$. This considers that $b_n = C \cdot c^n$. Yet $\sum c^n z^n = 0$ just if $c = 0$. This would indicate that $f(z) = z$ is cyclic yet a cyclic vector can not have an absolutely no in the device disk! What is incorrect below?

If I would certainly locate this $c$, after that I would certainly locate a factorization if the $c$ makes $f$ cyclic due to the fact that after that the internal function is $1$ and also the external function $f$. Just how to do it if $f$ is not cyclic?

2019-05-18 21:20:50
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Answers: 1

What is incorrect is the thinking where you created "But $\sum c^n z^n = 0$ just if $c=0$." What you require to recognize is for which selections of $c$ does the problem $b_{n+1}=c b_n$ for all $n$, with $b_0,b_1,\ldots$ being a square summable series, pressure every one of the $b_n$ is to be absolutely no. If $c=0$, this is not the instance ; simply take $h(z)=1$. The coefficients are not compelled to be absolutely no when $|c|<1$, where you can take $h(z)=1 + cz +c^2z^2 +c^3z^3 +\cdots$. Yet the problem $b_{n+1}=c b_n$ is difficult for $|c|\geq 1$ unless $h=0$, so your standard is $|c|\geq1$.

When $f$ is not cyclic, for internal component you can take the holomorphic automorphism of the device disk that swaps $c$ and also 0, and also what continues to be will certainly be an invertible component of $H^\infty$, therefore cyclic.

2019-05-21 03:40:35