spectral radius of a unilateral forward weighted shift

Question

Let $H$ be an infinite dimensional, separable complex Hilbert space with orthonormal basis $(e_n)_n$.

Let $T$ be the unilateral forward weighted shift. i.e. $Te_n=t_n e_{n+1}$. And $t_n = (gcd(n, 2^n))^{-1}$

Let $\rho(T)$ be the spectral radius of $T$.

Show $\rho(T)>0$.

Things I know:

I know that $\rho(T)= \sup \{ |\lambda|: \lambda \in \sigma(T)\} = \lim_{n \rightarrow \infty} \| T^n\|^{1/n}$.

Also, I showed that $\| T\| = \sup_n |t_n|$.

I was also given the hint to find $T^{2^k}$.

So I have that

$(Te_n)^{2^k}= (t_ne_n)^{2^k} = (t_n)^{2^k} (e_n)^{2^k} =(gcd(n, 2^n))^{-1})^{2^k} (e_n)^{2^k} $.

Not sure how to proceed from here.

Thank you!

Answer 1

Well your last equation does not make any sense at all, what is a power of a vector of a Hilbert space? Note that $T^{2^k}$ means the composition of $T$ with itself $2^k$ times.

As you said, by the Gelfand-Beurling formula we have that $\rho(T)=\lim_{n\to\infty}\|T^n\|^{1/n}$. Since this limit exists, we also have that $$\rho(T)=\lim_{n\to\infty}\|T^{2^n}\|^{\frac{1}{2^n}}.$$

Lemma: Fix an integer $p\geq1$ and let $T:H\to H$ be any bounded operator with $Te_n=c_n\cdot e_{n+p}$ for all $n$ where $c_n\in\mathbb{C}$. Then $\|T\|=\sup_{n\geq1}|c_n|$ (so $T$ is a bounded operator if-f $(c_n)$ is a bounded sequence). I leave this to you to prove, it's easy.

Now $T^me_k=T^{m-1}(t_ke_{k+1})=T^{m-2}t_k\cdot t_{k+1}e_{k+2}=\dots=(t_k\cdot t_{k+1}\cdots t_{k+m-1})e_{k+m}$ and this is true for all $k,m$. Therefore, for $m$ being a power of $2$ we have that $$T^{2^n}e_k=(t_k\cdot t_{k+1}\cdots t_{k+2^n-1})e_{k+2^n}$$

Another exercise: Show that for any $k,n\in\mathbb{N}$ we have that $$\require{cancel}\cancel{gcd(k,2^k)\cdot gcd(k+1,2^{k+1})\cdot gcd(k+2,2^{k+2})\cdots gcd(k+2^n-1,2^{k+2^n-1})=2^{2^n-1}}$$

Using our lemma and the above equation we conclude that $$\require{cancel}\cancel{\|T^{2^n}\|=2^{1-2^n}}$$ so $$\require{cancel}\cancel{\|T^{2^n}\|^{1/2^n}=2^\frac{1-2^n}{2^n}=2^{1/2^n}\cdot 2^{-1}\xrightarrow{n\to\infty}\frac{1}{2}}$$

and we conclude that $\require{cancel}\cancel{\rho(T)=1/2>0}$.

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Edit

The striked text is wrong, the exercise is not true as stated, obviously the RHS cannot be independent of $k$. I am sorry for such a big mistake. However, the following is true: Note that by our lemma we have that $$\|T^{2^n}\|=\sup_{k\geq1}\frac{1}{gcd(k,2^k)\cdot gcd(k+1,2^{k+1})\cdots gcd(k+2^n-1,2^{k+2^n-1})}\geq\frac{1}{gcd(1,2^1)\cdot gcd(2,2^2)\cdots gcd(2^n,2^{2^n})}$$

We will compute the RHS of this inequality with induction. Note that in general, an integer $k$ is uniquely written as $2^m\cdot(2l+1)$, where $m,l\geq0$ and then $gcd(k,2^k)=2^m$.

Claim: For all $n\geq1$ we have that $gcd(1,2^1)\cdot gcd(2,2^2)\cdots gcd(2^n,2^{2^n})=2^{2^n-1}$.

Proof of the claim: With induction on $n$. Obviously this is true for $n=1$. Suppose that this is true for some $n$, so $$\prod_{k=1}^{2^n}gcd(k,2^k)=gcd(1,2^1)\cdot gcd(2,2^2)\cdots gcd(2^n,2^{2^n})=2^{2^n-1}.$$ Now $$\prod_{k=1}^{2^{n+1}}gcd(k,2^k)=\prod_{k=1}^{2^n}gcd(k,2^k)\cdot\prod_{k=2^n+1}^{2^{n+1}}gcd(k,2^k)=2^{2^n-1}\cdot\prod_{k=2^n+1}^{2^{n+1}}gcd(k,2^k)=$$ $$=2^{2^n-1}\cdot\prod_{k=1}^{2^n}gcd(k+2^n,2^{2^n+k}).$$

Now let $1\leq k\leq 2^n$. Note that if $k=2^{m_k}\cdot(2l_k+1)$ where $m_k,l_k\geq0$, then $m_k\leq n$, because $1\leq 2l_k+1$, so $2^{m_k}\leq k\leq 2^n$. Also, if $m_k=n$, then $k=2^n$. Now obviously $gcd(2^n+2^n,2^{2^n+2^n})=2^{n+1}$. On the other hand, if $m_k<n$ then $n-m_k>0$, so $2^n+k=2^n+2^{m_k}(2l_k+1)=2^{m_k}(2^{n-m_k}+2l_k+1)$, so $gcd(2^n+k,2^{2^n+k})=2^{m_k}=gcd(k,2^k)$. Therefore, $$\prod_{k=1}^{2^n}gcd(k+2^n,2^{k+2^n})=gcd(2^n+2^n\cdot 2^{2^n+2^n})\cdot\prod_{k=1}^{2^n-1}gcd(k+2^n,2^{k+2^n})=$$ $$=2^{n+1}\cdot\prod_{k=1}^{2^n-1}gcd(k,2^k)=2^{n+1}\cdot\frac{1}{2^n}\cdot\prod_{k=1}^{2^n}gcd(k,2^k)=2^{n+1-n}\cdot2^{2^n-1}=2^{2^n}$$ so $$\prod_{k=1}^{2^{n+1}}gcd(k,2^k)=2^{2^n-1}\cdot2^{2^n}=2^{2^n+2^n-1}=2^{2^{n+1}-1}$$ as we wanted! ${\blacksquare}$

Now using the estimate on $\|T^{2^n}\|$, we have that $$\|T^{2^n}\|\geq\frac{1}{2^{2^n-1}}$$ so $$\rho(T)=\lim_{n\to\infty}\|T^{2^n}\|^\frac{1}{2^n}\geq\lim_{n\to\infty}\frac{1}{2^{1-\frac{1}{2^n}}}=\frac{1}{2}>0.$$

A final comment: To be honest I now think that more generally one can prove that for all $n,k\geq1$ we have that $$\prod_{j=1}^{2^n} gcd(k+j-1,2^{k+j-1})=2^{2^n+k-1}$$ and if this is true then all the inequalities in the edit actually become equalities so $\rho(T)=1/2$ indeed, but I think the above calculations were tiresome enough!