0
$\begingroup$

Let $H$ be an infinite dimensional separable Hilbert space and $\{e_n\}$ be a countable orthonormal basis for $H$.For a bounded sequence $\{a_n\}$ define $T(e_n)=a_ne_{n+1}$ and extend linearly to $\operatorname*{span}{\{e_n\}}$. Verify that $T$ extends to a bounded linear operator on $H$.Compute the adjoint of $T$ and polar decomposition of $T$. Find conditions on $\{a_n\}$ so that $T$ is a compact operator. I have verified that $T$ extends to $H$,but I am unable to do the other parts. Any help will be appreciated.

$\endgroup$
1
  • $\begingroup$ For reference: such operators are called "weighted shifts". See definition 27.1 here and the following pages. The answers to your questions follow. $\endgroup$
    – Julien
    Dec 4 '13 at 14:07
1
$\begingroup$

Let $T_{n}x = a_{n}(x,e_{n})e_{n+1}$. Then $$ (T_{n}x,y) = (a_{n}(x,e_{n})e_{n+1},y)=a_{n}(x,e_{n})(e_{n+1},y)=(x,\overline{a_{n}}(y,e_{n+1})e_{n}). $$ So $T_{n}^{\star}y=\overline{a_{n}}(y,e_{n+1})e_{n}$, and $T_{n}^{\star}T_{n}=|a_{n}|^{2}(x,e_{n})e_{n}$. The unique positive square root of $T_{n}^{\star}T_{n}$ is $|T_{n}|x = |a_{n}|(x,e_{n})e_{n}$. That should get your started.

$\endgroup$
1
0
$\begingroup$

Assuming that the inner products are linear with respect to the 2nd coordinate, let me introduce the bra-ket notation. Let $x \in H$, then $\left|x\right> \colon \mathbb{C} \rightarrow H$ if defined by $\left|x\right> \alpha = \alpha x$ for all $\alpha \in \mathbb{C}$, the operator $\left< x \right| \colon H \rightarrow \mathbb{C}$ is defined by $\left< x \right| y= \left<x, y \right>$ for all $y \in H$. Clearly, $\left<x\right|^* = \left|x\right>$.

Yout $T$ is given by the formula $T = \sum_{n \geq 0} a_n \left|e_{n+1}\right>\left<e_n\right|$, so we see that it is approximated by final rank operators, it is bounded if and only if $(a_n)$ is bounded it is compact if and only if $(a_n)$ is from $c_0$. The adjoint is given by $T^* = \sum_{n \geq 0} \overline{a_n} \left|e_{n}\right>\left<e_{n+1}\right|$. From here it is also easy to get the polar decomposition.

$\endgroup$

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.