1
$\begingroup$

Let $U$ be an operator defined on $l^{2}(\mathbb{Z})$ by $U(e_{n})=e_{n-1}$, where $e_{n}$ is an orthonormal basis of $l^{2}(\mathbb{Z})$. $U$ is a left shift operator. Since $U$ is unitary operator so spectrum is on $S^{1}$. What is spectral measure of $U$? What is its spectral decomposition with respect to multiplicity?

$\endgroup$

1 Answer 1

1
$\begingroup$

Under the canonical identification of $\ell^2(\mathbb Z)$ with $L^2(\mathbb T,m )$ (where $m$ is Lebesgue measure) and $e_n\longmapsto (z\longmapsto z^n)$, your $U$ is the multiplication operator $U=M_{\bar z}$.

So $\sigma(U)=\sigma(M_{\bar z})=\bar z(\mathbb T)=\mathbb T$. Using this answer, the spectral measure of $M_{\bar z}$ is given by $$ E(\Delta)f=1_{(\bar z)^{-1}(\Delta)}f=1_{\Delta}\,f. $$ The canonical unitary implementing the isomorphism $\ell^2(\mathbb Z)\simeq L^2(\mathbb T,m )$ is $V(\sum_n c_n e_n)=\sum _n c_n z^n$, and its inverse is $$ V^*f=\left\{\int_0^{2\pi} f(e^{i t})\,e^{in t}\,dt \right\}_n. $$ So the spectral measure of $U$ is $$ E_U(\Delta)x= V^*E(\Delta) Vx =V^*E(\Delta)\sum_n x_n z^n =V^* 1_{\Delta}\sum_n x_n z^n =\left\{\int_{\Delta}\,\sum_kx_k\,e^{ikt}\,dt \right\}_n $$

$\endgroup$
2
  • $\begingroup$ Thanks, Is the scalar valued spectral measure is Lebesgue. What is spectral multiplicity? $\endgroup$
    – mathlover
    Commented Feb 16, 2019 at 10:51
  • 1
    $\begingroup$ Your operator is already multiplication by the identity function, so the spectral multiplicity function is the constant function $\lambda\longmapsto \omega$. $\endgroup$ Commented Feb 16, 2019 at 15:02

Not the answer you're looking for? Browse other questions tagged .