# On spectral multiplicity of left shift operators [closed]

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?

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 $$E_U(\Delta)x= V^*E(\Delta) Vx =V^*E(\Delta)\sum_n x_n z^n =V^* 1_{\Delta}\sum_n x_n z^n =\left\{\int_{\Delta}\,\sum_kx_k\,e^{ikt}\,dt \right\}_n$$
• Your operator is already multiplication by the identity function, so the spectral multiplicity function is the constant function $\lambda\longmapsto \omega$. Commented Feb 16, 2019 at 15:02