Sheet4 B2

 
Let $X$ be a complex Hilbert space and $T ∈ ℬ(X)$ be normal (i.e. $T^* T=T T^*$). (a) Show that \[ r_σ(T)=‖T‖ \] Deduce that if $P$ is a polynomial, then \[ ‖P(T)‖=\sup_{λ ∈ σ(T)}|P(λ)| \] Solution. (6.7.1)Since $T$ is normal, $‖Tx‖=‖T^*x‖∀x∈X$ Apply this to $Tx$: $‖T^2x‖=‖T^*Tx‖$. By induction, $‖T^{2n}x‖=‖(T^*T)^nx‖∀x∈X∀n∈ℕ$, so $‖T^{2n}‖=‖(T^*T)^n‖∀n∈ℕ$. Since $T^*T$ is self-adjoint, $‖(T^*T)^n‖=‖T^*T‖^n$. By Gelfand's formula,$$r_σ(T)=\lim_{n→∞}‖T^{2n}‖^{\frac1{2n}}=\lim_{n→∞}‖T^*T‖^{\frac12}=‖T^*T‖^{\frac12}=‖T‖$$ For the second part: By the Spectral Mapping theorem, $σ(P(T))=P(σ(T))$. Since $P(T)$ is normal, $‖P(T)‖=\sup_{λ∈σ(P(T))}|λ|=\sup_{λ∈P(σ(T))}|λ|=\sup_{μ∈σ(T)}|P(μ)|$. (b) Let $P$ be a Laurent polynomial, i.e. $P(z)=\sum_k a_k z^k$ where the summation range is finite but may contains positive as well as negative powers. Show that if $T$ is unitary, then \[ ‖P(T)‖=\sup_{λ ∈ σ(T)}|P(λ)| . \] Solution. Let $P=\sum_{k=-n}^{n}a_kz^k,Q=\sum_{k=0}^{2n}a_{k-n}z^k$ so that $Q$ is polynomial and $P(z)=z^{-n}Q(z)$ Since $T$ is unitary, $σ(T)⊆\{λ∈ℂ∣|λ|=1\}$. By (a), we get \begin{align*} ‖P(T)‖&=‖T^{-n}Q(T)‖\\ &=\sup_{x∈X:‖x‖=1}‖T^{-n}(Q(T)(x))‖\\ &\overset{\text{isometry}}=\sup_{x∈X:‖x‖=1}‖Q(T)(x)‖\\ &=‖Q(T)‖\\ &\overset{\text{(a)}}=\sup_{λ∈σ(T)}|Q(λ)|\\ &\overset{|λ|=1}=\sup_{λ∈σ(T)}|P(λ)| \end{align*}