Invariant subspace for normal operators

[Boromir]

I have proved the spectral theorem for a normal operator T on an infinite dimensional hilbert space, and am now trying to deduce that T has non-trivial invariant susbspaces.

Case 1: If the spectrum of T consists of a single point: My book says that if this is the case then the set of continuous functions of the spectrum is isomorphic to the complex numbers. Don't know how they got that

case 2: If the spectrum of T has more than one point, then my book says that are disjoint borel subsets whose union is the whole spectrum but again not sure how they got that.

Thanks

 

[jvicens]

for sharing your proof and insights! The spectral theorem is a fundamental result in functional analysis and has many important applications in mathematics and physics.

To answer your questions, let's first review the statement of the spectral theorem for normal operators. The theorem states that for a normal operator $T$ on a complex Hilbert space $\mathcal{H}$, there exists a unique spectral measure $\mu$ on the Borel subsets of $\mathbb{C}$ such that $T$ is unitarily equivalent to the multiplication operator $M_\mu$ on $L^2(\mathbb{C}, \mu)$, where $M_\mu f(z) = zf(z)$ for all $z\in\mathbb{C}$.

Now, let's consider the two cases you mentioned:

Case 1: If the spectrum of $T$ consists of a single point $\lambda$, then the spectral measure $\mu$ is concentrated at this point, i.e. $\mu(\{\lambda\}) = 1$. In this case, $T$ is unitarily equivalent to the multiplication operator $M_\lambda$ on $L^2(\mathbb{C}, \mu)$, which means that $Tf = \lambda f$ for all $f\in\mathcal{H}$. This implies that $T$ has a one-dimensional invariant subspace spanned by the eigenvector $f_\lambda$ corresponding to the eigenvalue $\lambda$. Moreover, since $T$ is normal, it also has a unitary eigenvector $g_\lambda$ corresponding to the eigenvalue $\overline{\lambda}$, and the subspace spanned by $f_\lambda$ and $g_\lambda$ is also invariant under $T$. This shows that $T$ has non-trivial invariant subspaces when its spectrum consists of a single point.

Case 2: If the spectrum of $T$ has more than one point, then the spectral measure $\mu$ is a non-trivial measure on the Borel subsets of $\mathbb{C}$, which means that there are disjoint Borel subsets $E_1, E_2, \dots$ whose union is the whole spectrum $\sigma(T)$ and such that $\mu(E_j) > 0$ for all $j$. By the spectral theorem, $T$ is unitarily equivalent to the multiplication operator $M_\mu$ on $L^2