normal operator where the sum of eigenspaces is not equal to the entire Hilbert space.

Question

Hi I'm reading about Hilbert spaces and normal operators. In this i found the following result:

Let $H$ a Hilbert space of finite dimension and $T\in\mathcal{L}(H)$ a normal operator, then $$ H = \bigoplus_{i\in 1,\ldots,n}H_i $$ Where $(\lambda_i,H_i)_{i=1,\ldots,n}$ are the eigenvalue and eigenspace of $T$.

Indeed I'm looking for a example in infinite dimension that this result don't follow

Answer 1

Let $H=L^2[0,1]$ and $T$ the multiplication operator by the identity function, that is $$ (Tf)(x)=xf(x). $$ It is an easy exercise to show that $T$ is selfadjoint and that it has no eigenvalues.