Spectral theorem in terms of multiplication operators

Question

I have a problem attempting this task. Let $(\Omega , \mu)$ hasn’t got atoms. Let operator $A : L_2(\Omega , \mu) \to L_2(\Omega , \mu)$ be the operator of multiplication on real-valued function $\phi$. Let $\lambda \in \sigma_p(A)$ the question is which dimension could have space , generated by eigenvectors, corresponding this certain $\lambda$. I don’t know how to attempt this problem. I only have an assumption that the dimension can’t be higher than 1 (still I can’t provide it with formal explanations)

Answer 1

Well... what does it mean to lie in the point-spectrum? If $f$ is an eigenfunction of $A$, then $\phi(x)f(x)=\lambda f(x),$ implying for a.e. $x$ that $f(x)=0$ or $\phi(x)=\lambda$. Conversely, any $L_2$ function of the form $g 1_{\{\phi=\lambda\}}$ is, indeed, an eigenfunction. It follows that $$ \{ f\in L_2(\mu)|\lambda f=\phi f\}=L_2(\{\phi=\lambda\},\mu), $$ which is either trivial or infinite dimensional, since $\mu$ has no atoms.