Let $X= L^2([0,1],\#) $ ( where $\#$ is the measure that counts) $ g\in L^\infty([0,1],\#)$ and $A \in \mathcal{L} $ as $A(f)(x)=f(x)g(x)$.

i)Show $\sigma(A) = \overline{g([0,1])}$

ii) Determine the eigenvalues and their respective subspaces

iii)Show $A$ is compact iff there exists a sequence {$ a\_n$}$\_{n \in \mathbf{N} } $ s.t.$g(x)\neq 0$ iff $x = a\_n$ for some $n \in \mathbf{N} $ and $ g(a\_n) \rightarrow 0$ when $n\rightarrow \infty$

i)To see what values are in the spectrum , I have tryed to do $f(x)g(x)=\lambda f(x)$ so it tells me that the $\lambda = g(x)$ and $\lambda \in \sigma(A) $ and this is the spectrum if I am not wrong, but don't understand exatcly what the closure means .

ii) To determine the eigenvalues I have to take $\lambda $ and see if $A-\lambda I$ is injective so that $\lambda \in \sigma\_p(A) $ and their respective subspace would be $ ker(A-\lambda I)$
So I have to determine when $ f(x)g(x)-\lambda f(x)$ is injective . But I am a bit lost here.

iii) This one I don't know how to start it.