What are the eigenfunctions of the operators $M_1 f(x) = xf(x)$ and $M_2 f(x) = x^2f(x)$?

Question

Let $H = L^2([1,2])$ and consider the multiplication operators $$ \begin{align} M_1 f(x) &= xf(x) \\ M_2 f(x) &= x^2f(x). \end{align} $$ I know that $M_1$ has spectrum $\sigma_1 = [1,2]$ and $M_2$ has spectrum $\sigma_2 = [2,4]$ but what are the eigenfuctions that correspond to elements of the spectrums of $M_1$ and $M_2$?

Answer 1

For $M_1$:

if $ \mu \in \sigma_1$, $f \in H$ and $M_1f= \mu f$, then $(x- \mu)f(x)=0$ for all $x \in [0,1]$. This gives $f=0$ a.e.

Consequence: $M_1$ has no eigenvalues !