Statement: Let $M_x$ denote the multiplicative operator acting on $L^2([0,1], \, dx)$ by $M_x(f) = xf$. Show that $M_x$ has no eigenvalues
Attempt: Let $M_x(f) = xf$ then we should have $M_x(f) = \lambda f$
then we get $M_x(f) = \lambda f \Rightarrow xf = \lambda f \Rightarrow xf-\lambda f \Rightarrow (x-\lambda)f = 0$
Now $f \neq 0$ so that implies that $(x-\lambda) = 0$ but this is achieved when $x = \lambda$, so I don't see how it doesn't have eigenvalue, when we achieve the desired result.
If $\lambda$ were an eigenvalue of $M_x$, then $(x - \lambda) f = 0$ pointwise almost everywhere. From this, we can conclude that $f = 0$ almost everywhere on $\{\lambda\}^c$; then using the fact that a singleton set has measure $0$, it follows that $f = 0$ almost everywhere. This is the desired contradiction.