Spectrum of multiplication by $x$

Question

Suppose I have an operator on $L^2([0,1])$ defined by $f \longmapsto xf$. The eigenvalues occur at when $xf = \lambda x \implies \lambda = x$? Is this enough to conclude that $\sigma(T) = [0,1]$?

Answer 1

No! There are no eigenvalues, but the spectrum is indeed $[0,1]$.

EDIT To see that this is the spectrum, calling your operator $T$, note that if $\lambda \notin [0,1]$, $(T - \lambda I)^{-1}$ is multiplication by $1/(x-\lambda)$ (which is bounded since that is a bounded function on $[0,1]$), while if $\lambda \in [0,1]$ there is no $f \in L^2[0,1]$ for which $(x - \lambda) f(x) = 1$.

However, there are no eigenvalues: an eigenvalue for $\lambda$ would be a nonzero $f \in L^2[0,1]$ such that $T f - \lambda f = 0$, i.e. $(x - \lambda) f(x) = 0$ a.e. in $[0,1]$. But that makes $f(x) = 0$ a.e. in $[0,1]$.