Currently I'm self studying functional analysis, namely spectrum of bounded operators. In the text, the author gives the following example:
Example A: On $L_2[0,1]$ define $A\colon L_2[0,1]\to L_2[0,1]$ with $(Ax)(t):=t\cdot x(t)$. Then $[0,1]=\sigma_c(A)=\sigma(A)^{[1]}$.
My question is fairly straightforward: why is $[0,1]=\sigma_c(A)=\sigma(A)$? I'm not sure if this is correct, but thinking about this on my own I've come up with the following:
The set $\sigma(A)$ is all the $\lambda\in\mathbb{C}$ such that $A-\lambda\text{Id}$ is not invertible. As $A-\lambda\text{Id}$ is not invertible if $\text{ker}(A-\lambda\text{Id})\neq\{0\}$, we can find all the $\lambda\in\mathbb{C}$ for which $(A-\lambda\text{Id})(x(t))=0$ with $x(t)\neq0$. This is equivalent to finding all the $\lambda\in\mathbb{C}$ for which $t-\lambda=0$. Thus, the set $\sigma(A)$ is all the $\lambda\in\mathbb{C}$ such that $\lambda=t\in[0,1]$; hence, we have that $\sigma(A)=[0,1]$.
Now, if the above is correct, great! But I don't understand why $\sigma_c(A)=[0,1]$, as this would imply that $\sigma_p(A)^{[2]}=\emptyset$.
$^{[1]}$Note here that $\sigma(A)$ denotes the spectrum of $A$, and $\sigma_c(A)$ is the continuous spectrum of $A$.
$^{[2]}$Note here that $\sigma_p(A)$ denotes the point spectrum of $A$.
Best Answer
If $\lambda \in \sigma_p(A)$ the there exists $x \neq 0$ such that $tx(t)=\lambda x(t)$ for all $t$. But then $x(t)=0$ whenver $t \neq \lambda$ but this implies that $x(t)=0$ almost evrywhere which means it is the zero element of $L^{2}$. Hence the point spectrum is empty.