I am doing an assignment regarding the operator $T:L^2(-1,3)\rightarrow L^2(-1,3),\ Tx(t)=tx(t).$ More specifically, I am considering the operator $T-\lambda I$. I am supposed to show that for $\lambda\in [-1,3]$ that $T-\lambda I$ is invertible in the sense that it is injective (the definition on invertibility from the book we use), but I am having issues with that.
$\textbf{What I have done:}$ The operator $T-\lambda I$ is linear so it suffices to show that if $(T-\lambda I)x(t)=0\Rightarrow x(t)=0$. But since $(T-\lambda I)x(t)=x(t)(t-\lambda)=0$ this can only be the case if $\lambda\not\in [-1,3]$ since if both $t\in [-1,3]$ and $\lambda\in [-1,3]$ then it is possible for $t-\lambda=0$ hence $x(t)$ does not necessarily need to be $0$ thus I can not manage to show that the operator is injective.
Am I doing something wrong or is it perhaps a mistake that $\lambda\in [-1,3]$? Any help is highly appreciated. Thanks in advance.
Remember that $L^2([-1,3])$ is a function space and $T:L^2([-1,3])\to L^2([-1,3])$, therefore we says that $(T-\lambda I)f=0$ if the LHS of the previous equality is the zero function.
Then you need to prove that $(T-\lambda I)f=0$ if and only if $f$ is the zero function. Then you have that $$ (T-\lambda I)f=0\iff g(t):=f(t)(t-\lambda )=0 \text{ almost everywhere }\tag1 $$ for some chosen $f\in L^2([-1,3])$. Now note that the function defined by $h(t):=t-\lambda $ cannot be the zero function for any chosen constant $\lambda \in \Bbb C $ because the set $\{t\in [-1,3]:t\neq \lambda \}$ have full measure, that is, $\lambda (\{t\in [-1,3]:t\neq \lambda \})=4$. Hence $\rm(1)$ holds only if $f$ is the zero function (shortly written as $f=0$).