I am trying to solve the following problem:
Consider the operator
$$T:L^2(-1,3)\rightarrow L^2(-1,3),\ Tx(t):=tx(t).$$
Let $\lambda\not\in [-1,3]$. Prove directly that $T-\lambda I$ is invertible, and that $(T-\lambda I)^{-1}$ is bounded and defined on all of $L^2(-1,3)$.
$\textbf{My attempt:}$ Since $(T-\lambda I)x(t)=x(t)(t-\lambda)$ it’s clear that the inverse would be $$(T-\lambda I)^{-1}=\frac{1}{t-\lambda}.$$
Furthermore, since $t\in [-1,3]$ and $\lambda\not\in [-1,3]$ there are no issues with singularities of $(T-\lambda I)^{-1}$.
As for showing it is bounded and defined on all of $L^2(-1,3)$ I figured I would start showing that it is defined on $L^2(-1,3)$ since if I succeed with this boundedness should follow from that. $\textbf{Here is where I’m struggling:}$ I want to show that for $x\in L^2(-1,3)$ that
$$\int_{-1}^{3}|(T-\lambda I)^{-1}x(t)|^2\ dt<\infty$$
My idea is to consider some fixed $\lambda\in\mathbb{R}\setminus [-1,3]$. Then I would think that
$$\begin{align*} \int_{-1}^{3}|(T-\lambda I)^{-1}x(t)|^2\ dt &= \int_{-1}^{3}\frac{1}{|t-\lambda|^2}|x(t)|^2\ dt \\ &\leq\underset{t\in [-1,3]}{\sup}\left(\frac{1}{|t-\lambda|^2}\right)\int_{-1}^{3}|x(t)|^2\ dt \\ &<\infty \end{align*}$$
I’m not sure whether this approach is right. I assume that since $\lambda$ is fixed and $t\in [-1,3]$ then $\frac{1}{|t-\lambda|^2}$ should attain its supremum within the interval $[-1,3]$ and it should be finite, right? If this is correct, is there a more clever way of doing this?
Thanks in advance.
For $\lambda<-1$, then $|t-\lambda|\geq-1-\lambda>0$ for $t\in[-1,3]$. For $\lambda>3$, then $|t-\lambda|\geq\lambda-3>0$. So for $\lambda\notin[-1,3]$, there is a positive number $\eta_{\lambda}>0$ such that $|t-\lambda|\geq\eta_{\lambda}$, so $\sup_{t\in[-1,3]}1/|t-\lambda|^{2}\leq\eta_{\lambda}^{-2}<\infty$.