We define an operator $T$ on $X=L^2([0,1],\mathbb{R})$ as $T(f)(x)=min(2x,1)f(x)$ for all $x \in [0,1]$.
I'd like to find the spectrum of $T$. I computed the operator norm, which is $||T||=1$, so we know that $\sigma(T) \subset [-1,1]$. More specifically, I'd like to find the point spectrum $\sigma_p(T)$, i.e. all the $\lambda \in \mathbb{R}$ such that $\lambda Id_X-T$ is not injective.
Let's assume the opposite, meaning that there exists a non-zero function $f$ in $X$ such that $(\lambda Id_X-T)f(x)=0$. , i.e. $\lambda=min(2x,1)$
I think the correct answer is $\sigma_p(T)={1}$, but I don't understand why. My main issue is that $1$ does not seem to be in $\sigma_p(T)$ for $x \in (1/2,1]$, as we'd have $min(2x,1)=1 =\lambda$ (in the equality in the previous paragraph) and thus $(\lambda Id_X-T)$ would be injective. There must be something I'm not understanding correctly...
Finally, how does one find the continuous and residual spectrum? What is the procedure (a lot of detail would be very apprectiated)?