I'm trying to calculate the spectrum of the linear operator $T: L^2(0,1) \to L^2(0,1)$ given by $T(f) \to tf(t)$.
I've found a few facts about this operator but I'm still struggling to find the exact spectrum.
- The norm of $T$ is 1 and so we know the spectrum is contained in the unit disc.
- $T$ is self-adjoint and so the spectrum is real and also all the spectrum is approximate (the point spectrum is empty) – so I just need to look for approximate eigenvalues.
I know now that I should look for functions $f_n$ with unit norm such that $\int_0 ^1 |\lambda – t|^2 (f_n(t)^2) dt \to 0$ and from the above I just need to check $\lambda \in [-1,1]$.
I always find it difficult to find the approximate spectrum and I don't know how I can possibly go about finding such $f_n$ so I would really appreciate some tips on how to go about finding the approximate point spectrum more generally as well!
Thank you
Best Answer
You know that an operator is invertible if and only if it's injective and surjective. It's clear that $(\lambda-T)$ is always injective, hence the point spectrum is empty. Now try to figure out surjectivity. You should find that the spectrum is $[0,1]$.
Edit : If $(\lambda-T)$ is surjective, then for each $g\in L^2([0,1])$ there should be an $f\in L^2([0,1])$ such that $(\lambda-T)f(t)=g(t)$ a.e., equivalently $f(t)=\frac{g(t)}{(\lambda-t)}$ for almost all $t\in [0,1]$. Clearly when $\lambda\notin [0,1]$ this is not a problem.
When $\lambda\in [0,1]$, you can explicitly write down a function $g\in L^2([0,1])$ that is not in the image of $(\lambda-T)$, hence the spectrum is $[0,1]$.