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


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 $[0,1]$.

  • 1
    $\begingroup$ Thank you for your answer - perhaps it wasn't clear from my question but unfortunately that is exactly the bit I'm struggling with! $\endgroup$
    – Wooster
    Apr 4 '15 at 21:14
  • $\begingroup$ I edited my comment. Hint : To find the functions $f_n$, try to a multiple of an indicator function centered at $\lambda$ with smaller and smaller support but such that the norm of $f_n$ is one. $\endgroup$ Apr 4 '15 at 21:26
  • $\begingroup$ Okay great - I can see that works. Can I ask what was the intuition behind that choice? I can't see how I could have arrived at that $\endgroup$
    – Wooster
    Apr 4 '15 at 22:13
  • $\begingroup$ First of all, just looking at the integral, we see that $(\lambda-t)$ is small if $t$ is close to $\lambda$, so here $f_n$ can have some weight. On the other hand, if $t$ is not close to $\lambda$, $(\lambda-t)$ is large, hence $f_n$ should be small. Moreover, you can see that $f_n$ is not an eigenvector, but as $n$ rises it gets closer to something that does look like an eigenvector. $\endgroup$ Apr 5 '15 at 15:39
  • $\begingroup$ Be sure to mark this question as answered if you're satisfied with the response. $\endgroup$ Apr 5 '15 at 17:45

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