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


1 Answer 1


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
    Commented Apr 4, 2015 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. $\endgroup$ Commented Apr 4, 2015 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
    Commented Apr 4, 2015 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. $\endgroup$ Commented Apr 5, 2015 at 15:39
  • $\begingroup$ Be sure to mark this question as answered if you're satisfied with the response. $\endgroup$ Commented Apr 5, 2015 at 17:45

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