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I've nearly completed finding the spectrum of $(Tf)(t)=\max[0,\cos(t)]f(t)$ for $f\in L^2[-\pi,\pi]$, but I cannot seem to finish it off.

Here's what I've done so far:

  • I used the fact that $||T||=1$ and that the spectrum is closed and bounded by $||T||$ to limit my search to $\lambda \in (-1,1)$.

  • I then started with the point spectrum, which yielded $\sigma_p = (0,1)$ (by asking for which $\lambda$ there's an $f\neq 0$ such that $T-\lambda I$ is not injective and using the fact that there must be at least one point $t_1$ for which $f(t_1)\neq 0$. This also shows why negative $\lambda$ fail).

  • I'm now left with checking whether $\lambda \in (-1,0)$ are also in the spectrum (they're not in the point spectrum, but are still candidates according to my first point).

What am I missing to complete this?

I did think of finding the inverse, which, if I'm not mistaken, is $(Ah)(t) = \frac{h(t)}{\max[0,\cos(t)]-\lambda}$. This inverse is well-defined for all $\lambda \neq \max[0,\cos(t)]$, so it is clearly defined for $\lambda <0$. Since algebraic inverses in Hilbert spaces are bounded, this is indeed the inverse operator. Am I right to then conclude that $\lambda<0$ are NOT in the spectrum, and so $\sigma = [0,1]$? If so, is this the most efficient way to go about it?

Also - if there are any tips on how best to approach such spectrum-related problems more efficiently I would love to hear.

K.defaoite
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Anon
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  • It is self adjoint. – copper.hat Sep 23 '22 at 19:26
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    The operator is of the form $Tf=gf,$ where $g$ is a continuous function. There is no point spectrum if the function $g$ is not constant on any subinterval of $[-\pi,\pi].$ The spectrum is equal $g[-\pi,\pi]=[0,1]$ as $g(t)=\max{0,\cos t}.$ – Ryszard Szwarc Sep 23 '22 at 19:41
  • @RyszardSzwarc Is it compact? Otherwise I'm not sure I understand how you concluded that the spectrum is $[0,1]$. – Anon Sep 23 '22 at 20:43
  • The spectrum of a bounded operator is always a compact set. See this https://math.stackexchange.com/questions/2787066/spectrum-of-multiplication-operator – Ryszard Szwarc Sep 23 '22 at 20:44
  • @RyszardSzwarc I meant the operator, sorry (otherwise not sure how you made the jump from the point spectrum observation to the spectrum). How did you eliminate the negatives? – Anon Sep 24 '22 at 04:54
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    The set of values of the function $\max{0,\cos x}$ coincide with the interval $[0,1].$ Moreover the function is equal $0$ for ${\pi \over 2}\le |x|\le \pi.$ That's why the number $0$ in the point spectrum. – Ryszard Szwarc Sep 24 '22 at 11:07
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    The operator of the form $Tf=gf$ is never compact, if $g\neq 0.$ – Ryszard Szwarc Sep 24 '22 at 11:20

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The point spectrum $\sigma_p(T)$ consists of all $\lambda$ for which there exists $f\in L^2[-\pi,\pi]\setminus\{0\}$ such that $Tf=\lambda f$. It is not hard to see that $0$ is the only value in the point spectrum, and the corresponding eigenspace consists of all functions in $L^2[-\pi,\pi]$ that are supported a.e. outside $[-\pi/2,\pi/2]$; this eigenspace is infinite-dimensional. All other values in $[0,1]$ are in the continuous spectrum of $T$. Every $\lambda\notin [0,1]$ is in the resolvent set of $T$ because $1/(f-\lambda)$ is bounded on $[-\pi,\pi]$.

Disintegrating By Parts
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  • Is it not the case that $f$, where $f(0)=1$ and $f(t)=0$ otherwise, is an eigenvector for $\lambda = 1$ (and, using a similar argument for the other $\lambda \in [0,1]$, we can conclude that $[0,1]$ is the point-spectrum)? – Anon Sep 24 '22 at 06:16
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    @Anon : The function $f$ you describe is the $0$ element of $L^2[-\pi,\pi[$ because it is equal a.e. to $0$. So $f$ is not a valid eigenvector. – Disintegrating By Parts Sep 24 '22 at 06:35