Let $T: l_2(\mathbb{N}) \to l_2(\mathbb{N})$ be an operator given by

$$Tx = \left( \frac{n-1}{n+1} x_n \right)_{n \in \mathbb{N}} \text{ where } x = (x_n)_{n \in \mathbb{N}} \in l_2(\mathbb{N}).$$

My goal is it to determine the spectrum $\sigma(T)$, the point spectrum $\sigma_p(T)$, the continuous spectrum $\sigma_c(T)$ and the residual spectrum $\sigma_r(T)$.


  • Since $\langle Tx,y\rangle = \langle x,Ty\rangle$ for all $x,y \in l_2(\mathbb{N})$ we observe that $T=T^*$ is self-adjoint and hence closed. Therefore, we know that $\sigma(T)$ is the disjoint union of all the other spectra.

  • We define $M := \{ \frac{n-1}{n+1} | n \in \mathbb{N}\}$.

  • Consider the case $\lambda \in \mathbb{C} \setminus M$. Let $y \in l_2(\mathbb{N})$ be arbitrary. Then $$(T-\lambda)x = y \Longleftrightarrow \forall n \in \mathbb{N}: x_n = \left(\frac{n-1}{n+1} - \lambda \right)^{-1} y_n,$$ so we have a unique solution $x \in l_2(\mathbb{N})$. Therefore, $T-\lambda$ is bijective, so $M \subset \rho(T) := \mathbb{C} \setminus \sigma(T)$.

  • Now consider the case $\lambda \in M$, say $\lambda = \frac{k-1}{k+1}$ for a $k \in \mathbb{N}$. Then $$(T-\lambda)e^{(k)} = (T-\lambda)0 = 0$$ where $e^{(k)}$ denotes the $k$-th unit vector which is in $l_2(\mathbb{N})$. Since $e^{(k)} \neq 0$ we have $T-\lambda$ is not injective and we get $\sigma(T) = \sigma_p(T)=M$ and $\sigma_c(T) = \sigma_r(T) = \emptyset$.

Thank you!

  • 1
    $\begingroup$ Hint 1: operator $T$ is bounded, defined on $\ell^2$, hence closed. $\endgroup$ – Yurii Savchuk Feb 6 '18 at 16:14
  • 1
    $\begingroup$ Hint 2: the set $M$ is not closed, the spectrum must be a closed set. $\endgroup$ – Yurii Savchuk Feb 6 '18 at 16:14
  • $\begingroup$ Thanks for your response! Regarding Hint 1: Yeah, your solution is much shorter, thank you! But could you tell me if my approach with the self-adjointness is still correct? Regarding Hint 2: I think I have to choose $M = \{\frac{n-1}{n+1} | n \in \mathbb{N} \} \cup \{1\}$ instead, am I right? If $\lambda = 1$ the unique solution from above is not in $l_2(\mathbb{N})$, is that correct? $\endgroup$ – Diglett Feb 6 '18 at 16:29
  • $\begingroup$ Yes, that is the spectrum. For $\lambda=1$ the image of $(T-\lambda)$ is dense in $\ell^2$ but not equal to $\ell^2$. $\endgroup$ – Yurii Savchuk Feb 6 '18 at 16:33
  • $\begingroup$ Why is the image of $T-1$ not equal to $\ell^2$? Let me denote by $R(T-1)$ the image or range of $T-1$. Since $(T-1)^* = T-1$ (because $T$ and the identity are self-adjoint operators) we have $R(T-1) = \ker(T-1)^\perp$ and $\overline{R(T-1)} = \ker(T-1)^\perp$ too, so they are equal. Where is my mistake? $\endgroup$ – Diglett Feb 6 '18 at 17:51

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