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This question is from Kreyszig 7.3, #4-6:

Let $T: \ell^2 \rightarrow \ell^2$ be defined by $y = Tx, x = (x_i), y = (y_i), y_i = a_ix_i$ where $(a_i)$ are dense in [0,1]. Find $\sigma_p(T)$ and $\sigma(T)$. Moreover, show that if $\lambda \in \sigma-\sigma_p$ then $R_\lambda(T)$ is unbounded. (The definitions of spectrum can be found here)

Extending the foregoing problem: Find a linear operator $T: \ell^2 \rightarrow \ell^2$ whose eigenvalues are dense in a given compact set $K \subset \mathbf{C}$ and $\sigma(T) = K$.

So far: My only steps so far were recognizing that $(a_i)$ should be in $\sigma_p$. I'm having trouble exhibiting a sequence that's in the spectrum but not in the point spectrum. I tried to use the second part of the first question by thinking about what would make $(T-\lambda I)$ unbounded but to no avail.

Any thoughts? Thanks!

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1 Answer 1

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Let $A$ stand for the (dense) set of $(a_n)_n$. Since $\sigma(T)$ is closed and $A\subseteq\sigma(T)$, it follows that $[0,1]=\overline{A}\subseteq\sigma(T)$. On the other hand, if $\lambda\notin[0,1]$, then $S$ defined by $Se_j=\frac{1}{\lambda-a_j}e_j$ for any $j\in\mathbb{N}$, is a bounded (diagonal) operator whose inverse is $\lambda I-T$.Thus $\sigma(T)=[0,1]$.

If $\mu$ is an eigenvalue, then we have $Tx=\mu x$ for some non-zero $x=\sum b_je_j$. Then $Tx=\sum a_j b_j e_j=\sum \mu b_j e_j$. Hence, for any $j\in\mathbb{N}$, $a_j b_j=\mu b_j$, which only can happen if $\mu=a_k$ for some $k$. Thus $\sigma_p(T)=A$.

The generalization has almost identical proof. Take $A$ to be a countable, dense subset of $K$, and define a diagonal operator as above.

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  • $\begingroup$ Thx! - last part: Why is $R_\lambda(T)$ unbounded? So far - $ ||R_\lambda(T)x||<c||x||$ is ($\sum{(\frac{x_i}{\lambda_i-a_i})^2})^{\frac{1}{2}}<c(\sum{x_i}^2)^{\frac{1}{2}}$. Why is this unbounded with $\lambda_i$'s are irrational in [0,1]? $\endgroup$
    – yoshi
    Feb 20, 2017 at 11:30
  • $\begingroup$ o wait, i see :-) $\endgroup$
    – yoshi
    Feb 20, 2017 at 11:35

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