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It is well known that if $K$ is any compact set in $\mathbb{C}$, then there exist a bounded linear operator $T:l_2\to l_2$ such that $\sigma(T)=K$. My questions are:

Q1) Does there exist $T$, a bounded linear operator on $l_2$ such that the point spectrum $\sigma_p(T)$ is the given $K$? That is, the only eigenvalues are the complex numbers in $K$?

Q2) Does there exist $T$, a bounded linear operator on $l_2$ such that $\sigma(T)=\sigma_p(T)=K$. That is, there are no other points in the spectrum except the eigenvalues in $K$?

Clearly a positive answer to Q2) implies a positive answer to Q1).

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This doesn't answer the question, but you can get bounded open sets as point spectra by taking adjoints of multiplication operators on Hilbert spaces of analytic functions like Bergman spaces (which are unitarily equivalent to operators on $\ell_2$). An example on $\ell_2$ is the backward shift $(x_0,x_1,x_2,\cdots)\mapsto (x_1,x_2,x_3,\ldots)$ with point spectrum $\{z\in\mathbb C:|z|<1\}$. –  Jonas Meyer May 9 '12 at 3:14
    
@JonasMeyer Can any bounded open set be a point spectra this way? –  Theo May 9 '12 at 15:14
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If $G$ is your bounded open set, and $G^*=\{\overline z:z\in G\}$, then take $T$ to be the adjoint of multiplication by $z$ on the Bergman space of $G^*$. Another way to see that you can get any bounded open set as point spectrum is to take (countable) direct sums of rescaled and translated backward shift operators. If $S$ is the backward shift, then $rS+aI$ has point spectrum equal to the open disk of radius $r$ centered at $a$, and you can write any open set as a countable union of such disks. –  Jonas Meyer May 9 '12 at 15:35
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@Neal: Let $\{a_0,a_1,a_2,\ldots\}$ be a countable dense subset of $K$, and define $T$ by $T(x_0,x_1,x_2,\ldots)=(a_0x_0,a_1x_1,a_2x_2,\ldots)$. Then $\sigma(T)=K$. Or let $\mu$ be a regular Borel measure with support $K$, and let $T$ be (unitarily equivalent to) multiplication by $z$ on $L^2(\mu)$ (assuming $K$ is infinite). (See also math.stackexchange.com/q/28670/1424, math.stackexchange.com/q/35279/1424, math.stackexchange.com/q/35623/1424) –  Jonas Meyer May 12 '12 at 17:09
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@Jonas: I think I have a positive answer to Q2! It is rather involved, I'm writing it up at the moment... –  George Lowther May 19 '12 at 22:16
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2 Answers

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To answer Q1, we can say exactly when a subset of $\mathbb{C}$ can be realized as the point spectrum of a bounded linear operator on $\ell^2$.

Theorem: A set $K\subseteq\mathbb{C}$ is equal to the point spectrum of a bounded linear operator $T\colon \ell^2\to \ell^2$ if and only if $K$ is bounded and is the union of countably many closed sets (an Fσ set).

In fact, we can go a little bit further. If $K$ is contained in the open ball $B_R(0)$ for some $R > 0$ then it is possible to choose $T$ such that $\Vert T\Vert=R$.

Before constructing $T$ with the given point spectrum, I'll quickly prove the converse. If $K=\sigma_p(T)$ then $K$ is a bounded Fσ set (this will use the fact that $\ell^2$ is a Banach space whose dual is separable). First, $K$ is contained in the spectrum $\sigma(T)$, which is contained in the closed ball $\bar B_R(0)$ for $R=\Vert T\Vert$ (by standard properties of bounded operators), so is bounded. Next, let $B=\lbrace x\in \ell^2\colon\Vert x\Vert\le 1\rbrace$ be the closed unit ball in $\ell^2$ with the weak topology, which is compact. Then, $B^0\equiv B\setminus\lbrace0\rbrace$ is σ-compact. In fact, choosing any dense sequence $u_1,u_2,\ldots$ in $\ell^2$, we can explicitly write $B^0$ as the union of compact sets $$ B^0=\bigcup_{n=1}^\infty\lbrace x\in B\colon\vert\langle u_n,x\rangle\vert\ge1\rbrace. $$ Set $C=\lbrace (x,\lambda)\in B^0\times\mathbb{C}\colon Tx=\lambda x\rbrace$ with the product topology which, being a closed subset of $B^0\times\mathbb{C}$, is σ-compact. Defining the continuous map $\pi\colon C\to\mathbb{C}$ by $\pi(x,\lambda)=\lambda$, then, $\sigma_p(T)=\pi(C)$ is σ-compact and, hence, is an Fσ set.

Now, let's move on to constructing $T$ with norm $\Vert T\Vert=R$ and point spectrum a given bounded Fσ set $K\subseteq B_R(0)$. The general case follows from the case where $K$ is closed, since if we write $K=\bigcup_{n=1}^\infty K_n$ for closed $K_n$ then, choosing operators $T_n$ with $\Vert T_n\Vert=R$ and $\sigma_p(T_n)=K_n$, the direct sum $T=\oplus_n T_n$ is an operator on $\oplus_n\ell^2\cong\ell^2$ with $\Vert T\Vert=R$ and point spectrum $K$ as required.

So, suppose that $K\subseteq B_R(0)$ is closed. By scaling, we can assume that $R > 1$ and that $K\subseteq B_1(0)$. I'll construct a linear operator $T$ on the Hilbert space $H=\ell^2(\mathbb{N}^2)\cong\ell^2$ (I'm using $\mathbb{N}=\lbrace0,1,2,\ldots\rbrace$). Recall that $H$ consists of the functions $f\colon\mathbb{N}^2\to\mathbb{C}$ with finite norm $\Vert f\Vert^2=\sum_{m,n}\vert f(m,n)\vert^2$ and inner product $\langle f,g\rangle=\sum_{m,n}\overline{f(m,n)}g(m,n)$. Now, choose sequences of positive real numbers $r_i,\epsilon_i$ and complex $z_i$ ($i=1,2,\ldots$) such that $\bar B_{r_i}(z_i)$ is disjoint from $K$ and such that $K=\bar B_1(0)\setminus\bigcup_iB_{r_i}(z_i)$. It doesn't matter exactly what values $\epsilon_i$ take at the moment, I'll choose these in more detail in a moment. Define $T\colon H\to H$ by $$ Tf(m,n)=\begin{cases} f(m,n+1),&\textrm{if }m=0,\cr r_mf(m,n-1)+z_mf(m,n),&\textrm{if }m,n > 0,\cr \epsilon_m r_m f(0,0)+z_mf(m,n),&\textrm{if }m > 0, n=0. \end{cases} $$ The idea is that $T$ behaves as a left-shift (on $m=0$), so that the point spectrum is contained in $\bar B_1(0)$, joined to a set of right-shifts (on each $m > 0$) to remove the balls $\bar B_{r_i}(z_i)$ from the point spectrum. It can be seen that $T$ has norm $$ \Vert T\Vert\le\max_m\left(1,\vert z_m\vert+r_m\right)+\left(\sum_m\epsilon_m^2r_m^2\right)^{1/2}. $$ This satisfies $\Vert T\Vert\le R$ so long as we choose $\bar B_{r_i}(z_i)\subseteq B_R(0)$ and choose $\epsilon_i$ so that $\sum_i\epsilon_i^2r_i^2$ is small enough. And (if you like), by scaling $\epsilon_i$ up, we can ensure that $\Vert T\Vert=R$.

Now, a complex number $\lambda$ is in the point spectrum $\sigma_p(T)$ if and only if there is a nonzero $f\in H$ such that $Tf=\lambda f$. Plugging in $Tf$ from the definition above and solving the linear equations shows that, up to a scaling factor, $f$ must be given by $$ f(m,n)=\begin{cases} \lambda^n,&\textrm{if }m=0,\cr \epsilon_m\left(r_m/(\lambda-z_m)\right)^{n+1},&\textrm{if }m > 0. \end{cases} $$ If $\vert\lambda\vert\ge1$ then $\sum_n\vert f(0,n)\vert^2=\infty$, so we must have $\vert\lambda\vert < 1$. Similarly, if $\lambda\in\bar B_{r_i}(z_i)$ then $\sum_n\vert f(i,n)\vert^2=\infty$. So, in order that $f\in H$ we must have $\lambda\in K$, in which case, $$ \Vert f\Vert^2=\frac{1}{1-\vert\lambda\vert^2}+\sum_{m=1}^\infty\frac{\epsilon_m^2}{\vert\lambda-z_m\vert^2/r_m^2-1}. $$ Now, as $\bar B_{r_i}(z_i)$ is disjoint from $K$, the term $(\vert\lambda-z_i\vert^2/r_i^2-1)^{-1}$ is a continuous positive function of $\lambda\in K$, so is bounded above by some $d_i^2 > 0$. Then, $$ \Vert f\Vert^2\le\frac{1}{1-\vert\lambda\vert^2}+\sum_{m=1}^\infty\epsilon_m^2d_m^2. $$ So long as we choose $\epsilon_i$ so that $\sum_m\epsilon_m^2d_m^2 < \infty$, this implies that $f\in H$ for all $\lambda\in K$, so $\sigma_p(T)=K$.

Note: The operator $T$ I constructed here does not give a positive answer to Q2. For any $\omega\in S^1$ and $\epsilon > 0$, setting $f_\epsilon(m,n)=1_{\lbrace m=0\rbrace}\epsilon\omega^n(1+\epsilon)^{-n-1}$ gives $\Vert f_\epsilon\Vert=1$ and $\Vert(T-\omega)f_\epsilon\Vert\to0$ as $\epsilon\to0$, so $\omega\in\sigma(T)$. Therefore, $S^1\subseteq\sigma(T)$.

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I'm impressed. Thank you! Is there a reference for this Theorem? If it is new and you came up with it I think it would make a nice short paper. Thanks again! –  Theo May 14 '12 at 23:31
    
@Theo: It's new to me, and is just something I came up with after thinking about your problem for a bit. Once I realized that the point spectrum has to be F-sigma, so that the general case follows from the case with $K$ closed, it was just a matter of joining together shift maps to try and get the required closed point spectrum. I'm going to think some more about Q2, but its getting late so that's all for me for tonight. –  George Lowther May 14 '12 at 23:44
    
If I could vote this up twice, I would. –  Neal May 15 '12 at 13:33
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This question has been bugging me for a while but, now, I can finally give a positive answer to question 2. In fact, we can say exactly which sets occur as the spectrum and point spectrum of a linear operator on $\ell^2$.

Theorem. Let $K_1,K_2$ be subsets of $\mathbb{C}$. Then, there exists a bounded linear operator $T\colon\ell^2\to\ell^2$ with $\sigma_p(T)=K_1$ and $\sigma(T)=K_2$ if and only if (i) $K_2$ is a nonempty closed and bounded set and (ii) $K_1\subseteq K_2$ is an Fσ set (a countable union of closed sets).

First, these conditions on $K_1,K_2$ are necessary. By standard properties of bounded operators, $\sigma(T)=K_2$ must be closed, bounded and nonempty. Also, the point spectrum must be contained in the spectrum and, as shown in my other answer to this question (the answer to Q1), is an Fσ set.

Let's start with the case where $K=K_1=K_2$ is a nonempty closed and bounded set. I'll construct an example with the following properties.

$H$ is a separable Hilbert space and $T\colon H\to H$ is a bounded operator whose point spectrum includes the closed ball $\bar B_{1/2}(0)$. For any closed nonempty $K\subseteq\bar B_{1/2}(0)$, let $H_K$ be the closed subspace of $H$ generated by the eigenvectors of $T$ with eigenvalues in $K$ and let $T_K\colon H_K\to H_K$ be the restriction of $T$ to $H_K$. Then, $\sigma_p(T_K)=\sigma(T_K)=K$.

So, we can find examples with spectrum $\sigma(T)=\sigma_p(T)=K$ by restricting to subspaces $H_K$ of a single given separable Hilbert space $H$. Then, $H_K$ is a separable Hilbert space so is either isomorphic to $\ell^2$ or is finite dimensional (in which case the direct sum of a countable set of copies of $H_K$ is isomorphic to $\ell^2$). We can then find examples for any closed bounded nonempty set $K$ by scaling.

First, I'll set up a bit of notation. Choose a finite sequence $z_0,z_1,\ldots,z_{N-1}\in B_{1/2}(0)$ such that $\bigcup_{n=0}^{N-1}\bar B_{1/4}(z_n)\supseteq\bar B_{1/2}(0)$. Let $[N]=\lbrace0,1,\ldots,N-1\rbrace$ and define $[N]^\ast$ to be the disjoint union of $[N]^n$ over $n\in\mathbb{N}$ (I'm using $\mathbb{N}=\lbrace0,1,\ldots\rbrace$). Note that $[N]^\ast$ is countable. For each $a=(a_0,\ldots,a_{n-1})\in[N]^n$ set $z_a=z_{a_0}+2^{-1}z_{a_1}+\cdots+2^{-(n-1)}z_{a_{n-1}}$ and $r_a=2^{-n}$. Also set $a\cdot k=(a_0,\ldots,a_{n-1},k)\in[N]^{n+1}$ for each $k\in[N]$. By construction we have, $$ \begin{align} &B_{r_{()}}(z_{()})=B_1(0),\cr &\bar B_{\frac{r_a}{2}}(z_a)\subseteq\bigcup_{k\in[N]}\bar B_{\frac{r_{a\cdot k}}{2}}(z_{a\cdot k}) \end{align} $$ Now define a separable Hilbert space $V=\bigoplus_{a\in[N]^\ast}\ell^2$. Elements $f\in H$ are maps $f\colon[N]^\ast\to\ell^2$ with finite norm $\Vert f\Vert^2=\sum_{a\in[N]^*}\Vert f(a)\Vert^2$, and inner product $\langle f,g\rangle=\sum_{a\in[N]^\ast}\langle f(a),g(a)\rangle$. Define the shift map $S\colon\ell^2\to\ell^2$ by $S f(n)=f(n+1)$. Then, $\Vert S^n\Vert=1$ for all $n\ge 0$ in which case, for all $\lambda\in\mathbb{C}$ with $\vert\lambda\vert > 1$, $$ \begin{align} &(\lambda-S)^{-1}=\sum_{n=0}^\infty\lambda^{-n-1}S^n,\cr &\Vert(\lambda-S)^{-1}\Vert\le\sum_{n=0}^\infty\vert\lambda\vert^{-n-1}=(1-\vert\lambda\vert)^{-1}. \end{align} $$ So, $\sigma(S)\subseteq\bar B_1(0)$. Also, for $\vert\lambda\vert < 1$ set $x_{\lambda}=(1,\lambda,\lambda^2,\ldots)\in\ell^2$ so that $S x_\lambda=\lambda x_\lambda$. In fact, it is easily seen that every eigenector of $S$ with eigenvalue $\lambda$ is a multiple of $x_\lambda$. We also have $\Vert x_\lambda\Vert^2=(1-\vert\lambda\vert^2)^{-1}$.

Define the linear map $T\colon V\to V$ by $$ Tf(a) = (r_aS+z_a)f(a). $$ As $B_{r_a}(z_a)\subseteq B_1(0)$ this is a bounded map with norm $\Vert T\Vert\le1$. Note that $f\in V$ is an eigenvector of $T$ with eigenvalue $\lambda$ if and only if $f(a)$ is proportional to $x_{(\lambda-z_a)/r_a}$ whenever $\lambda\in B_{r_a}(z_a)$ and $f(a)=0$ otherwise.

Now, for any $\alpha,\beta\in\mathbb{C}$ with $\vert\alpha\vert+\vert\beta\vert < 1$ define $U_{\alpha,\beta}\colon\ell^2\to\ell^2$ by $$ U_{\alpha,\beta}f(n)=\sum_{k=0}^n\binom{n}{k}\alpha^{n-k}\beta^kf(n-k). $$ This is a bounded linear operator with norm $$ \begin{align} \Vert U_{\alpha,\beta}\Vert&\le\sum_{k=0}^\infty\sup_{n\ge k}\binom{n}{k}\vert\alpha^{n-k}\beta^k\vert\le\sum_{k=0}^\infty\sum_{n=k}^\infty\binom{n}{k}\vert\alpha^{n-k}\beta^k\vert\cr &=\sum_{n=0}^\infty\sum_{k=0}^n\binom{n}{k}\vert\alpha^{n-k}\beta^k\vert=\sum_{n=0}^\infty(\vert\alpha\vert+\vert\beta\vert)^n\cr &=\left(1-\vert\alpha\vert-\vert\beta\vert\right)^{-1}. \end{align} $$ It can be seen that this moves the eigenvectors of $S$ according to $U_{\alpha,\beta}x_\lambda=x_{\alpha\lambda+\beta}$ and, also, satisfies the commutation relation $SU_{\alpha,\beta}=U_{\alpha,\beta}(\alpha S+\beta)$ (you just need to check that this holds on the eigenvectors $x_\lambda$). Define the map $U\colon V\to V$ by $$ Uf(a)=\sum_{k\in[N]}U_{1/2,z_k}f(a\cdot k). $$ This has finite norm, $$ \Vert U\Vert\le\sum_{k\in[N]}\Vert U_{1/2,z_k}\Vert < \infty. $$ Furthermore, from the commutation relations for $S,U_{\alpha,\beta}$, it can be seen that $T$ and $U$ commute. Fixing any $c\in\mathbb{C}$ with $0 < \vert c\vert < 1$, let $H$ be the set of $f\in V$ with $Uf=cf$, which is a closed subspace of $H$. Let $T_H\colon H\to H$ be the restriction of $T$ to $H$.

With the (rather lengthy!) setup done, we can now show that $T_H$ has the required properties. Choosing any $\lambda\in\bar B_{1/2}(0)$ we can show that there is an $f\in H$ with $Tf=\lambda f$, so that $\bar B_{1/2}(0)\subseteq\sigma_p(T_H)$. By construction, there is a sequence $a_0,a_1,a_2,\ldots\in[N]$ such that, setting $a_{(n)}=(a_0,a_1,\ldots,a_{n-1})\in[N]^n$ then $\lambda\in\bar B_{\frac{r_{a_{(n)}}}{2}}(z_{(n)})$ for each $n\in\mathbb{N}$. Define $f\in H$ by $f(a_{(n)})=c^nx_{(\lambda-z_{a_{(n)}})/r_{a_{(n)}}}$ and $f(b)=0$ for all $b\in[N]^\ast$ not equal to any of the $a_{(n)}$. We need to check that this does have finite norm, so is really in $H$. As $\lambda\in\bar B_{\frac{r_{a_{(n)}}}{2}}(z_{(n)})$, we have $\Vert f(a_{(n)})\Vert^2\le\vert c\vert^{2n}(1-1/4)^{-1}$. So, $$ \Vert f\Vert^2\le\sum_n \frac43\vert c\vert^{2n}=\frac{4}{3(1-\vert c\vert^2)} < \infty. $$ Also, $T f=\lambda f$ so $\sigma_p(T_H)\supseteq\bar B_{1/2}(0)$ as required.

Now choose any nonempty closed $K\subseteq\bar B_{1/2}(0)$, let $H_K$ be the closed subspace of $H$ generated by eigenvectors of $T_H$ and $T_K\colon H_K\to H_K$ be the restriction of $T$ to $H_K$. We automatically have $K\subseteq\sigma_p(T_K)\subseteq\sigma(T_K)$, so it just needs to be shown that $\sigma(T_K)\subseteq K$. That is, fixing $\lambda\in\mathbb{C}\setminus K$, we need to show that $\lambda-T_K$ is invertible. Let $d$ be the distance from $\lambda$ to the set $K$, and choose $n\ge0$ so that $d > 2^{1-n}$. Define the linear map $W\colon V\to V$ by $$ Wf(a)=1_{\lbrace a\in[N]^m,m\ge n\rbrace}1_{\lbrace B_{r_a}(z_a)\cap K\not=\emptyset\rbrace}(\lambda-r_aS-z_a)^{-1}f(a). $$ Note that if $B_{r_a}(z_a)\cap K\not=\emptyset$ and $a\in[N]^m$ for some $m\ge n$ then $z_a$ is within $r_a$ of $K$. So, $\vert\lambda-z_a\vert\ge d-r_a$ and, hence, $$ \Vert(\lambda-r_aS-z_a)^{-1}\Vert\le r_a^{-1}(\vert\lambda-z_a\vert/r_a-1)^{-1}\le(d-2r_a)^{-1}\le(d-2^{1-n})^{-1}. $$ So $W$ is a bounded operator with $\Vert W\Vert\le(d-2^{1-n})^{-1}$.

If $f\in H$ is an eigenvalue of $T$ with eigenvector $\mu\in K$ then we have $f(a)=0$ whenever $B_{r_a}(z_a)\cap K=\emptyset$. So, $(\lambda-T)Wf(a)=W(\lambda-T)f(a)=f(a)$ for all $a\in[N]^m$ with $m\ge n$. So, if we also have $f\in H_K$, then $$ c^{-n}U^nW(\lambda-T)f=f. $$ So, $L\Vert(\lambda-T)f\Vert\ge\Vert f\Vert$ for all $f\in H_K$, where I have set $L=\vert c\vert^{-n}\Vert U^n\Vert$. We can define $(\lambda-T_K)$ first by defining on the eigenvectors with eigenvalues $\mu\in K$ (which is just multiplication by $(\lambda-\mu)^{-1}$, so is well defined), noting that this is a linear map on the space generated by the eigenvalues with norm bounded by $L$ and, hence, has a unique continuous extension to $H_K$. So, $\lambda\not\in\sigma(T_K)$, showing that $\sigma(T_K)\subseteq K$ as required.


Now, consider the case where $K_1$ is an Fσ set and $K_2$ is its closure (assumed to be nonempty and bounded). Then, we can write $K_1=\bigcup_{n=1}^\infty K_{1,n}$ for nonempty closed sets $K_{1,n}$. By scaling, we suppose that $K_2\subseteq\bar B_{1/2}(0)$. Let $H$ be the separable Hilbert space and $T\colon H\to H$ be the bounded linear map described above. Then, $H_{K_{1,n}}$ is the closed subspace of $H$ generated by the eigenvectors of $T$ with eigenvalues in $K_{1,n}$ and $T_{K_{1,n}}\colon H_{K_{1,n}}$ is the restriction of $T$ to $H_{K_{1,n}}$. As explained above, $\sigma_p(T_{K_{1,n}})=\sigma(T_{K_{1,n}})=K_{1,n}$. Also, $\Vert T_{K_{1,n}}\Vert\le\Vert T\Vert$. Set $\tilde H=\oplus_n H_{K_{1,n}}$ which is a separable Hilbert space, so is isomorphic to $\ell^2$. Define $\tilde T\colon\tilde H\to\tilde H$ by $\tilde T\oplus_n T_{K_{1,n}}$. This is a bounded linear map with $\Vert\tilde T\Vert\le\Vert T\Vert$ and point spectrum $\bigcup_n K_{1,n}=K_1$. It just needs to be shown that $\sigma(\tilde T)\subseteq K_2$ or, equivalently, for any $\lambda\in\mathbb{C}\setminus K_2$ that $\lambda-\tilde T$ is invertible. If its inverse exists, it must be of the form $(\lambda-\tilde T)^{-1}=\oplus_n(\lambda-T_{K_{1,n}})^{-1}$ which has norm $\sup_n\Vert(\lambda-T_{K_{1,n}})^{-1}\Vert$. It needs to be shown that this norm is finite. Note that, as $\lambda\not\in K_2$ then $\lambda-T_{K_2}$ is invertible, and $\lambda-T_{K_{1,n}}$ is just its restriction to $H_{K_{1,n}}$. So, $\Vert(\lambda-T_{K_{1,n}})^{-1}\Vert\le\Vert(\lambda-T_{K_2})^{-1}\Vert$ is bounded independently of $n$, as required.


Finally, consider the general case where $K_1$ is an Fσ set and $K_2\supseteq K_1$ is closed, bounded and nonempty. In the case where $K_1$ is empty then we need to find a bounded operator with empty point spectrum and given spectrum $K_2$. If $K_1$ is nonempty, then an operator of the form $T_1\oplus T_2$ on $\ell^2\oplus\ell^2$ will work, where $T_1$ has point spectrum $K_1$ and spectrum $K_2$. In either case, it reduces to finding a bounded linear operator $T$ with empty point spectrum and spectrum a given closed bounded and nonempty set $K$. Define the right-shift $S\colon\ell^2\to\ell^2$ by $Sf(n)=1_{\lbrace n\ge1\rbrace}f(n-1)/n$. Then, $S^kf(n)=1_{\lbrace n\ge k\rbrace}f(n-k)/(n(n-1)\cdots(n-k+1))$. So, $\Vert S^k\Vert=1/k!$. Therefore, for any $\lambda\in\mathbb{C}$ with $\lambda\not=0$ we have $$ \begin{align} &(\lambda-S)^{-1}=\sum_{n=0}^\infty\lambda^{-1-n}S^n\cr &\Vert(\lambda-S)^{-1}\Vert\le\sum_{n=0}^{\infty}\vert\lambda\vert^{-1-n}/n!=\vert\lambda\vert^{-1}\exp(\vert\lambda\vert^{-1}). \end{align} $$ So, $S$ has spectrum $\lbrace0\rbrace$. It is easily seen to have no eigenvalues, so empty point spectrum. Now choose a sequence $z_1,z_2,\ldots$ dense in $K$ and set $T=\oplus_n(S+z_n)$. This has empty point spectrum and spectrum containing $z_n$, so $\sigma(T)\supseteq K$. On the other hand, if $\lambda\not\in K$ then, letting $d$ be the distance of $\lambda$ from $K$, the operators $(\lambda-s-z_n)^{-1}$ have norm bounded by $d^{-1}\exp(d^{-1})$. So, $(\lambda-T)^{-1}=\oplus_n(\lambda-S-z_n)^{-1}$ is a bounded linear operator, and $\lambda\not\in\sigma(T)$. Therefore, $\sigma(T)=K$.

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This answer is so long now, that I guess there won't be too many people who read through in detail to check the correctness. I'm quite confident that it is all correct up to minor typos though. –  George Lowther May 20 '12 at 20:29
    
I don't have time to read it carefully right now, but I will will soon. Thank you for taking the time to think about these questions! –  Theo May 21 '12 at 15:55
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