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Let $K$ be a non-empty compact subset of $\mathbb{C}$, the complex field. Does there exist an operator $A\in \mathscr{B}(C[0,1])$ such that $\sigma(A)=K$?

$A$ is a multiplication operator iff $K$ is non-empty compact, connected and locally connected set in $C$. As the Hahn-Mazurkiewicz theorem states that:

A non-empty Hausdorff topological space is a continuous image of the unit interval if and only if it is a compact, connected, locally connected second-countable space.

How about the other kinds of compact sets? Any help would be appreciated!

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David Lee
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1 Answers1

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Yes. The reason is that $C[0,1]$ is isomorphic as a Banach space to $c_0(C[0,1])$, the $c_0$-sum of countably many copies of itself. Let $E=c_0(C[0,1])$ and denote by $Q_n$ the projection onto the $n^{{\rm th}}$ copy of $C[0,1]$ in $E$.

Let $K$ be a non-empty compact subset of the complex plane and let $(a_n)_{n=1}^\infty$ be a sequence whose entries are dense in $K$. Define an operator $T\colon E\to E$ such that for each $x\in Q_n(E)$ we have $$Tx = a_n x.$$ All $a_n$ ($n\in \mathbb{N}$) are eigenvalues of $T$. Moreover, they are dense in the spectrum, so $\sigma(T)=K$.

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Tomasz Kania
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  • Thank you for your help. Could you explain more clear about the first sentence of your answer? What is $c_0$-sum? Why is $C[0,1] \cong c_0(C[0,1])$? – David Lee Feb 21 '16 at 07:29
  • @LiJingyang, this follows from Milutin's theorem (see https://www.ma.utexas.edu/users/rosenthl/pdf-papers/92.pdf); it says that for each uncountable compact metric space $K$, $C(K)$ is isomorphic to $C[0,1]$. Here $c_0(C[0,1])\cong c(C[0,1])) \cong C( [0,\omega]\times [0,1])$. – Tomasz Kania Feb 21 '16 at 11:57
  • ,$C[0,1]$ is isomorphic to $c_0(C[0,1])$, but not isometric. right? – David Lee Feb 25 '16 at 04:27
  • @LiJingyang, of course but $B(X)$ is Banach-algebra isomorphic to $B(Y)$ if and only if $X$ is Banach-space isomorphic to $Y$. – Tomasz Kania Feb 25 '16 at 08:15
  • @Tomek.But,I think that spectra is not invariant under Banach-algebra isomorphic. – David Lee Feb 25 '16 at 11:14
  • @LiJingyang why not? Let $A$ and $B$ be unital Banach algebras and let $f$ be an isomorphism from $A$ to $B$. Take $a\in A$ and a complex number $\lambda$. Then $\lambda 1_A - a$ is invertible if and only if $f(\lambda 1_A - a)=\lambda 1_B - f(a)$ is invertible. – Tomasz Kania Feb 25 '16 at 11:27
  • @Tomek.One more question: $X$ is Banach-space isomorphic to $Y$ means there exist a invertible operator $T\in \mathbb{B}(X,Y)$. right? – David Lee Feb 25 '16 at 11:54
  • @LiJingyang, yes. – Tomasz Kania Feb 25 '16 at 12:56
  • @Tomek.thank you again. – David Lee Feb 25 '16 at 12:58