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I was reading Kreyszig’s Introductory Functional Analysis with Applications some time ago in which the author discusses bounded linear operators $T$ on complex Banach spaces $X$, proving results such as that the spectrum $\sigma(T)$ ($\lambda\in\mathbb{C}$ such that $(T-\lambda I)^{-1}$ is not a bounded operator defined on a set dense in $X$) is closed and (using complex analysis) that it is nonempty and has spectral radius $\sup{|\lambda\in\sigma|}=\lim_\limits{n\rightarrow\infty}{||T^n||^{\frac{1}{n}}}$ (where the norm of an operator is defined in the usual way as $||T||=\sup_\limits{||x||=1}{||Tx||}$). The use of complex analysis and bounding within a disk made me think for some reason of the Mandelbrot set and wonder whether there could exist a bounded linear operator whose spectrum ‘happened’ to be the Mandelbrot set, and if so how it would be found. Specifically are there higher-order constraints on the shape of the spectrum of these operators? I could not think of any immediate way of doing the problem in reverse to try to define an operator whose resolvent would not exist at specified locations, or perhaps this is actually trivial (the book did not touch on this topic too deeply)? I could find nothing relevant when I searched for this sort of thing.

Thus my question is the following: can bounded linear operators on complex Banach spaces have relatively arbitrary shapes in the complex plane and if not what are the constraints? Finally, within those constraints is it possible to find possible operators and Banach spaces (since we constrain neither) for given spectral shapes?

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  • $\begingroup$ In the case we are working with bounded operators on a Hilbert space, given an arbitrary compact set $\Omega$ in $\mathbb{C}$ I think one should be able to construct a projection-valued measure on the Borel sets of $\Omega$ and then define a bounded operator by integrating the identity function against this measure. Let me think and see if this construction can be made precise. $\endgroup$ Commented Nov 11, 2016 at 0:07
  • $\begingroup$ That looks like an interesting way to do it. $\endgroup$ Commented Nov 11, 2016 at 0:20
  • $\begingroup$ @MartinArgerami's answer is essentially this in the case of $\ell^2$. Given the construction $\{e_n\}\leftrightarrow c_n$ and a Borel set $U\subset K$ associate to $U$ the orthogonal projection onto the span of the $e_n$ contained in $U$. $\endgroup$ Commented Nov 11, 2016 at 0:24
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    $\begingroup$ Also, if I may be pedantic for a moment, the answer is negative in the case of a finite-dimensional Banach space. :) $\endgroup$ Commented Nov 11, 2016 at 0:30

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The specific answer is very easy. The spectrum of a bounded operator is always a compact subset of $\mathbb C$. Conversely, if we consider for instance bounded operators on a Hilbert space, for any compact $K\subset\mathbb C$ there exists $T\in B(H)$ with $\sigma(T)=K$. In particular, since the Maldelbrot set is compact, it is the spectrum of a certain operator $T\in B(H)$.

The construction is not very exciting, actually. Consider $H=\ell^2(\mathbb N)$ and fix an orthonormal basis $\{e_n\}$. Given $K\subset\mathbb C$ compact, let $\{c_n\}$ be a countable dense subset of $K$. Then define $T\in B(H)$ by $$ Te_n=c_ne_n $$ and extend by linearity. Each $c_n$ is an eigenvalue of $T$ by construction. As the spectrum is closed, the closure of $\{c_n\}$ is in $\sigma(T)$, so $K\subset \sigma(T)$. And if $c\not\in K$, then there exists $\delta>0$ with $|c-c_n|>\delta$ for all $n$. Then, given $x=\sum x_ne_n\in H$, $$ \|(T-cI)x\|^2=\left\|\sum_n x_n(c_n-c)e_n\right\|^2=\sum_n|x_n|^2\,|c_n-c|^2\geq\delta^2\,\sum_n|x_n|^2=\delta^2\|x\|^2. $$ Thus $T-cI$ is bounded below; it is easy to show that it is surjective, and so it is invertible. Thus $\sigma(T)=K$.

Note also that $T$ is normal, so we don't need $T$ to be very exotic.

The following argument is actually the same as above, but presented in a different way. Given $K\subset \mathbb C$ compact, we may consider the Banach algebra $C(K)$ (complex-valued continuous functions over $K$). We can see these as linear operators acting (by multiplication) on $L^2(K)$. Then the spectrum of each function is the closure of its range. In particular, the spectrum of the identity function $f(z)=z$ is $K$.

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  • $\begingroup$ +1 I see. It's interesting that it is so simple. I was wondering about your last statement though; I'm not so familiar with normal operators, but if the $c_n$ are dense for instance in the Mandelbrot set then I would have expected $T$ would have to be reasonably complicated even if not in any way elegant? $\endgroup$ Commented Nov 11, 2016 at 0:15
  • $\begingroup$ Well, you can see the argument as saying that normal operators can be complicated :) Actually, that reminded me of something that I will add to the answer. $\endgroup$ Commented Nov 11, 2016 at 0:18
  • $\begingroup$ This is a construction on $\ell^2$ but what if we relax to considering an arbitrary Banach space $X$? I think then we should take a Schauder basis of $X$ and the argument should still carry through. $\endgroup$ Commented Nov 11, 2016 at 0:20
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    $\begingroup$ @Neal: sounds natural, but I see two problems: first, not every Banach space has a Schauder basis; and, second, even if it has one, it is not immediately obvious to me that defining an operator in the way above will give a bounded operator. $\endgroup$ Commented Nov 11, 2016 at 0:24
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    $\begingroup$ To say that $T$ is bounded, you should be able to relate the norms of $Tx$ and of $x$. What you want to do is define $T(ax+by)=acx+bdy$. To say that $T$ is bounded, you should prove that there exists a constant $k$ (not dependent on $x$ and $y$) such that $\|acx+bdy\|\leq k\|ax+by\|$. There is no general way of doing this. It works in a Hilbert space when you have an orthonormal basis, and there might be other ad-hoc tricks in other spaces, but there is no general result. $\endgroup$ Commented Nov 11, 2016 at 1:10

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