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?
Best Answer
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 $K$.