Analysis Problem of the Day 9

The following exercise is Problem 1 on Stanford’s Fall 2022 Analysis Qual:

Problem 1: Let $Y \subseteq \mathbb{R}$ be a non-empty compact subset. Show that there exists a Hilbert space $H$ and a self-adjoint bounded linear operator $T: H \to H$ such that the spectrum of $T$ is exactly $Y.$

Solution: As with a lot of problems regarding (separable) Hilbert spaces, the prototypical example that one should be working with is $H=l^2.$ Our goal is to make the spectrum of $T$ equal precisely to $Y,$ that is, make $T-\lambda I$ invertible if and only if $\lambda \not \in Y.$ Now, one easy way to make an operator non-invertible is to force it to send non-trivial elements to zero, that is, have a non-empty kernel. However, with $l^2,$ one is working with at most countably many terms – so how does one ensure the kernel is nonempty for all of the (possibly uncountably many) elements of $Y?$

Well, one doesn’t. Instead, we will approximate all elements of $Y$ by a countable subset. In fact, since $Y$ is compact, $Y$ is separable, so by definition, it has a countable dense subset $Y' \subseteq Y.$ Let $\lambda_1, \lambda_2,...$ be an enumeration of elements of $Y'.$ Then, define the operator $T: l^2 \to l^2$ by

$T(a_1,a_2,...) = (\lambda_1 a_1, \lambda_2 a_2,...)$

Note that $T$ is bounded since

$\|Ta\|_{l^2} = \sum_i |\lambda_i a_i|^2 \leq (\sup_i |\lambda_i|) \sum_i |a_i|^2 \leq (\sup_{\lambda \in Y} |\lambda|) \|a\|_{l^2}.$

Claim 1: $Y' \subseteq \sigma(T)$ i.e. $Y'$ is contained in the spectrum of $T.$ This is true since the non-zero element $a \in H$ given by $(a)_j = \delta_{ij}$ satisfies $Ta =0$ and thus belongs to $\ker(T-\lambda_i I),$ so $T-\lambda_i I$ is not invertible.

Claim 2: $Y \subseteq \sigma(T).$ This can be seen in two ways. The easy way is to recall that the spectrum of a bounded operator is always compact, so if $Y' \subseteq \sigma(T),$ then the closure is also contained in the spectrum, i.e. $\overline{Y'} = Y \subseteq \sigma(T).$ More explicitly, if one were to attempt to construct an inverse $S=(T-\lambda)^{-1}$ for $\lambda \in Y \setminus Y',$ then

$S(a_1,a_2,...) = ((\lambda_1-\lambda)a_1,(\lambda_2-\lambda)a_2,...).$

However, this operator is not bounded, i.e. an inequality of the form $\|Sa\| \leq C\|a\|$ cannot hold for any $C>0.$ This is because for a sequence of elements $a_n \in l^2, (a_n)_i = \delta_{im},$ where $m=m(n)$ is chosen so that $\lambda_m \to \lambda$ as $n \to \infty$ (which is possible since we $Y'$ is dense in $Y$ ), we have that $\|Sa_n\| =1$ but $\|a_n\| \to 0$ as $n \to \infty.$ Thus, while $S$ would exist as an unbounded operator, it is still not bounded and therefore $\lambda \in \sigma(T).$

Claim 3: $Y^c \subseteq \rho(T).$ This can be seen by directly constructing the inverse $S=(T-\lambda I)^{-1}$ for $\lambda \in Y^c$ as

$S(a_1,a_2,...) = \left(\frac{a_1}{\lambda_1-\lambda},\frac{a_2}{\lambda_2-\lambda},...\right)$

Since $\inf_i |\lambda_i - \lambda| >0,$ we obtain that $S: l^2 \to l^2$ is bounded, so $\lambda \in \rho(T).$ Finally, $T$ is self-adjoint since it is a bounded normal operator with real spectrum.

Remark: The same argument easily extends to showing that for any compact $Y \subset \mathbb{C},$ there exists an operator $T: l^2 \to l^2$ with $\sigma(T)=Y.$

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