1
$\begingroup$

I have been studying functional analysis lately, specifically spectrum of operators. I know how to find the spectra of an operator, but what if I have the spectra and I want to find an operator with such spectra?

Let's say I have $\delta(T) = \{3i\}$ as the spectra of some linear and bounded operator T. I guess that T could be something like this:

$ T: l^2 \to l^2$, such that $\{x\}_j \to 3i\{x\}_j$ could be a trivial candidate because $Tx = \lambda x$. However what if I have $\delta(T) = \{3i,5i,7i\}$ for example?

$\endgroup$
  • 1
    $\begingroup$ You can take an orthonormal basis $e_1,e_2,...$ for $l^2$. Multiply $e_1$ by $3i$, $e_2$ by $5i$ and all the rest by $7i$. In other words $(x_1,x_2,x_3,x_4,...)\mapsto (3i x_1,5ix_2,7ix_3, 7ix_4,...)$. $\endgroup$ – Yanko Feb 25 at 13:56
  • $\begingroup$ @Yanko Can I also just take $(3ix_1, 5ix_2, 7ix_3, x_4,x_5,x_6,...)$? $\endgroup$ – qcc101 Feb 25 at 14:07
  • 1
    $\begingroup$ No, because then $1$ is also in the spectrum. $\endgroup$ – Yanko Feb 25 at 14:07
1
$\begingroup$

It is simple to do so because the multiplication operator $T: L^2(\mathbb{R}) \to L^2(\mathbb{R})$, $f \mapsto g f$, with some bounded function $g: \mathbb{R} \to \mathbb{C}$, has as spectrum the closure of the range of $g$.

So for your first example, you take the constant function $g: g(x) = 3i$ and get the spectrum $\{ 3i\}$. Similarly you can choose any function with $g(\mathbb{R}) = \{3i,5i,7i\}$ for your second example (take something peacewise constant for example) or similarly in any example you can think of.

$\endgroup$
  • $\begingroup$ (+1) I think $l^2$ in this context means $l^2(\mathbb{N})$ or $l^2(\mathbb{Z})$. However the idea is the same. $\endgroup$ – Yanko Feb 25 at 13:58
  • $\begingroup$ Oh, I was actually thinking about $l^2$ as sequences, is that the same? This one: en.wikipedia.org/wiki/Sequence_space $\endgroup$ – qcc101 Feb 25 at 14:01
  • $\begingroup$ @qcc101 yes the idea is the same. Just consider $g$ as a function from $\mathbb{N}$ to $\mathbb{C}$ instead. $\endgroup$ – Yanko Feb 25 at 14:06
  • 2
    $\begingroup$ On $L^2(\mathbb{R})$, the spectrum of the multiplication operator is not the closure of the range of $g$, but the essential range (you can change $g$ on a set of measure zero without changing the operator). $\endgroup$ – MaoWao Feb 25 at 14:06
  • $\begingroup$ @MaoWao: Yes, that's right. $\endgroup$ – Luke Feb 28 at 11:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.