# An example of an operator whose spectrum is the unit square

Is there an example of an operator, the spectrum of which is a unit square $[0, 1] \times [0, 1] \subset \mathbb{C}$?

• And I always thought the spectrum was a subset of $\Bbb C$. – Angina Seng Dec 18 '17 at 21:22
• You are right. There is a square in the complex plane. – Victor Dec 18 '17 at 21:27

Take any $K\subset\mathbb C$, compact. Let $\{q_j\}_{j\in\mathbb N}\subset K$ be dense. Now for $T\in B(\ell^2(\mathbb N))$ by $$Te_j=q_je_j,$$ where $\{e_j\}$ is the canonical basis.
Then all $q_j$ are eigenvalues of $T$. The spectrum is closed, so $K\subset \sigma(T)$. If $\lambda\in \mathbb C\setminus K$, the compactness of $K$ guarantees that there exists $\delta>0$ with $|\lambda-t|\geq\delta$ for all $t\in \sigma(T)$. Then the operator $S$ given by $$Se_j=\frac1{q_j-\lambda}\,e_j$$ is bounded with $\|S\|\leq\frac1\delta$, and $S(T-\lambda I)=(T-\lambda I)S=I$. So $\mathbb C\setminus K\subset\mathbb C\setminus\sigma(T)$, which implies $\sigma(T)\subset K$.
Thus $\sigma(T)=K$.
Let $Mf = zf(z)$ be the multiplication operator on $L_{\mu}^2(S)$ where $S$ is the square $S=\{ z \in \mathbb{C} : 0 \le \Im z \le 1, 0 \le \Re z \le 1 \}$, and where $\mu$ is Lebesgue measure on $S$. Then the spectrum of $M$ is $S$. Using various measures on $\mathbb{C}$ and the operator $M$, you can design $M$ to have any closed compact set as its spectrum.