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Let $\Lambda : X \to X$ be a bounded linear operator on a Banach space $X$. My question is whether the set $$ \{\lambda \in \mathbb C: \lambda I - \Lambda \quad\text{is surjective} \} $$ is necessarily open. The above set is similar to the resolvent set of $\Lambda$, which is defined to be the set of all $\lambda \in \mathbb C$ such that $\lambda I - \Lambda$ is invertible; I know that the resolvent set is always open. However, what about the set above?

For reference, it was a problem on a past qualifying exam (see problem 6) to prove that the set is in fact open. I'm not sure if they meant to indicate the resolvent set, or if the problem is correct as stated.

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A related thread proving the stronger result that the space of surjective operators is open. – t.b. Mar 20 '12 at 10:20
@t.b. Thanks for pointing that out, the argument is very nice. – Nick Strehlke Mar 20 '12 at 14:17
up vote 2 down vote accepted

Let $\lambda$ such that $T := \lambda - \Lambda$ is surjective. Then $\bar T: X/\ker T \to X/\ker T$, $\bar T(x + \ker T) = Tx + \ker T$ is invertible, hence there is some $\varepsilon > 0$ such that $\bar T + \mu$ is invertible for $|\mu| < \varepsilon$.

Let $0 < |\mu| < \varepsilon$ and $y \in X$. Then there is some $x \in X$ such that $(\bar T + \mu)(x + \ker T) = y + \ker T$, which means that $Tx + \mu x + \ker T = y + \ker T$. So there is some $z \in X$ with $Tz = 0$ and $Tx + \mu x + z = y$. Then \begin{align*} (T + \mu)\left(x + \frac 1\mu z\right) &= Tx + \mu x + z\\ &= y. \end{align*} As $y$ was arbitrary, $T + \mu$ is surjective and the set in question is open.


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Thanks martini, this is very helpful. – Nick Strehlke Mar 20 '12 at 14:15

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