# Confusing definition of resolvent set

1. Aug 7, 2011

### AxiomOfChoice

The definition I have found in a couple of places is the following: We have $\lambda \in \rho(T)$ for a bounded linear operator $T$ on a Banach space $X$ iff $(T-\lambda)$ is bijective with a bounded inverse. (This seems to be equivalent to just saying that $(T-\lambda)$ is both injective and surjective, since this implies $(T-\lambda)^{-1}$ is bounded by the inverse mapping theorem.) But in class, my professor said that $\lambda \in \rho(T)$ iff $(T-\lambda)^{-1}$ exists and $\mathcal R(T-\lambda) = X$.

I don't see why my professor's definition doesn't contain superfluous information. Obviously, if $\mathcal R(T-\lambda) = X$, then $(T-\lambda)$ is surjective. But doesn't saying "$(T-\lambda)^{-1}$ exists" contain the surjectivity statement? I mean, why not just say: "We need (1) $(T-\lambda)$ injective and (2) $(T-\lambda)$ surjective" and be done with it? It just seems he's said more than he really needs to.

Of course, if $\mathcal R(T-\lambda) \neq X$, I guess that $(T-\lambda)^{-1}$ only makes sense as a map $\mathcal R(T-\lambda) \to X$...so you can't really just say "$\lambda \in \rho(T)$ iff $(T-\lambda)^{-1}$ exists as a bounded operator" without there being some ambiguity...because can't $(T-\lambda)^{-1}: \mathcal R(T-\lambda) \to X$ be bounded without having $\mathcal R(T-\lambda) = X$?

2. Aug 7, 2011

### micromass

Staff Emeritus
You are absolutely correct, when dealing with the classical functional analysis.
However, your professor most likely likes to work with unbounded operators. With unbounded operators, it's normal to have operators that are only densely defined (that is: only defined on a dense subset). So it's ok if the inverse of an operator is only defined on a dense subset. What your professor asks is that the operator is actually defined on the entire set, and not only on the dense subset.

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