# Non-empty resolvent set, then operator closed?

On Hilbert spaces, the following is true:

Let $T$ be a densely-defined linear operator with non-empty resolvent set, then $T$ is closed.

The obvious proof I see to show this uses explicitly the Hilbert space structure which is why I would like to ask:

Is the same result true for operators on Banach spaces?

What I would consider the obvious proof uses only the Banach space structure.

If $\lambda$ is in the resolvent set, the graph $G(T)$ of $T$ maps in an obvious way to the graph of $(T-\lambda I)^{-1}$: $G(T) = f^{-1}(G((T-\lambda I)^{-1}))$ where $$f:\;(x, y) \mapsto (y-\lambda x, x)$$ Since $f$ is continuous from $X \times X$ to itself and $G((T-\lambda I)^{-1})$ is closed, $G(T)$ is closed.

(This is really a very long comment...)

I think maybe the actual question comes about because some of the terminology in this area is hazy. Let $T:X\supseteq D(T)\rightarrow X$ be a linear operator on a Banach space $X$. For example, Resolvent set, wikipedia defines $\lambda\in\mathbb C$ to be in the resolvent if:

1. $T-\lambda I$ injects;
2. $(T-\lambda I)^{-1}$ is bounded; and
3. $(T-\lambda I)^{-1}$ is densely defined

Under this definition, it is not true that having non-empty resolvent set implies closed. For example, let $X=\ell^2$, let $D(T) = c_{00}$ be the space of eventually 0 sequences, and define $T((x_n)) = (nx_n)$. Set $\lambda=0$ and check the conditions: $T$ is bijective between $c_{00}$ and $c_{00}$; $T^{-1}((x_n)) = (n^{-1}x_n)$ is bounded; $c_{00}$ is dense in $\ell^2$. But $T$ is not closed; only closable.

Let's be a bit more precise. For any $T:X\supseteq D(T)\rightarrow X$, if $T$ is injective then we may define $T^{-1}:D(T^{-1})\rightarrow X$ by setting $D(T^{-1})$ to be the image of $T$, and defining $T^{-1}(T(x)) = x$ for $x\in D(T)$. This is well-defined as $T$ is injective. Let's compare the graphs: $$\mathcal{G}(T) = \{ (x,T(x)) : x \in D(T) \}, \quad \mathcal{G}(T^{-1}) = \{ (T(x),x) : x\in D(T) \},$$ so clearly $T$ is closed if and only if $T^{-1}$ is. As Robert Israel observes, it is also true that $T$ is closed if and only if $T-\lambda I$ is closed, and hence $(T-\lambda I)^{-1}$ closed does imply that $T$ is closed.

There are further definitions of "resolvent set". E.g. Engel, Nagel define $\lambda$ to be in the resolvent set if $T-\lambda I:D(T)\rightarrow X$ is bijective, that is, $T-\lambda I$ has range the whole of $X$. They also by definition already assume $T$ is closed, so that then $(T-\lambda I)^{-1}$ is closed and hence by the Closed Graph Theorem, bounded.

STANDING ASSUMPTIONS: Let $$T:D_T\rightarrow X$$ be a linear operator, where $$X$$ is a normed space (or metrizable TVS) and $$D_T\subset X$$.

DEFINITION: Then $$\lambda$$ belongs to the resolvent set $$\rho(T)$$ iff $$T-\lambda I$$ is one-to-one and onto and its inverse is bounded.

LEMMA. A linear operator $$T$$ in a normed space $$X$$ is closed if$$^\dagger$$ it has a non-empty resolvent set.

PROOF. Let $$D_T\owns x_n\rightarrow x$$, $$Tx_n\rightarrow y$$.$$^\ddagger$$ Let $$\lambda$$ be in the resolvent set. Then $$x=\lim_n x_n = \lim_n (T-\lambda)^{-1}(T-\lambda)x_n=(T-\lambda)^{-1}(y-\lambda x)$$ (+). Therefore, $$x\in (T-\lambda)^{-1}(X)=D_T$$. By (+), $$(T-\lambda)x=(T-\lambda)(T-\lambda)^{-1}(y-\lambda x)=(y-\lambda x)$$;  hence $$Tx=y$$, so $$T$$ is closed, QED.

Note: Robert Israel's shorter proof on this page shows the same for any TVS (if you require that the inverse is a continuous operator). The above proof (that works for all metrizable TVSs) might be easier for some readers.

$$\dagger$$) The "if" in the lemma cannot be reversed, as some closed operators (even on $$\ell^2$$) have an empty resolvent set, by https://math.stackexchange.com/questions/3262168/closed-operator-with-trivial-resolvent-set

$$\ddagger$$) In a metrizable space, a set is closed iff it is sequentially closed. So the graph $$G_T:=\{(x,Tx):\, x\in D_T\}$$ is closed iff $$G_T\owns (x_n,Tx_n)\rightarrow (x,y)\ \Rightarrow\ (x,y)\in G_T$$ (i.e., $$x\in D_T$$ and $$y=Tx$$).

Note that your assumption that the operator is densely defined was not needed. This is a strict improvement:

EXAMPLE. Let $$L$$ denote the inverse of the right-shift $$R:(x_1,x_2,\cdots)\rightarrow (0,x_1,x_2,\cdots)$$. Then dom$$L=R(\ell^2)$$ is not dense, as $$(1,0,0,\cdots)$$ is not in its closure, but yet the (onto-)resolvent set $$\rho(L)$$ is nonempty, as $$(L-0)^{-1}=R$$ and hence $$0\in\rho(L)$$.

This "onto" definition seems to be standard: Rudin: F.A., 13.26 & https://en.wikipedia.org/wiki/Spectrum_(functional_analysis)#Spectrum_of_an_unbounded_operator

Your claim is true only for this standard definition ("onto"), not for the older definition ("dense range" in place of "onto"), by Matthew Daws' example. For closed operators, the two definitions are equivalent. For a further discussion on this and the two definitions, see: Standard definition of a resolvent: A-zI must be onto, not merely have a dense range?