STANDING ASSUMPTIONS: Let $T:D_T\rightarrow X$ be a linear operator, where $X$ is a normed space and $D_T\subset X$.
DEFINITION 1 ("onto"): A complex number $\lambda$ belongs to the resolvent set $\rho(T)$ iff $T-\lambda I$ is one-to-one and onto and its inverse is bounded.
DEFINITION 2 ("dense range"): A complex number$\lambda$ belongs to the dresolvent set $\rho(T)$ iff $T-\lambda I$ is one-to-one and has a dense range and its inverse is bounded.
So the only difference is that one requires a dense range instead of "onto".
I believe that the onto-definition is now the standard one and the dense-definition now rare (when not equivalent).
QUESTION: Is this really the case?
To me it seems that, in current papers, usually the resolvent is not defined, but when you dig deeply in the references, you find that the onto-definition is used. (But I have now checked this just for 2 papers.) This is dangerous, as old classics may use the dense-definition. Often there is no difference:
LEMMA: For closed operators (hence for bounded operators) the two definitions are equivalent.
PROOF: Set $A:=T-\lambda I$. Let $R=A^{-1}$.Given any $y\in X$, density implies $y=\lim Ax_n$ for some $\{x_n\}\subset$Dom$\,T$. But $\lim x_n=\lim RA x_n=R\lim A x_n=Ry$, hence $(Ry,y)\in$graph$(A)$, as $A$ is closed, QED.
FACT: The definitions are not equivalent.
PROOF: any of Example 1, Example 2 and Example 3 below. QED.
EXAMPLE 1. Let $T$ be the identity operator $I$ restricted to a proper dense subset of $X$. (This is possible even for $X=\ell^2$.) Then $I=(T-0)^{-1}$, hence $T$ has the dresolvent $I$ at $0$ but not a resolvent, QED.
EXAMPLE 2: Let $X=\ell^2$. Define $h\in\ell^2$ by $h_i=2^{1-i}$. Define $R\in\mathcal L(X)$ by $Rg=g-g_1 h$. Then $Rg=0$ iff $g\in {\mathbb C} h$. Set $Y:=c_c$ (sequences that are zero outside a finite set). Then $R$ is 1-1 on $Y$. Set $D_A:=RY$, $A:=R^{-1}$ to define a linear operator $A$ whose dresolvent at $0$ is $(A-0I)^{-1}=R$ (as range$(A)=Y$ is dense, $A$ is 1-1 and $R$ is bounded) but which has no resolvent (proof: not onto) and is not closable (proof: if $Y\owns y_n\rightarrow h$, then $x_n:=Ry_n\rightarrow Rh=0$ but $Ax_n=y_n\rightarrow h\ne0$, QED).
EXAMPLE 3: Here $$T:\mathcal{D}(T)\subseteq X\to Y:\text{ injective, dense range, continuously invertible, not closable!}$$ Therefore, $0$ lies in its spectrum but not in its dspectrum. Details: https://math.stackexchange.com/questions/819526/operator-not-closable
(BTW, the $T$ of Example 1 is closable and its closure $I$ does have a resolvent at $0$. The $A$ of Example 2 is not densely defined, so Example 3 is better in that sense.)
The difference is often important:
LEMMA. A linear operator $T$ in a normed space $X$ is closed if it has a non-empty resolvent set. However, the reverse is not true, and a non-closed operator may have a non-empty dresolvent set.
PROOF: Non-empty resolvent set, then operator closed?
The difference between the two definitions is discussed here: https://math.stackexchange.com/questions/815377/resolvent-definition But my question is the above, i.e., is the dresolvent definition fairly obsolete and does the spectrum nowadays usually mean the onto-spectrum?
N.B. Many books define (unbounded) linear operators invertible when they are 1-1 and onto and have a bounded inverse (but some other books, e.g., Goldberg, define an inverse for any 1-1 operator). At least in many of them, $0\in\rho(T)$ is defined analogously, leading to the onto-resolvent definition.
N.B. C-Star-W-Star calls the onto-definition "usual" here after another commenter (D.F.) suggested the opposite: https://math.stackexchange.com/questions/1345613/definition-of-resolvent-set#comment2737222_1345626
Non-normed spaces: Instead of "bounded" one should write "the restriction of a continuous operator that is defined on the whole $X$" to generalize this to all TVSs. For normed spaces this makes no difference to either definition (and the non-equivalence remains).
"ONTO"-DEFINITION IS USED IN:
- Rudin: Functional Analysis, 13.26
- https://en.wikipedia.org/wiki/Spectrum_(functional_analysis)#Spectrum_of_an_unbounded_operator & https://en.wikipedia.org/wiki/Decomposition_of_spectrum_(functional_analysis) (proof: continuous spectrum consists of non-eigenvalues for which the range of T-z is dense but not onto).
- Alessandra Lunardi: Analytic Semigroups and Optimal Regularity in Parabolic Problems, 1995. (Proof: p. 33 (below Def. 2.0.1) says that nonempty resolvent implies closedness.)
- Pazy: Semigroups of ..., 1983. (Proof: Requires $z-A$ boundedly "invertible", which must mean the onto-definition, as the alternative would be that not even denseness was required, merely 1-1.)
"DENSE"-DEFINITION IS USED IN:
- Goldberg: Unbounded Linear Operators (1966), Definition II.5.2, p. 71.
- Hille-Phillips ("Revised Edition", p. 54, Def. 2.16.1)
- https://en.wikipedia.org/wiki/Resolvent_set
