Given a Banach space $E$.
Consider linear operators:
$$T:E\supset\mathcal{D}(T)\to E:\quad T(\kappa x+\lambda y)=\kappa T(x)+\lambda T(y)$$
(No other assumptions on the operator!)
Denote for shorthand:
$$R_\lambda:\mathcal{R}_\lambda\to\mathcal{D}_\lambda:\quad R_\lambda:=(\lambda-T)^{-1}$$
The standard definition for the resolvent set:
$$\rho(T):=\{\lambda\in\mathbb{C}:\mathcal{N}_\lambda=(0),\mathcal{R}_\lambda=E,\|R_\lambda \|<\infty\}$$
An alternative definition for the resolvent set:
$$\rho(T):=\{\lambda\in\mathbb{C}:\mathcal{N}_\lambda=(0),\overline{\mathcal{R}_\lambda}=E,\|R_\lambda\|<\infty\}$$
(See Werner or Weidmann resp. Kubrusly or Kreyszig.)
Do these definitions really agree?
Clearly for closed operators they do.
Best Answer
Here's an idea. Let $X=l^{2}$ be the space of sequences $\{ x_{n}\}_{n=1}^{\infty}$ of square-summable sequences of complex numbers. Let $e_{n}$ the standard basis element which is a sequence with all 0's except for a $1$ in the n-th place. Let $M$ be the subspace spanned by finite linear combinations of $\{ e_{1}+\frac{1}{n}e_{n}\}_{n=2}^{\infty}$. The closure $M^{c}$ includes $e_{1}$ and, because of that, also includes every $e_{n}$ for $n \ge 2$. So $M^{c}=X$. Define $S$ to be the restriction of the left shift to $M$. $S$ is bounded on $M$, and $\mathcal{N}(S)=\{0\}$ because $e_{1} \notin M$. Let $T$ be the inverse of $S$. Then $T : \mathcal{D}(T) \rightarrow X$ has domain $\mathcal{D}(T)=S(M)$, and the range of $T$ is $M$ which is dense in $X$. And $S=T^{-1}$ is bounded by $1$, which gives $\|Tx\| \ge \|x\|$ for all $x \in \mathcal{D}(T)$. So $0 \in \rho(T)$ using the second definition of resolvent, which is very wrong from my point of view. However, $T$ is not closable because $(e_{1},0)$ is in the closure of the graph of $S$ and, so, $(0,e_{1})$ is in the closure of the graph of $T$. I think this works, but please check my details.