Is there an operator between Banach spaces with the following properties: $$T:\mathcal{D}(T)\subseteq X\to Y:\text{ injective, dense range, continuously invertible, not closable!}$$ (Note that the ambient spaces should be really Banach spaces - otherwise this is trivial.)
Obviously this operator must automatically satisfy:
The question is related to the alternative definition of: Resolvent: Definition
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 the vectors $e_{1}+\frac{1}{n}e_{n}$ for $n \ge 2$.
The closure $M^{c}$ of the linear space $M$ contains $e_{1}$ because $\lim_{n} (e_{1}+\frac{1}{n}e_{n})=e_{1}$. $M^{c}$ also contains $e_{n}=n(e_{1}+\frac{1}{n}e_{n})-ne_{1}$ for $n\ge 2$. So the subspace $M$ is dense in $X$ because it contains all of the standard basis elements.
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 $x \in \mathcal{D}(T)$.
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 $T$. Also notice that $\mathcal{D}(T)$ is dense because the range of $M$ under $S$ includes all finite linear combinations of the standard basis elements.
I think this works, but please check my details.
