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?



Non-empty resolvent set, then operator closed?
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.


