Background
For a (closed) linear operator on a complex Hilbert space $T:D(T) \to H$ the resolvent set is defined as
$\rho (T) = \{z \in \mathbb C: (T - zI) $ has a bounded (two-sided) inverse $(T - zI)^{-1}: H \to D(T)\}$
Then for $z \in \rho(T)$ the resolvent operator $R_T (z) = (T - zI)^{-1}$
The First Resolvent Identity states that for $w, z \in \rho(T)$ then
$R_T (w) - R_T (z) = (w - z)R_T (w) R_T (z)$
The proof I'm looking at goes as follows.....
Consider the expression, $(T – w)^{-1} – (w – z)(T – w)^{-1}(T – z)^{-1} $
= $(T – w)^{-1} (1 – (w – z)(T – z)^{-1} )$
= $(T – w)^{-1} (1 – (w - T + T – z)(T – z)^{-1} )$ ......(1)
which then simplifies to $(T – z)^{-1}$ and rearranging gives the result.
Question:
If $T$ is an unbounded operator then the expression in (1) would only seem to be valid on $D(T)$.
If $D(T)$ is dense then knowing $R_T (w) - R_T (z) = (w - z)R_T (w) R_T (z)$ on $D(T)$ and that the resolvents are bounded means (by BLT) that the result holds everywhere.
But, is this result still true if $D(T)$ is not dense ?