Resolvent identity, densely defined operator

Question

Let $L:D \rightarrow H$ be a densely defined, closed, and symmetric operator. Then we know we may define bounded inverse $R(z): im (L-z) \rightarrow D$, where $im(L-z)$ is closed subset of $H$.

Prove:

If $z,w$ are distinct, strictly complex numbers, with $R(z)$ and $R(w)$ are defined everywhere, then prove that $$ (z-w)R(z)R(w)= R(z)-R(w)$$

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Reference: Tao's blog, Exercise 8(i) I do not even understand the given hint (what does $H$ mean in this case?)

Answer 1

I think I got this.

$$\begin{align} R_z - R_w &= R_z(L-w)R_w - R_z(L-z)R_w \\ &= R_zLR_w-R_zLR_w +R_z(z-w)R_w\\ &= (z-w)R_zR_w \end{align} $$