Let $X$ be a Banach space, $T$ a strongly continuous semigroup generated by $X$ with generator $A:D(A)\rightarrow X$, where $D(A)$ is the domain of $A$. Let $\beta$ be in the resolvent set of $A$. We define $$X_1=(D(A), \Vert \cdot \Vert_1)\qquad \Vert x\Vert_1:= \Vert (A-\beta)x\Vert_X.$$
The first claim is that this norm is equivalent to the graph norm on $D(A)$, i.e. to $\Vert x\Vert= \Vert x \Vert_X + \Vert Ax \Vert_X$. The second claim is that the embedding $i: X_1 \rightarrow X$ is continuous and dense.
For the first claim: one direction I can easily do with the triangle inequality: $$\Vert (A-\beta)x\Vert_X \leq \Vert Ax\Vert_X +\vert \beta \vert \Vert x \Vert_X + \vert \beta \vert \Vert A x\Vert_X \leq (1+\vert \beta\vert)(\Vert x \Vert_X + \Vert Ax \Vert_X)$$
For the other direction I can only obtain $$\Vert x \Vert_X + \Vert Ax \Vert_X \leq (1+\vert \beta\vert)(\Vert (A-\beta)x\Vert_X +\Vert x\Vert_X).$$ Can anybody help me out?
For the second claim: I think that from the first claim follows that $X_1$ is complete, but where does the densness come from?
The resolvent $(A-\beta)^{-1}$ is automatically bounded (it is closed because $A$ is closed and it is defined on the whole Banach space, so the closed graph theorem applies), hence $$ \Vert x\Vert_X=\Vert(A-\beta)^{-1}(A-\beta)x\Vert_X\le \Vert (A-\beta)^{-1}\Vert\cdot \Vert (A-\beta)x) \Vert_X=:C\Vert x\Vert_1 $$ Use this together with your last inequality, to get the other direction.