# how to prove the generator of semigroup is a Banach space

I am not familiar with semigroup theory, so please stand with my dummy question.

Say, $A$ is the generator of a semigroup, consider space $X_{n} = D(A^{n})$ with graph norm, $\|f\|_{A^{n}}:=\|f\| + \|A^{n}f\|$.

Now, for $n \in \mathbb{N}$, define $\||x\||:=\|x\|+\|Ax\|+...+\|A^{n}x\|$. I need to prove $\||.\||$ and the standard norm(graph norm given above) are equivalent and furthermore, the space is Banach.

Regarding equivalence: $\||x\||\geq \|x\|_{A^{n}}$ is obvious, but how to prove the other direction?

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By the Hille-Yosida theorem, $A$ is closed. Hence $A^n$ is closed and so $X_n$ is Banach under $\|\cdot\|_{A^n}$. – user12014 Nov 3 '11 at 3:53
@PZZ hmm, indeed it is a short cut. How about to prove the equivalence of norms? – newbie Nov 3 '11 at 12:56

To see that the norms are equivalent, you have to know a few results from the theory of maximal monotone operators. First, $I+A$ is surjective (by assumption) and you can show that $(I+A)^{-1}$ and $A(I+A)^{-1}=(I+A)^{-1}A:H \to H$ are bounded and of course linear. Then you can get interpolation results, like the following:
$\|Ax\| = \|A(I+A)(I+A)^{-1}x\| = \|A(I+A)^{-1}x + (I+A)^{-1}A^2x\| \leq C(\|x\| + \|A^2x\|)$
Then you can repeat these arguments for $n\geq 2$.