Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

Any comments are welcome. Cheers.

share|improve this question
    
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

1 Answer 1

up vote 0 down vote accepted

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 $n\geq 2$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.