# Relation between a norm and norm of Besov spaces

Let $(H, \|\cdot\|)$ be a Hilbert space, $A \colon D(A)\subset H \longrightarrow H$ generates an analytic semigroup $T(t)$ on $H$. We define the following Banach space with the respect norm $$F=\{x\in H : \int_0^{+\infty} \|AT(t)x\|^2 dt <\infty\},$$ $$\|x\|_F =\|x\|+ \left(\int_0^{+\infty} \|AT(t)x\|^2 dt\right)^{1/2}.$$

If we consider $H=L^2(\Omega)$ and $D(A)=H^2(\Omega)\cap H_0^1(\Omega)$ ($\Delta$ for example). Is there any relation between the space $F$ and Besov spaces, or can we prouve the interpolation $F=(H,D(A))_{1/2,2}=H_0^1(\Omega)$ ?

Thank you for any help.

Yes, your identities are correct. Theorem 1.14.5 in Triebel's book [T] says that $$F = (H,D(A))_{1/2,2},$$ and $(1/2,2)$-real interpolation spaces between Hilbert spaces are in fact exactly the $1/2$-complex interpolation spaces, so $$(H,D(A))_{1/2,2} = [H,D(A)]_{1/2}$$ as proven for example in [P]. If the operator $A$ now admits bounded imaginary powers, then you have $$[H,D(A)]_{1/2} = D(A^{1/2}),$$ see e.g. [T, Theorem 1.15.3]. The latter is indeed known for the Laplacian and there you will also indeed have $D(A^{1/2}) = H^1_0(\Omega)$. See also this related question.