Differential operator continuity

Question

Let $$A: = {\partial _x}:{L^2}(0,L) \to {H^{ - 1}}(0,L)$$ and $$B = :{(1 - \partial _x^2)^{ - 1}}:{H^{ - 1}}(0,L) \to H_0^1(0,L)$$ How can I show that $A$ and $B$ are continuous? I know that$${\partial _x}:{H^1}(0,L) \to {L^2}(0,L)$$ is continuous, but I have no idea about how treat this problem with dual space.. Thnaks.

Answer 1

Boundedness of $A$. We want to prove that there exists a constant $C$ such that $$\|Af\|_{H^{-1}}\leq C\|f\|_{L^2},\quad\forall \ f\in L^2(0,L).$$

Let $f\in L^2(0,L)$. Then $\partial_xf\in H^{-1}(0,L)$ with $$\langle\partial_x f,\phi\rangle=-\int_0^Lf(x)\phi_x(x)\;dx,\quad\forall\ \phi\in H_0^1(0,L)$$ and thus, by the Hölder's inequality, $$|\langle\partial_x f,\phi\rangle|\leq\|f\|_{L^2}\|\phi\|_{H_0^1}, \quad\forall\ \phi\in H_0^1(0,L).$$ This implies that $$\|Af\|_{H^{-1}}=\sup_{\phi\in L^2\\\phi\neq 0} \frac{|\langle \partial_x f,\phi\rangle|}{\|\phi\|_{H_0^1}}\leq \|f\|_{L^2}$$ which proves the desired result.

Boundedness of $B$. As above, we can show that the linear bijection $$(1-\partial_x^2):H_0^1(0,L)\to H^{-1}(0,L)$$ is continuous. So, the desired result follows from the Bounded Inverse Theorem (also called Banach Isomorphism Theorem).