# Brezis' exercise 8.17: is $D(A^*) \subset H^2 (I)$?

Let $$I$$ be the open interval $$(0, 1)$$. I'm trying to solve below exercise in Brezis' Functional Analysis, i.e.,

Exercise 8.17 Let $$H=L^2(I)$$ and $$A: D(A) \subset H \rightarrow H$$ be the unbounded operator defined by $$A u=u''-xu'$$, whose domain $$D(A)$$ will be made precise below. Determine $$A^{*}, D\left(A^{*}\right), N(A)$$, and $$N\left(A^{*}\right)$$ in the following cases:

1. $$D(A)=\left\{u \in H^2 (I) : u(0)=u(1)=0\right\}$$.
2. $$D(A)=H^2(I)$$.
3. $$D(A)=\left\{u \in H^2(I) : u(0)=u(1)=u'(0)=u'(1)=0\right\}$$.
4. $$D(A)=\left\{u \in H^2(I) : u(0)=u(1)\right\}$$.

Let $$v \in D(A^*)$$. Then $$\int_I v (u''-xu') = \int_I (A^*v)u, \quad \forall u \in C^\infty_c (I).$$ and thus $$\left | \int_I v u'' \right | \le \| A^*v \|_{L^2} \| u \|_{L^2} + \| v \|_{L^2} \| u' \|_{L^2} \quad \forall u \in C^\infty_c (I).$$

I think that $$v \in H^2 (I)$$ but don't know how to proceed. Could you provide me with some hints?

We have $$\int_I v (u''-xu') = \int_I (A^*v)u, \quad \forall u \in C^\infty_c (I)$$ implies $$A^*v = (v' + vx)'$$ in the distributional sense. It follows from $$A^*v \in L^2 (I)$$ that $$v' + vx \in H^1 (I)$$ and thus $$v' + vx \in L^2 (I)$$. It follows from $$v \in L^2$$ that $$vx \in L^2$$ and thus $$v' \in L^2 (I)$$. This implies $$v \in H^1 (I)$$ and thus $$vx \in H^1 (I)$$. This implies $$v' \in H^1(I)$$ and thus $$v \in H^2 (I)$$.
Let $$\varPhi$$ be a distribution on $$I$$. If $$\varPhi' \in L^2 (I)$$ then $$\varPhi \in L^2 (I)$$.