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?


1 Answer 1


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 $v \in H^2 (I)$.

In above reasoning, we use several times the lemma

Let $\varPhi$ be a distribution on $I$. If $\varPhi' \in L^2 (I)$ then $\varPhi \in L^2 (I)$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .