Let $a$ be a positive continuous function nowhere differentiable on $[0,1]$. The operator $T$ in $H:=L^2(0,1)\oplus L^2(0,1)$ defined by
$$T(u\_1,u\_2) := (u\_1' + au\_2',0)$$
on $\textrm{Dom} \,T := \{u=(u\_1,u\_2) \in H \ \vert\ u\_j \in C^1[0,1],j=1,2\}$ is supposed to be non-closable, since the domain of the adjoint operator is $\{0\} \oplus L^2(0,1)$, not dense in $H$.


I am having trouble proving that the domain of the adjoint is the one mentioned above, could you please help me?



Non-closability of an operator
From the definition of the adjoint, we have that $(v\_1,v\_2)\in D(T^\*)$ precisely if there are $y\_1,y\_2$ such that
$$
\langle v\_1, u'\_1+au'\_2 \rangle = \langle y\_1, u\_1 \rangle + \langle y\_2, u\_2 \rangle
$$
for all $(u\_1,u\_2)\in D(T)$. Since we can vary $u\_1, u\_2$ independently, this means that
$$
\langle v\_1, u'\_1 \rangle = \langle y\_1, u\_1 \rangle , \quad\quad \langle v\_1, au'\_2\rangle = \langle y\_2, u\_2 \rangle ;
$$
in other words, $v\_1\in D(T\_1^\*)\cap D(T\_2^\*)$, where $T\_1u=u'$, $T\_2u=au'$. Now an operator of the type $Su=pu'$ on the domain you specified has adjoint $S^\*u=-(pu)'$ on the obvious domain (I need $pu$ to have a derivative in $L^2$, and I also have boundary conditions, but these don't matter here).


In particular, $v\_1$ as above would have to satisfy $v'\_1, (av\_1)'\in L^2(0,1)$, which cannot work since $a$ is nowhere differentiable. More explicitly, if we had such an absolutely continuous function $v\_1$ that is not the zero element of $L^2(0,1)$, then, since $v\_1, av\_1$ are differentiable in the classical sense almost everywhere, we could find an $x\_0\in (0,1)$ such that $v\_1(x\_0)\not= 0$ (for the unique continuous representative) and $v\_1, av\_1$ are differentiable at $x\_0$. But then $a=(av\_1)/v\_1$ would also be differentiable at $x\_0$.


