# Non-closability of an operator

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 $H$.

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 $x_0$.
• Could you please develop on why $(av_{1})'$ cannot belong to $L^{2}(0,1)$ given $v_{1} \in L^{2}(0,1)$? – Geno Whirl Mar 4 '15 at 16:53
• It can; what I'm trying to say is that you cannot have both $v_1$ and $av_1$ absolutely continuous for a non-differentiable $a>0$. – Christian Remling Mar 4 '15 at 17:07