Unable to see how convergence in $L^p$ norm is being used to derive an expression? I'm reading notes on the closeability of differential operators where the author says let $\Omega\subset \mathbb{R}^n$, and then defines the operator $A:C_0^\infty(\Omega)\to L^p(\Omega)$ by
$$
Au := \sum_{|\alpha|\le m}a_\alpha \partial^\alpha u,
$$
and its formal adjoint $B:C_0^\infty(\Omega)\to L^q(\Omega)$ by
$$
Bv := \sum_{|\alpha|\le m}(-1)^{|\alpha|}\partial^\alpha (a_\alpha v),
$$
He then says that integration by parts shows that
$$
\int_\Omega v(Au) = \int_\Omega (Bv) u, \quad \quad \text{for all} \ u,v \in C_0^\infty(\mathbb{R}^n).
$$
Now let $u_k\in C_0^\infty(\Omega)$ be a sequence of smooth functions with compact support and let $v\in L^p(\Omega)$ such that
$$
(*) \quad \lim_{k\to\infty} ||u_k||_{L^p} = 0, \quad \lim_{k\to\infty} ||v - Au_k||_{L^p} = 0.
$$
Then, for every test function $\phi \in C_0^\infty(\Omega)$, we have
$$
(**) \quad \int_\Omega \phi v = \lim_{k\to \infty} \int_\Omega \phi (Au_k) = \lim_{k\to \infty} \int_\Omega (B\phi) u_k = 0.
$$
Since $C_0^\infty(\Omega)$ is dense in $L^q(\Omega)$, this implies that $\int_\Omega \phi v = 0$ for all $\phi \in L^q(\Omega)$.


*

*I don't see how the fact that $u_k$ converges to zero, and $v$ converges to $Au_k$ in the $L^p$ norm allows us to derive $(**)$ from $(*)$? The integrals in $(**)$ do not even use the $L^p$ norm, so how is statement $(**)$ valid?

*The author says that since $C_0^\infty(\Omega)$ is dense in $L^q(\Omega)$, this implies that $\int_\Omega \phi v = 0$ for all $\phi \in L^q(\Omega)$. How can I show this fact?

 A: *

*By Holder's inequality : $$\int_\Omega |\phi( v-Au_k)|=\|\phi( v-Au_k)\|_{L^1} \leq \| \phi \|_{L^{q}}\|v-Au_k \|_{L^{p}} \underset{k \to \infty}\to 0.$$

*Let $\psi$ any function of $L^q(\Omega)$ and $(\psi_k)$ a sequence of $C_0^\infty(\Omega)$ s.t. $\|\psi-\psi_k\|_{L^q}\underset{k \to \infty}\to 0$.
$$\int_{\Omega}\psi v\overset{\delta}=\lim_{k\to \infty} \int_{\Omega}\psi_kv=0.$$
The equality $\delta$ comes from the same reasonning as before :
$$\int_{\Omega}|(\psi-\psi_k) v|=\|(\psi-\psi_k) v\|_{L^1} \leq\|\psi-\psi_k\|_{L^q} \|v\|_{L^p} \underset{k \to \infty}\to 0.$$

A: Strong convergence in a Banach space implies weak convergence. It means that if
${f}_{k} \rightarrow  f$ in ${L}^{p} \left({\Omega}\right)$ and if $g \in  {L}^{q} \left({\Omega}\right)$, then
$\left\langle g , {f}_{k}\right\rangle  \rightarrow  \left\langle g , f\right\rangle $ where $\left\langle \ \right\rangle $ is the duality bracket.
The reason is that
$$\left|\left\langle g , f\right\rangle -\left\langle g , {f}_{k}\right\rangle \right| = \left|\left\langle g , f-{f}_{k}\right\rangle \right|  \leqslant  {\left\|g\right\|}_{L ^q(\Omega)} {\left\|f-{f}_{k}\right\|}_{L ^p(\Omega)}$$
It turns out that the duality bracket between ${L}^{p} \left({\Omega}\right)$ and ${L}^{q} \left({\Omega}\right)$ is given
by the integral
$$\left\langle g , f\right\rangle  = \int_{{\Omega}}^{}g f d x$$
The (**) formula can be rewritten
$$\left\langle {\phi} , {\nu}\right\rangle  = {\lim }_{k \rightarrow  \infty } \left\langle {\phi} , A {u}_{k}\right\rangle  = {\lim }_{k \rightarrow  \infty } \left\langle B {\phi} , {u}_{k}\right\rangle  = 0$$
Now the duality bracket is continuous, that is to say ${\phi} \rightarrow  \left\langle {\phi} , {\nu}\right\rangle $
is continuous in ${L}^{q} \left({\Omega}\right)$. If it is zero on the dense subspace ${\mathscr{C}}_{0}^{\infty } {(\Omega)} \subset  {L}^{q} \left({\Omega}\right)$,
then it must be zero on all of ${L}^{q} \left({\Omega}\right)$.
