Unable to see how convergence in $L^p$ norm is being used to derive an expression?

Question

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 $\phi \in L^q(\Omega)$.

1. 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?
2. 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?

Answer 1

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 ${L}^{q} \left({\Omega}\right)$.

Answer 2

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