Let $\Omega \subset \mathbb R^n$ open, bounded with Lipschitz boundary.
I have a theorem that says that:
If $u_n\rightharpoonup u$ in $L^p$ and $v_n\to v$ in $L^{p'}$ then $u_nv_n\rightharpoonup uv $ in $L^1$ where $\frac{1}{p}+\frac{1}{p'}=1.$
Now I have function $f_n\rightharpoonup f$ and $g_n\rightharpoonup g$ in $L^p$ and that $\varphi_n\to \varphi$ in $L^\infty $.
Why the theorem above tell us that $\varphi_nf_n\rightharpoonup \varphi f$ and $\varphi_n g_n \rightharpoonup \varphi g$ in $L^p$ ? There is something I don't understand here.
There is an isometric embedding $L^\infty\to B(L^p)$ given by $\psi\mapsto (f\mapsto \psi\cdot f)$. It is a general statement that if you have a Banach space $X$ and a a norm convergent sequence of operators $A_n$ in $B(X)$ and a weakly convergent sequence $f_n\in X$ that then $A_nf_n$ converges weakly.
To do it let $h\in X^*$ and let $A$ be the limit of $A_n$, $f$ the limit of $f$:
$$\langle h, A_n f_n-A f\rangle=\langle h,(A_n- A) f_n\rangle+\langle h,A( f_n- f)\rangle = \underbrace{\langle h,(A_n- A)f_n\rangle}_{(1)} +\underbrace{\langle A^* h, f-f_n\rangle}_{(2)}$$ Now term $(2)$ converges to zero, since $f_n$ converges weakly to $f$. Term $(1)$ can be bounded by $$(1)≤\|h\|\,\|A-A\|\,\|f_n\|$$ if we see that $\|f_n\|$ is a bounded sequence we find that $(1)$ must also converge to zero. Well $f_n$ can be viewed as a sequence of functionals $X^*\to\Bbb C$ so that $f_n(h)$ converges for any $h\in X^*$, this means $$\sup_{n\in\Bbb N}|f_n(h)|<\infty$$ for any $h\in X^*$ by application of Banach Steinhaus you find that $\sup_{n\in\Bbb N}\|f_n\|<\infty$.