[Math] Product of weak/strong converging sequences

Question

Let's consider two sequences $u_n$ and $v_n$ such that
$$u_n\to u\,,\,\,\,\rm{in}\,\,\,L^\infty(\mathbb{R}^n)$$
and
$$v_n\rightharpoonup v\,,\,\,\,\rm{in}\,\,\,L^2(\mathbb{R}^n)$$
What can I say of the convergence of the product $u_nv_n$? In particoular I want that $u_nv_n$ converges weakly in $L^2_{loc}(\mathbb{R}^n)$. I procced as follows: since $u_n$ converges strongly in $L^\infty$ it converges strongly in $L^2_{loc}$ and so we have a product of a strongly convergent sequence and a weakly convergent one in $L^2_{loc}$ which gives a weakly convergent sequence. Is it right?

Answer 1

Let's check by definition: Let $\phi \in L^2(\mathbb R^n)$. Then

$$\bigg| \int u_nv_n \phi dx - \int uv \phi dx\bigg| \leq \bigg|\int (u_nv_n - uv_n) \phi dx\bigg| + \bigg| \int (uv_n - uv) \phi dx \bigg|$$

$$\leq ||u-u_n||_{L^\infty} \int |v_n|\cdot |\phi| dx + \bigg| \int (v_n - v) (u\phi) dx \bigg|$$

The second term tends to zero as $v_n$ converges weakly to $v$ and $u \phi \in L^2(\mathbb R^n)$. For the first term, use

$$\int |v_n| |\phi|dx \leq ||v_n||_{L^2} ||\phi||_{L^2}$$

and the fact that weakly convergent sequences are bounded (Consequence of uniform boundedness principle in the general case). Then the first term also tends to zero. Thus $u_nv_n$ converges to $uv$ weakly in $L^2(\mathbb R^n)$.