# Convergence of the product of three sequences

Let $$Q=(0,T)\times \Omega$$, $$\Omega$$ being a bounded subset of $$\mathbb R^d$$, sufficiently smooth. Consider three sequences $$u_n$$, $$v_n$$, and $$w_n$$ such that:

1. $$u_n$$ is bounded in $$L^\infty(Q)$$ and converges strongly to $$u$$ in $$L^2(0, T; H^1(\Omega))$$.
2. $$v_n$$ converges strongly to $$v$$ in $$L^2(Q)$$.
3. $$w_n$$ converges weakly to $$w$$ in $$L^2(Q)$$.

Could us expect that: $$\lim_{n \to \infty} \int_Q u_n v_n w_n \, dx \, dt = \int_Q u v w \, dx \, dt?$$

I was trying to show that the sequence $$u_nv_n$$ converges strongly to $$uv$$ in $$L^2(Q)$$ and to use the weak strong convergence for the rest of the proof. I proceed as follows: $$u_nv_n-uv=u_n(v_n-v)+v(u_n-u).$$ But the fact is that by taking the $$L^2$$-norm on both sides, I need $$v$$ to be in $$L^\infty(Q)$$. Could you please give me some insights on how to proceed or suggest an alternative approach?

• Your second displayed formula (which you want to prove) is identical to your first displayed formula, which you seem to assume to hold. If so, then what is the problem? Commented May 7 at 12:05
• I conjecture in the first sentence "such that" has to be replaced by "for which we want to prove that" Commented May 8 at 6:28

Your approach seems correct to me; here is how to complete it. As you observed, the sequence $$u_n$$ converges in particular in the norm of $$L^2(I,L^2(\Omega))\sim L^2(I\times \Omega)$$, so summarising and abstracting: we have three sequences on a measure space $$(Q,\mathcal S,\mu)$$ such that
• $$u_n$$ is bounded $$L^\infty(Q)$$, converges strongly to $$u$$ in $$L^2(Q)$$
• $$v_n$$ converges strongly to $$v$$ in $$L^2(Q)$$
• $$w_n$$ converges weakly to $$w$$ in $$L^2(Q)$$.
Up to extracting subsequences, $$u_n$$ and $$v_n$$ converge a.e., and $$v_n$$ is dominated in $$L^2(Q)$$, so $$u_nv_n$$ is dominated in $$L^2$$ and converges a.e. to $$uv$$, thus in $$L^2(Q)$$ by the $$L^2$$- dominated convergence theorem. So $$\langle u_nv_n,w_n\rangle=\int_Q u_nv_nw_nd\mu$$ converges to $$\langle uv,w \rangle=\int_Quvw d\mu$$ (and by the usual sub-sub-sequence argument, the convergence is true for the whole initial sequences too).
• (Alternatively, we may directly use the assumption $\int_Qu_nv_nw_n\to\int_Quvw$, and the proof is even shorter because there is nothing to prove) Commented May 7 at 20:47