If $\Omega\subset\mathbb{R}^3$ is bounded, $$f_n\to f\mbox{ in }L^q(0,T;L^p(\Omega)),\,1\leq q<\infty,\,1\leq p<2 $$ and $$f_n\to g\mbox{ weak-star in } L^\infty(0,T;L^2(\Omega)),$$ then $f=g$ in $L^\infty(0,T;L^2(\Omega))$.


For all $\varphi\in C^\infty_c([0,T]\times\Omega)$ $$\int_0^T\int_\Omega(f-g)\varphi\,dxdt=\int_0^T\int_\Omega(f-f_n)\varphi\,dxdt+\int_0^T\int_\Omega(f_n-g)\varphi\,dxdt.$$ Thus we have $$\int_0^T\int_\Omega(f-g)\varphi\,dxdt=0$$ as $n\to\infty$.

Therefore, $f=g$ in $L^\infty(0,T;L^2(\Omega))$.$\blacksquare$

Is it correct?


Yes, your proof is correct. It is a common way to prove that two limiting processes have the same result:

  1. Show that each notion of convergence implies distributional convergence (i.e., integrals against test functions converge)
  2. Appeal to the fact that distributional limits are unique.

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