interpolation between Bochner spaces involving $H^{-1}$

Question

I am reading a paper in which they say that: Since $u_n$ is bounded in $L^2(0,T;H^2(\Omega))$ and due to the strong convergence of $u_n$ in $L^2(0,T;H^{-1}(\Omega))$ we conclude with an interpolation argument that $$u_n\to u\mbox{ in }L^2(0,T;H^1(\Omega)).$$ Here $\Omega$ is a bounded domain with smooth boundary.
So I really do not know that interpolation argument they meant, because I did not see interpolation involving the dual space before. Does anyone know this argument or know any reference about this?
Besides, I have a further question, if I have a sequence $f_n$ which is strong convergent in $L^{8/7}(0,T;W^{-1,4}(\Omega))$ and is bounded in $L^{\infty}(0,\infty;L^2(\Omega))$ and is bounded in $L^2(0,\infty;H_0^1(\Omega))$, may I conclude with a similar argument that $$f_n\to f\mbox{ in }L^2(0,T;L^2(\Omega)).$$ Here $W^{-1,4}(\Omega)$ is the dual space of $W_0^{1,4}(\Omega)$.

Answer 1

It is not about dual spaces, it is just that the interpolation of Sobolev spaces gives Sobolev spaces. If $\Omega$ is a $C^\infty$ bounded set, then $$ H^s(\Omega) = [H^{-1}(\Omega),H^2(\Omega)]_{\theta} $$ if $\theta = \frac{s+1}{3}$ (see e.g. Triebel, Theory of Function Spaces, Theorem in Sec. 3.3.6, together with the fact that $F^s_{2,2}(\Omega) = B^s_{2,2}(\Omega) = H^s(\Omega)$). Taking $s=1$, it yields $$ \|u\|_{H^1(\Omega)} \leq C\,\|u\|_{H^{-1}(\Omega)}^{1/3}\,\|u\|_{H^2(\Omega)}^{2/3}. $$ (Of course, this can also be proved more directly, if $\Omega=\Bbb R^d$, then this comes easily from the Fourier characterization of $H^s$). Hence, taking the $L^2$ norm in time, using Hölder's inequality and replacing $u$ by $u-u_n$, $$ \|u_n-u\|_{L^2_tH^1(\Omega)} \leq C\,\|u_n-u\|_{L^2_tH^{-1}(\Omega)}^{1/3}\,\|u_n-u\|_{L^2_tH^2(\Omega)}^{2/3}. $$

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Your new question is more delicate, but it indeed also works! By interpolation of $L^2_tH^1_0(\Omega)$ and $L^{8/7}_tW^{-1,4}(\Omega)$, you can get convergence in $L^{16/11}_tL^{8/3}(\Omega)$ (since $11/16 = 1/2(1/2+7/8)$ and $3/8 = 1/2(1/4+1/2)$). Since $\Omega$ is bounded, it implies convergence in $L^{16/11}_tL^{2}(\Omega)$. Combining that with boundedness in $L^\infty_tL^{2}(\Omega)$, it implies convergence in $L^2_tL^{2}(\Omega)$. More precisely $$\begin{align} \|u\|_{L^2_tL^{2}(\Omega)} &\leq C\, \|u\|_{L^{16/11}_tL^{2}(\Omega)}^{8/11}\|u\|_{L^\infty_tL^{2}(\Omega)}^{3/11} \\ &\leq C\, \|u\|_{L^{8/7}_tW^{-1,4}(\Omega)}^{4/11} \|u\|_{L^2_tH^1(\Omega)}^{4/11} \|u\|_{L^\infty_tL^{2}(\Omega)}^{3/11}. \end{align}$$