Aubin-Lions Lemma

By | April 16, 2018

In this article, we prove the celebrated compactness lemma which will be used to show the global existence of weak solution of Navier-Stokes equation.

Lemma 1. Let X_{0},X and X_{1} be three Banach spaces with

    \[ X_{0}\hookrightarrow\hookrightarrow X\hookrightarrow X_{1}. \]

Suppose X_{0} is reflexive. Then for each \varepsilon>0, there is a constant C_{\varepsilon}>0 such that

    \[ \norm uX\le\varepsilon\norm u{X_{0}}+C_{\varepsilon}\norm u{X_{1}}\quad\text{for all }u\in X_{0}. \]

Proof. If the statement were not true, then there exists a number \varepsilon>0
and a sequence \left\{ u_{k}\right\} in X_{0} such that

    \[ \norm{u_{k}}X>\varepsilon\norm{u_{k}}{X_{0}}+k\norm{u_{k}}{X_{1}}\quad\text{for all }k\in\mathbb{N}. \]

Note u_{k}\neq0 for each k. So we may assume \norm{u_{k}}{X_{0}}=1. By Banach-Alalogu theorem, there exists a subsequence of \{u_k\} and we still denote it by the same indices. Since X_{0}\hookrightarrow\hookrightarrow X, we have u_{k}\rightarrow u in X for some u\in X. Moreover, since
X\hookrightarrow X_{1}, it follows that u_{k}\rightarrow u in
X_{1}. But

    \[ \norm{u_{k}}X>\varepsilon\quad\text{and\quad}\frac{1}{k}=\frac{1}{k}\norm{u_{k}}X>\norm{u_{k}}{X_{1}} \]

for all k. Hence, letting k\rightarrow\infty, we obtain

    \[ \norm uX\ge\varepsilon\quad\text{and}\quad\norm u{X_{1}}=0, \]

which is a contradiction. This completes the proof.

Theorem (Aubin-Lions). Let X_{0},X and X_{1} be three Banach spaces with

    \[ X_{0}\hookrightarrow\hookrightarrow X\hookrightarrow X_{1}. \]

Suppose that X_{0},X_{1} are reflexive. Then for 0<T<\infty and 1<r,s<\infty, we have

    \[ \Leb r\left(0,T;X_{0}\right)\cap\Sob 1s\left(0,T;X_{1}\right)\hookrightarrow\hookrightarrow\Leb r\left(0,T;X\right). \]

Proof. Since X_{0} and X_{1} are reflexive, so are \Leb r\left(0,T;X_{0}\right) and \Sob 1s\left(0,T;X_{1}\right). It suffices to show that if u_{k}\rightarrow0 weakly both in \Leb r\left(0,T;X_{0}\right) and \Sob 1s\left(0,T;X_{1}\right), then u_{k}\rightarrow0 strongly in \Leb r\left(0,T;X\right).

Observe that it suffices to show that u_{k}\rightarrow0 strongly in \Leb r\left(0,T;X_{1}\right). Indeed, if u_{k}\rightarrow0 strongly in \Leb r\left(0,T;X_{1}\right), then by Lemma 1, for each \varepsilon>0, we have

    \[ \int_{0}^{T}\norm{u_{k}\left(t\right)}X^{r}dt\le\varepsilon\int_{0}^{T}\norm{u_{k}\left(t\right)}{X_{0}}^{r}dt+C_{\varepsilon}\int_{0}^{T}\norm{u_{k}\left(t\right)}{X_{1}}^{r}dt. \]

Since u_{k}\rightarrow0 weakly in \Leb r\left(0,T;X_{0}\right), \left\{ u_{k}\right\} is bounded in \Leb r\left(0,T;X_{0}\right). Thus,

    \[ \limsup_{k\rightarrow\infty}\int_{0}^{T}\norm{u_{k}\left(t\right)}X^{r}dt\le\varepsilon\sup_{k}\int_{0}^{T}\norm{u_{k}\left(t\right)}{X_{0}}^{r}dt. \]

Since \varepsilon>0 was arbitrary chosen, u_{k}\rightarrow0 strongly in \Leb r\left(0,T;X_{1}\right).

Suppose thus that u_{k}\rightarrow0 weakly both in \Leb r\left(0,T;X_{0}\right) and \Sob 1s\left(0,T;X_{1}\right). Since u_{k}\rightarrow0 weakly in \Sob 1s\left(0,T;X_{1}\right), \left\{ u_{k}\right\} is bounded in \Sob 1s\left(0,T;X_{1}\right). Set

(1)   \begin{equation*} M=\sup_{k}\norm{u_{k}}{\Sob 1s\left(0,T;X_{1}\right)}<\infty. \end{equation*}

Since \Sob 1s\left(0,T;X_{1}\right)\hookrightarrow C\left(\left[0,T\right];X_{1}\right), \left\{ u_{k}\right\} is bounded in C\left(\left[0,T\right];X_{1}\right). Hence, to prove that

    \[ \lim_{k\rightarrow\infty}\int_{0}^{T}\norm{u_{k}\left(t\right)}{X_{1}}^{r}dt=0, \]

we will show that

    \[ \lim_{k\rightarrow\infty}\norm{u_{k}\left(t\right)}{X_{1}}=0\quad\text{for a.a. }t\in\left[0,T\right]. \]

Then the conclusion follows from the dominated convergence theorem.

Let 0<t_{0}<T be fixed. Since u_{k} is continuous on \left[0,T\right], it is uniformly continuous and thus it is absolutely continuous. So for all t\in\left[t_{0},T\right],

    \[ u_{k}\left(t_{0}\right)=u_{k}\left(t\right)-\int_{t_{0}}^{t}u_{k}^{\prime}\left(\tau\right)d\tau. \]

Integrating this over \left[t_{0},t_{1}\right], we also have

    \[ \left(t_{1}-t_{0}\right)u_{k}\left(t_{0}\right)=\int_{t_{0}}^{t_{1}}u_{k}\left(t\right)dt-\int_{t_{0}}^{t_{1}}\int_{t_{0}}^{t}u_{k}^{\prime}\left(\tau\right)d\tau dt \]

and so integration by part gives

    \[ u_{k}\left(t_{0}\right)=\frac{1}{t_{1}-t_{0}}\int_{t_{0}}^{t_{1}}u_{k}\left(t\right)dt-\frac{1}{t_{1}-t_{0}}\int_{t_{0}}^{t_{1}}\left(t_{1}-\tau\right)u_{k}^{\prime}\left(\tau\right)d\tau \]

for all t_{1}\in\left(t_{0},T\right). By (1) and Hölder’s inequality, we obtain

    \begin{align*} \norm{\int_{t_{0}}^{t_{1}}\left(t_{1}-\tau\right)u_{k}'\left(\tau\right)d\tau}{X_{1}} & \le\left(\int_{t_{0}}^{t_{1}}\left(t_{1}-\tau\right)^{s'}d\tau\right)^{\frac{1}{s'}}\norm{u_{k}'}{\Leb s\left(0,T;X_{1}\right)}\\ & \le M\left(t_{1}-t_{0}\right)^{1+1/s'}. \end{align*}

Hence given \varepsilon>0, we can choose t_{1}\in\left(t_{0},T\right) such that

    \[ \sup_{k}\norm{\int_{t_{0}}^{t_{1}}\left(t_{1}-\tau\right)u_{k}'\left(\tau\right)d\tau}{X_{1}}<\varepsilon. \]

For each k, define

    \[ \Phi_{k}=\frac{1}{t_{1}-t_{0}}\int_{t_{0}}^{t_{1}}u_{k}\left(t\right)dt. \]

Then note that \Phi_k\rightharpoonup 0 weakly in X_0. Indeed, let f\in X_0^*=X_0. Then we have

    \[   \action{f,\Phi_k} = \frac{1}{t_1-t_0} \int_{t_0}^{t_1} \action{f,u_k(t)} dt. \]

Since u_k \rightharpoonup 0 weakly in \Leb{r}(0,T;X_0) and f can be regarded as a function in \Leb{r'}(0,T;X_0), the right hand side converges to 0.
Thus, \{\Phi_k\} converges to 0 strongly in X_1. Hence,

    \begin{align*}  & \limsup_{k\rightarrow\infty}\norm{u_{k}\left(t_{0}\right)}{X_{1}}\\  & \le\limsup_{k\rightarrow\infty}\norm{\Phi_k}{X_{1}}+\frac{1}{t_{1}-t_{0}}\limsup_{k\rightarrow\infty}\int_{t_{0}}^{t_{1}}\left(t_{1}-\tau\right)u_{k}^{\prime}\left(\tau\right)d\tau\\  & \le\varepsilon. \end{align*}

This completes the proof.

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