## A proof on the invariance of attractors.

Consider the semigroup $S(t)$ corresponding to the reaction diffusion equation

$u_t - \Delta u + f(u) = g(x), u(0) = u_0\in L^2$

and $u$ is vanished on the boundary of the domain.

It is noticed that $S(t)$ is continuous in $L^2$.

In the previous post, we wonder that: can the invariance of the attractor in $H^1$ be implied from the invariance of the attractor in $L^2$? In this post, we give a short proof of it. To this end, we need the following lemma.

Lemma 1. Let $\{x_n\}$ be a sequence in $H^1$ such that $x_n\rightarrow x_0$ strongly in $L^2$ and $x_n \rightarrow y_0$ in $H^1$. Then $x_0 = y_0$.

PROOF. Since $H^1 \subset L^2$ continuously, we can see that $x_n \rightarrow y_0$ strongly in $L^2$. Due to the uniqueness of the limit, we conclude that $x_0 = y_0$.

We also state a well-known result.

Theorem 1. The semigroup $S(t)$ has an absorbing set $B_2$ in $L^2$ and is asymptotically compact in $L^2$. This implies that $S(t)$ has a global attractor $A_2$ in $L^2$ and

$A_2 = \cap_{s\geq 0}\overline{\cup_{t\geq s}S(t)B_2}^{L^2}$.        (1)

Now, we state our main result.

Theorem. Assume that $S(t)$ has an absorbing set $B$ in $H^1$ and $S(t)$ is asymptotically compact in $H^1$. Then there exists a unique global attractor $A$ in $H^1$ for $S(t)$, and

$A = \cap_{s\geq 0}\overline{\cup_{t\geq s}S(t)B}^{H^1}$.         (2)

PROOF.

The compactness and the attracting property of $A$ are implied from the asymptotic compactness of $S(t)$ and definition of $A$. Thus, we only have to prove the invariance of $A$ under $S(t)$.

Let $B_0 = B\cap B_1$, then $B_0$ is an absorbing set for $S(t)$ in both $L^2$ and $H^1$.

Since $S(t)A_2 = A_2$ for all $t\geq 0$. We will prove that $A \equiv A_2$.

Indeed, first, taking any $x\in A$. From definition (2) (with $B_0$ in place of $B$), there exist a sequence $t_n\rightarrow +\infty$ and $x_n\in B_0$ such that $S(t_n)x_n \rightarrow x$ strongly in $H^1$. Because $H^1 \subset L^2$ continuously, we get that $S(t_n)x_n \rightarrow x$ strongly in $L^2$. Hence, $x\in A_2$ because of the definition of $A_2$. Thus, $A\subset A_2$.

On the other hand, let $x\in A_2$. Then there exist sequences $t_n\rightarrow +\infty$ and $x_n\in B_0$ satisfying $S(t_n)x_n \rightarrow x$ in the strong topology of $L^2$. Since $S(t)$ is asymptotically compact in $H^1$, there exists a subsequence $\{n_k\}$ of $\{n\}$ such that $S(t_{n_k})x_{n_k} \rightarrow y$ strongly in $H^1$ for some $y\in A$. It implies that $S(t_{n_k})x_{n_k} \rightarrow y$ strongly in $L^2$. Due to the uniqueness of the limit, we obtain that $x = y$. Thus, $A_2 \subset A$.

This completes the proof.