Weak convergence of solution I'm trying to figure out why in this paper:
Asymptotic convergence of semigroups
In the proof of theorem 1, I don't understand why $\{x(t):t\geq0\}$ is weakly sequentially precompact and why this should mean that for two $(*)$-sequences $(t_n)$ and $(s_n)$, $x(t_n)$ and $x(s_n)$ converge to some weak limit.
Could someone explain in detail? Thank you 
 A: Since $x : [0,\infty) \to H$ is continuous, and it was shown that $\lim_{t \to \infty} \|x(t)-y_0\|$ exists, then in particular, $\{x(t) : t \ge 0\}$ is bounded in norm.  So by Alaoglu's theorem, it is weakly compact.  
This also implies weak sequential compactness.  Indeed, if $H$ is separable then any bounded set is weakly metrizable, so then weak compactness and weak sequential compactness are equivalent.  If $H$ is not separable, then given a sequence $\{x(t_n)\}$, consider the separable subspace $H' \subset H$ consisting of the closed linear span of $\{x(t_n)\}$.  Then by the previous case there is a subsequence converging weakly in $H'$, which thus also converges weakly in $H$.
After that, I think you're misreading the sentence around (1.5).  What is being asserted in this sentence (and what the rest of the proof goes on to show) is the conditional statement that if $x(t_n)$ and $x(s_n)$ both converge weakly then their limits must be equal.  
To see why this assertion (call it A) implies the claim that $x(t_n)$ converges weakly, use the "double subsequence trick".  Suppose it did not.  Weak compactness says there is a subsequence $t_{n_k}$ (which is also a (*)-sequence) with $x(t_{n_k})$ converging weakly to some $x^*$.  Since the original sequence $x(t_n)$ did not converge weakly to $x^*$ or anything else, there is a weakly open neighborhood $U$ of $x^*$ and a subsequence $t_{m_k}$ such that all $x(t_{m_k})$ are outside $U$.  This subsequence has a further subsequence converging weakly to some $x^{**}$, which is necessarily outside $U$ and in particular not equal to $x^*$.  But this contradicts the assertion A.
