# Existence of convergent sequence implies convergence of function as $t\to\infty$

Let $$\Omega \subset \mathbb{R}$$ be an unbounded domain and $$u \in C([0,\infty);H_{0}^{1}(\Omega))$$. Assume there exists a sequence $$\{t_{n}\}_{n\in\mathbb{N}}$$ such that $$t_{n}\to\infty$$ and $$||u(t_{n})||_{H_{0}^{1}(\Omega)}\to 0$$ as $$n\to\infty$$. Is it possible to show that $$||u(t)||_{H_{0}^{1}(\Omega)}\to 0$$ as $$t\to\infty$$? I am not sure how to show this if this is possible. If this is not possible, then is there any counter example? Moreover, can I still obtain similar result if I replace $$H_{0}^{1}(\Omega)$$ with $$X$$ a Banach space?

Any hint is much appreciated! Thank you!

Let $$u(t)=\sin(t\pi)f$$ where $$f$$ is a fixed non-zero element of $$H_0^{1}(\Omega)$$. Take $$t_n=n$$.
• what if I add one more condition such that $\forall n \in \mathbb{N}, u(t_{n})\neq 0$? – Evan William Chandra Mar 25 at 8:02
• Just add $\frac 1 t$ to $\sin(t\pi)$. – Kavi Rama Murthy Mar 25 at 8:09