# Is a bounded sequence of $H^1(\Omega)$ tight?

Assume $$\Omega$$ is a bounded subset of $$\Bbb R^d$$ and $$(u_n)_n$$ is a bounded sequence of the Sobolev space $$H^1(\Omega)$$.

Question: Can we say that $$(u_n)_n$$ is tight in $$L^2(\Omega)$$ namely: For very $$\varepsilon>0$$ there exists a compact set $$K_\varepsilon\subset \Omega$$ such that $$\sup_{n}\int_{\Omega\setminus K_\varepsilon }|u_n(x)|^2dx<\varepsilon$$

I failed to find a counter example. Any help or reference in which I can find related topic is welcome.

• If $H^1(\Omega)$ is compactly embedded into $L^2(\Omega)$ in your setting, shouldn't the Vitali convergence theorem and tightness of the Lebesgue measure do the trick? (.. if you are fine with a subsequence ..) – Hannes Jun 27 at 15:16

The answer is yes if your domain has compact embedding $$H^1(\Omega)\to L^2(\Omega)$$ (that is, for any regular enough domain), or you consider $$H^1_0(\Omega)$$, but no in general. Counterexample is as follows:
Let $$\Omega = (\frac{1}{2}, 1)\cup (\frac{1}{8}, \frac{1}{4})\cup (\frac{1}{32}, \frac{1}{16})\cup \ldots = I_1\cup I_2\cup \ldots \subset \mathbb{R}$$ and consider functions $$f_n = \frac{1}{\sqrt{|I_n|}}\chi_{I_n}$$. They all are bounded in $$H^1(\Omega)$$ (indeed, they has norm $$1$$), but you can't find a compact $$K\subset \Omega$$ such that $$K\cap I_n \neq \varnothing$$ for all $$n$$ because such a $$K$$ has to contain $$0$$ which is not in $$\Omega$$.
If you do not like nonconnected examples, consider $$\Omega_1 = \Omega \times [\frac{1}{2}, 1) \cup (0, 1)\times (0, \frac{1}{2})$$ (it looks like a comb), choose a smooth function $$g$$ with support in $$(\frac{3}{5}, \frac{4}{5})$$ and consider $$g_n(x, y) = f_n(x)g(y)$$.
On the other hand, if you have a compact embedding then the answer is yes. First of all let us consider a sequence of compact sets $$K_n\subset \Omega$$ such that $$K_n \subset Int(K_{n+1})$$ and $$\bigcup K_n = \Omega$$ (such a sequence of compacts exists for any manifold and a fortiori for any open subset of $$\mathbb{R}^d$$). Assume that there exists $$\varepsilon > 0$$ such that for any $$n$$ there exists $$u_n\in H^1(\Omega)$$, $$||u_n||_{H^1(\Omega)} = 1$$ such that $$||u_n||_{L^2(\Omega \backslash K_n)} \ge \varepsilon$$. By compactness of embedding (passing to a subsequence, but for notational simplicity I will assume that it is the same sequence) we may assume that $$u_n\to u$$ in $$L^2(\Omega)$$. But functions $$v_n = |u|^2 \chi_{\Omega \backslash K_n}$$ has a common $$L^1$$-majorant $$|u|^2$$ and converges pointwise to $$0$$. Therefore $$||v_n||_{L^2(\Omega)} < \varepsilon$$ for some $$n$$ by Lebesgue Theorem. On the other hand $$w_m = u_m \chi_{\Omega \backslash K_n}$$ converges to $$v_n$$ in $$L^2(\Omega)$$. Thus, for big enough $$m$$ $$||w_m||_{L^2(\Omega)} < \varepsilon$$. But on the other hand $$||w_m||_{L^2(\Omega)} \ge ||u_m||_{L^2(\Omega \backslash K_m)}\ge \varepsilon$$ for $$m > n$$ -- a contradiction.