# Show a non-compact embedding

compactnessfunctional-analysisreal-analysissobolev-spaces

We know over bounded domains smoothness yields pre-compactness by Sobolev embeddings, and this is not true over unbounded domains. I am trying to make an example to show that the embedding
$$H^1(\mathbb{R}) \hookrightarrow L^2(\mathbb{R})$$ is not compact. I chose this sequence (which I think will work)
$$\{ u_n=e^{-(x + n)^2}\}_{n \in \mathbb{N}}$$,and showing it is bounded in $$H^1(\mathbb{R})$$, as

\begin{align} \|u_n\|_{H^1}^2 &= \int_\mathbb R (1 + |k|^2) |\mathcal{F}({e^{-(x + n)^2})}|^2 dk\\ &=\int_\mathbb R (1 + |k|^2) |\frac{1}{\sqrt{2}} e^{\frac{-k^2}{4} -i n k}|^2 dk \\ &\leq \int_\mathbb R (1 + |k|^2) e^{\frac{-k^2}{2}} dk\\ &= 2 \sqrt{2 \pi} \end{align}

I am trying to show that $$u_n$$ does not have a convergent subsequence in $$L^2(\mathbb{R})$$. Could you please show a systematic way to tackle such tasks.

A systematic way to show that a sequence has no convergent subsequence is to recall that every convergent sequence has to be a Cauchy sequence. Hence, if you can show that there is some $$\delta>0$$ such that $$\lVert u_n-u_m\lVert\ge\delta$$ for every $$n\ne m$$, you are done. (The condition is actually even necessary in complete spaces if the sequence elements are pairwise different, but this is harder to see.)