Is it true (is there a commonly known theorem) that says:
$f \in H^1(\mathbb R ^n)$ $\Rightarrow$ $\displaystyle \lim_{|x| \to \infty} f(x) = 0$ pointwise (where $H^1$ denotes the Sobolev space $W^{1,2}$)?
This can be inferred from the Sobolev embeddings for dimension sufficiently small, but is it true in general?
On
This is true for $n = 1$, but false for $n \ge 2$. Recall, that for $n \ge 2$, there is a positive function $h \in H^1(\mathbb R^n) \setminus L^\infty(\mathbb R^n)$. Now, take a sequence $\{x_i\}$ with $|x_i| \to \infty$ and define $$ f(x) = \sum_{i=1}^\infty 2^{-i} \, h(x - x_i). $$ Then you have $f(x_i) = +\infty$ for all $i$ (if you want finite function values, take $\tilde x_i$ in a neighbourhood of $x_i$).
See also here, where I used a similar construction.
I don't think so. You can imagine an infinite sequence of pyramids of height 1 with ever tinier hypersquare bases of area $A_m$, which will clearly be in $L^2$, so long as $\sum |A_m| < \infty$. The question is whether there's enough room for the gradient to remain square integrable. The gradient is like the height over the distance, which is like $|A_m|^{-\frac{1}{n}}$ where $n$ is the spatial dimension, so the requirement is for $\sum |A_m|^{-\frac{2}{n} + 1} < \infty$ as well. When $n = 1$ or $n=2$ this is clearly impossible. For higher dimensions, it is doable.