A Sobolev embedding type question My question is, if a function is in $H^1(\mathbb{R})$, must it necessarily vanish at infinity?
If not, what are some additional criteria required? Thanks.

Strictly speaking, the elements of $H^1(\mathbb R)$ are not  functions, but rather equivalence classes. But since each of them admits a continuous representative, we can focus on that representative. And with that interpretation the answer is yes, $H^1(\mathbb R)\subset C_0^1(\mathbb R)$. 
Proof "from scratch". Let $u\in H^1(\mathbb R)$. Given $\epsilon>0$, pick $M$ such that $$\int_{|x|>M}u^2<\epsilon^2\tag1$$ and $$\int_{|x|>M}(u')^2<\epsilon^2\tag2$$  Suppose $|x|>M+1$. By (1), $u^2$ cannot be greater than $\epsilon^2$ everywhere on the interval $[x-1,x+1]$. Let $y\in [x-1,x+1]$ be a point with $|u(y)|\le \epsilon$. Use the Fundamental Theorem of Calculus, Cauchy-Schwarz, and (2):
$$\begin{split}
|u(x)|&\le \epsilon+\int_{[x,y]}|u'| \\&\le \epsilon+\int_{x-1}^{x+1}|u'| \\& \le \epsilon+\sqrt{2}\,\sqrt{\int_{x-1}^{x+1}(u')^2} \\& < (1+\sqrt{2})\epsilon
\end{split}
$$
Done.
