If $f \in H^{1}(\mathbb{R})$ does this imply that $f(x) \to 0$ as $|x| \to \infty$?
I know that if $f $ is uniformly continuous then the answer is yes, but i'm not so sure about this case..
2026-04-07 03:16:41.1775531801
$H^{1}(\mathbb{R})$ functions vanish at infinity?
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By the fundamental theorem of calculus and Cauchy-Schwarz, for $x>y$: $$ |f(x)-f(y)|\leq \int_y^x |f'(t)|dt \leq \Vert f'\Vert_{L^2}|x-y|^{1/2}. $$ Therefore, $f\in H^1$ implies $f$ uniformly continuous.