Asymptotics of a Sobolev function on unbounded interval Suppose $f : (0,\infty) \to \mathbb{R}$ is locally $H^1$ and $\int_0^\infty (|f'(t)|^2 + |f(t)|^2) e^{-t} dt$ is finite. Is then $\lim_{t \to \infty} e^{-t} |f(t)|^2 = 0$?
 A: Yes. Suppose there is $c>0$ such that $|f(t_k)|>ce^{t_k/2}$ for some sequence $t_k\to \infty$. We may assume that the intervals $I_k=[t_k-1,t_k+1]$ are disjoint. For each $k$, one of the following holds:


*

*If $\min_{I_k} |f | \ge \frac12 ce^{t_k/2}$, then $\int_{I_k} e^{-t}|f(t)|^2\,dt\ge c^2/2$.   

*If $\min_{I_k } |f | < \frac12 ce^{t_k/2}$, then by the Fundamental theorem of calculus (which applies because $f$ is locally absolutely continuous) $\int_{I_k} |f'(t)|\,dt \ge  \frac12 ce^{t_k/2}$. By the   Cauchy-Schwarz inequality $$\int_{I_k} e^t\,dt\  \int_{I_k} e^{-t}|f'(t)|^2 \,dt \ge \left(\int_{I_k} |f'(t)|\,dt \right)^2\ge \frac14 c^2 e^{t_k}$$
which gives a uniform lower bound on  $\int_{I_k} e^{-t}|f'(t)|^2$. 


Thus,  $\int_{I_k} e^{-t}(|f(t)|^2+|f'(t)|^2)\,dt$ diverges.
A: Yes, it is. To show this, denote
$$
M=\int_0^\infty \bigl(|f'(t)|^2 + |f(t)|^2\bigr) e^{-t} dt\tag{1}
$$
and observe that function $g(t)=f(t)e^{-t/2}$ satisfies inequality
$$
\|g\|^2_{L^2(0,\infty)}=\int_0^{\infty}|g(t)|^2dt=\int_0^{\infty}|f(t)|^2e^{-t}dt
\leqslant M, \tag{2}
$$
while $|g'(t)|^2= \bigl(f'(t)e^{-t/2}-g(t)/2\bigr)^2\leqslant 
2|f'(t)|^2e^{-t}+|g(t)|^2/2$,
whence by $(1),(2)$ follows
$$
\|g\|^2_{H^1(0,\infty)}=\|g\|^2_{L^2(0,\infty)}+\|g'\|^2_{L^2(0,\infty)}
\leqslant 2M.
$$
Sobolev space $H^1(0,\infty)$ is known to consist of uniformly absolutely 
continuous on $[0,\infty)$ functions vanishing at infinity, which immediately 
implies
$$
\lim_{t\to\infty}e^{-t/2}f(t)=\lim_{t\to\infty}g(t)=0.
$$
Remark. It is not difficult to verify that all functions $g\in H^1(0,\infty)$ 
vanish at infinity. Indeed, denote by $\widetilde{g}$ an even extension of $g$ from 
$(0,\infty)$ to $\mathbb{R}$, and notice that $\widetilde{g}\in H^1(\mathbb{R})$.
By virtue of Plancherel's theorem, Fourier transform $\widehat{g}=F[\widetilde{g}]$ 
satisfies inequality
$$
\|\widehat{g}\|^2_{L^1{(\mathbb{R})}}\leqslant 
\int_{\mathbb{R}}\bigl(1+{\xi}^2\bigr)|\widehat{g}(\xi)|^2 d\xi
\cdot\!\!\int_{\mathbb{R}}\frac{d\xi}{1+{\xi}^2}
=2{\pi}^2\|\widetilde{g}\|^2_{H^1(\mathbb{R})}=
4{\pi}^2\|g\|^2_{H^1(0,\infty)}\,,
$$
i.e.,   $\|\widehat{g}\|_{L^1{(\mathbb{R})}}\leqslant 2\pi \sqrt{M}<\infty$.  
And finally, the Riemann-Lebesgue lemma
(http://en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma)   implies that 
$$
\lim_{t\to\infty}g(t)=\lim_{t\to\infty}\widetilde{g}(t)=0.
$$
