Let $H^1(\mathbb{R})=\{f\in L^2(\mathbb{R}): f\in AC([a,b])\,\, \text{for all}\,\, [a,b]\subseteq\mathbb{R}\,\,\text{and}\,\, f^{\prime}\in L^2(\mathbb{R}) \}$, where $AC([a,b])$ denotes the set of all absolutely continuous functions on $[a,b]$, namely all functions such that there is $h\in L^1([a,b])$ such that $f(x)=f(a)+\int_{[a,x]} h(t)dt$ for all $x\in[a,b]$. Now I want to prove that for all $f\in H^1(\mathbb{R})$ one has $\lim_{b\to\infty} f(b)=0$ and $\lim_{a\to-\infty} f(a)=0$(which can later be used in a proof that the momentum operator is in fact self-adjoint on $H^1(\mathbb{R})$). I was given the hint to use the integration by parts formula for $f$ and $f^{\prime}$ which yields
$\langle f,f^{\prime}\rangle+\langle f^{\prime},f\rangle=f(b)\overline{f(b)}-f(a)\overline{f(a)}=\lvert f(b)\rvert^2-\lvert f(a)\rvert^2$.
I tried using the fact that $f$ is absolutely continuous and hence admitting such a representation as above but it did not work. I suppose you have to use the integration by parts formula on different intervals because it is valid on all $[a,b]\subseteq\mathbb{R}$ but I don't see how.
Can someone please help me?
First note that $f$ is uniformly continuous: if $x<y$ then$$\begin{aligned}|f(x)-f(y)|&=\left|\int_x^yf '\right|\le\int_x^y|f'|=\int f'\chi_{[x,y]}\\&\le ||f'||_{L^2(\Bbb R)}||\chi_{[x,y]}||_{L^2(\Bbb R)}=||f'||_2(y-x)^{1/2}.\end{aligned}$$ (So given $\epsilon>0$ if you choose $\delta>0$ so $||f'||_2\delta^{1/2}<\epsilon$ you're set.)
Now if $f$ does not tend to $0$ at $\infty$ then there is a sequence $x_n\to\infty$ and a number $c>0$ with $|f(x_n)|\ge c$. Now since $f$ is uniformly continuous this implies that $\int|f|^2=\infty$, because...
Hint for that last because: Since $f$ is uniformly continuous there exists $\delta>0$ such that $|f|\ge c/2$ on $[x_n-\delta, x_n+\delta]$ for every $n$...