About limit at $\infty$ of $f\in L^2(\mathbb R)\cap AC(\mathbb R)$ with $f'\in L^2(\mathbb R)$ If $f\in L^2(\mathbb R)\cap AC(\mathbb R)$ and $f'\in L^2(\mathbb R)$ is it true that $\lim_{t\rightarrow\infty}f(t)=0$? If true, proof? If false, counterexample?
Thanks
PS: note that the standard proof that works with $f\in L^1(\mathbb R)\cap AC(\mathbb R)$ such that $f'\in L^1(\mathbb R)$, use the fact that if $x\in\mathbb R$ then $\int_x^{+\infty}|f'(t)|dt<+\infty$ to show that the limit is Cauchy and then exists. Here we no longer have that $\int_x^{+\infty}|f'(t)|dt<+\infty$, at least in principle, so, answering the question about $L^1(\mathbb R)$ is not the same of answering the question about $L^2(\mathbb R)$.
 A: Suppose $x<y.$ Then $$|f(y)-f(x)|= |\int_x^yf'\,| \le (\int_x^y|f'|^2)^{1/2}(y-x)^{1/2}\le \|f'\|_2(y-x)^{1/2}.$$ This shows $f \in \text {Lip}_{1/2}(\mathbb {R)}.$ Thus $f$ is uniformly continuous. Now a uniformly continuous function on $\mathbb {R}$ that doesn't $\to 0$ at $\pm \infty$ cannot belong to any $L^p(\mathbb {R}), 0 < p <\infty.$ Since we know $f\in L^2,$ we must have $f\to 0$ at $\pm \infty.$
A: Here's a more direct proof. Since $f$ is absolutely continuous, we can write:
$$
f(t)=f(s)+\int_s^tf'(\xi)\,d\xi \quad \forall t,s \in \mathbb{R}.
$$
Using the fact that
$$
(a+b)^2 \le 2(a^2+b^2) \quad \forall a,b \in \mathbb{R},
$$
we deduce that
$$\tag{1}
f^2(t)\le 2\left[f^2(s)+\left(\int_s^tf'(\xi)\,d\xi\right)^2\right] \quad \forall t,s \in \mathbb{R}.
$$
Integrating (1) over $t-1/2\le s\le t+1/2$, it follows from Jensen's inequality
and Fubini's theorem that:
\begin{eqnarray}
f^2(t)&\le& 2\int_{t-1/2}^{t+1/2}\left[f^2(s)+\left(\int_s^tf'(\xi)\,d\xi\right)^2\right]\,ds\\
&\le& 2\left[\int_{t-1/2}^{t+1/2}f^2(s)\,ds+\int_{t-1/2}^{t+1/2}\int_{t-1/2}^{t+1/2}(f')^2(\xi)\,d\xi\,ds\right]\\
&=&2\left[\int_{t-1/2}^{t+1/2}f^2(s)\,ds+\int_{t-1/2}^{t+1/2}(f')^2(\xi)\,d\xi\right]
\end{eqnarray}
i.e.
$$\tag{2}
f^2(t)\le 2\int_{t-1/2}^{t+1/2}\left[f^2(\xi)+(f')^2(\xi)\right]\,d\xi \quad \forall t\in \mathbb{R}.
$$
It follows from (2) that
$$\tag{3}
f^2(t)\le 2\int_{t-1/2}^\infty\left[f^2(\xi)+(f')^2(\xi)\right]\,d\xi \quad \forall t\in \mathbb{R}.
$$
Since $f,f'\in L^2(\mathbb{R})$ we have
$$
\lim_{t\to\infty}\int_{t-1/2}^\infty\left[f^2(\xi)+(f')^2(\xi)\right]\,d\xi=0,
$$
and thanks to (3) we have $\lim_{t\to\infty}f^2(t)=0$. Similarly, we have $\lim_{t\to-\infty}f^2(t)=0$.
