Do square integrable function and its derivative imply the function tends to $0$ for $|x|$ large? I know a square integrable function does not necessarily tend to $0$ for $|x|$ large. Now suppose $f\colon\mathbb{R}\to\mathbb{R}$ is a smooth function satisfying
\begin{equation}
f, f' \in L^2(\mathbb{R}).
\end{equation}
Then is it true that
\begin{equation}
\lim_{x\to\pm\infty} f(x) = 0 \quad \text{ and } \quad \lim_{x\to\pm\infty} f'(x) = 0 \, ?
\end{equation}
Is it also true for any derivative of $f$?
 A: It is true that $\lim_{x\to\infty} f(x) = 0$.  To see this, let $\chi_R(x)$ be a smooth function with $\chi_R(x) = 0$ when $|x|<R$ and $\chi_R(x) = 1$ for $|x|>2R$.  Given an $\epsilon>0$ we can find large enough $R$ so that
$$
\|\chi_R f\|_{L^2} + \|(\chi_R f)'\|_{L^2} < \epsilon.
$$
Together these imply that $\|\chi_R f\|_{L^\infty} < C \epsilon$ (this is a Sobolev embedding, but a simpler case).  This implies the first limit is true (at least once you modify $f$ on a set of measure zero). 
Let me explain that point a bit more in detail.  Let $g \in L^2(\mathbb{R})$ be a square integrable function with $g'\in L^2(\mathbb{R})$.  We want a bound on
$|g(x)|$.  Momentarily assume that $g\in C_c^\infty(\mathbb{R})$ is smooth and compactly supported.  Then by the fundamental theorem of calculus, and the Cauchy-Schwarz inequality,
\begin{align*}
|g(x)^2| &\leq \int_{-\infty}^x |(g(x)^2)'|\,dx \\
&= \int_{-\infty}^x 2|g(x) g'(x)|\,dx  \\
&\leq 2 \left(\int_{-\infty}^x |g(x)|^2\right)^{1/2}\left(\int_{-\infty}^x |g'(x)|^2\,dx\right)^{1/2}.
\end{align*}
Taking square roots we obtain the expression
$$
|g(x)| \leq \sqrt{2} \|g\|_{L^2} \|g'\|_{L^2}.
$$
Now that we know this is true for smooth compactly supported functions, it remains true for functions with $f,f'\in L^2(\mathbb{R})$ by a limiting argument.  
The second limit, $\lim_{x\to\infty} f'(x)$ does not need to exist or tend to $0$ because all we know about $f'$ is that $f'\in L^2$.  We construct an example here for completeness.  Let $h(x)$ be a triangular "hat" function centered at $0$, with support in $[-1,1]$, and with slope $\pm 1$ on either side of the origin.  Observe that $2^{-n} h(2^nx-4^{n})$ is centered at $2^n$, has width $2^{1-n}$, and slope $\pm 1$ on either side of its peak at $2^n$.  Thus
$$
f(x) = \sum_{n=0}^\infty 2^{-n} h(2^nx - 4^{n})
$$
is square integrable and has square integrable derivative, but the derivative $f'(x)$ does not tend to $0$ as $x\to\infty$.
