Let $f: \mathbb{R} \to \mathbb{C}$ be a $C^1$-differentiable function such that for all $n\in \mathbb{N} \cup \{0\}$,
$\int_{-\infty}^{\infty}|f(x)|^2(1+|x|)^ndx < \infty$
Are the following two assertions true?
$\lim_{|x| \to \infty}f(x)=0$
$\lim_{|x| \to \infty}f'(x)=0$
Let $\psi$ be the Bump Function which is $1$ on $[1/3,2/3]$ and $0$ outside $(0,1)$ and set $f=\sum_{n\ge 0} \psi(2^n (x-n))$. Note that $\text{supp}|f|^2=\text{supp}f$ because each $\psi(2^n(x-n))$ has disjoint support and $|f|^2\le 1$. Then $$\begin{align} \int_{\Bbb R} |f|^2(1+|x|)^k \text{d}x &= \int_0^{\infty} \sum_{n\ge 0} \psi^2(2^n (x-n))(1+x)^k\text{d}x \overset{(1)}=\sum_{n\ge 0}\int\psi^2(2^n(x-n))(1+x)^k\text{d}x \\&\le \sum_{n\ge 0}\int_{(n,n+2^{-n})}(1+x)^k\text{d}x\le \sum_{n\ge 0}\int_{(n,n+2^{-n})}(2+n)^k\text{d}x \\ &\le \sum_{n\ge 0}2^{-n}(2+n)^k<\infty \quad \text{by the Ratio Test} \end{align} $$ Where $(1)$ is justified because $\psi(2^n(x-n))(1+x)^k$ are all measurable. Yet $f(2^{-n}/3+n)=1$ so $\lim_{|x|\to \infty} f(x)\neq 0$. Similarly, the MVT shows there exists some point $c\in (0,1/3)$ where $\psi'(c)=1$, so $\lim_{|x|\to \infty} f'(x)\neq 0$ either.