limit of function in Sobolev space

Question

Let function $f(x)$ and $f(x)g(x)$ belong to $\mathcal{W}^{s+1}$ with $s\ge 1$, where $\mathcal{W}^{s+1}$ be the Sobolev space of regularity $s+1$ in $\mathbb{L}^2$-norm. We also have $g\in \mathbb{L}^{\infty}(\mathbb{R_+})$ with $\mathbb{R}_+ = [0,\infty)$ and for all $k\ge 0$, $\int x^kf(x)dx <\infty$ and $0 <\int f(x)g(x)dx <\infty$. Can we conclude that: $$ \lim_{x\rightarrow \infty} xf(x)g(x) = 0 $$

Answer 1

No, that's not true. Take $$ g(x) = 2^{-x}. $$ Then consider a function $\phi\in C^\infty$ which has compact support in $[-1/2,1/2]$ and such that $0\le \phi(x)\le 1$ and $\phi(0)=1$. Define $$ f(x) = \sum_{k=1}^\infty \phi((2k^22^k)^k(x-k^2 2^k)). $$

Notice that for $x$ fixed the previous sum has at most one term which is different from zero. Hence the function $f$ has infinitely many thin spikes centered in $x=k^22^k$ where the function takes value $1$. For the rest of the time it is $0$.

Notice that for $x_k=k^2 2^k$ one has $$ \lim_{k\to \infty} x_k f(x_k) g(x_k) = \lim_k k^2 2^k \phi(0) 2^{-k} = +\infty $$ so your limit will not exist.

However $f$ and $g$ have the requested properties. In particular $$ \int x^k f(x)\, dx \le \sum_{k=1}^\infty (k^2 2^k+1/2)^k \frac{1}{(2k^22^k)^k} < +\infty. $$