Unnecessary context: I'm trying to show equivalence between two conditions for convergence in Schwartz space $\mathcal{S}(\mathbb{R}^d) = \{f \in C^\infty(\mathbb{R}^d) | x^\alpha D^\beta f \text{ is bounded for any multi-indices } \alpha,\beta \}$, and I've reduced it to the following problem (in one dimension):
If $(f_n)_{n=1}^\infty \subset C^\infty(\mathbb{R})$ satisfies $\forall l \in \mathbb{N}$ $f_n^{(l)} \to 0$ uniformly and $\forall k,l \in \mathbb{N}$ $\exists C_{k,l}>0$ such that $\forall x \in \mathbb{R}, n \in \mathbb{N}$ we have $|x^kf_n^{(l)}(x)| \leq C_{k,l}$, then $xf_n(x) \to 0$ uniformly (on $\mathbb{R}$).
In the above, $\mathbb{N}$ includes $0$. The fairly strong hypotheses are remnants of the more general condition I'm trying to show-- they're needed for the generalization, but I doubt we'll need the conditions on all derivatives or all powers of $x$ for the above result (if it holds). I've been considering this for a while and haven't been able to come up with a proof. Noting that we trivially have the needed bounds on $|xf_n(x)|$ for $x \in [-1,1]$, my thought is to consider the integral representation of $xf_n(x)$ for $x \neq 0$ as
$$ xf_n(x) = \int_{-\infty}^{x^2} f_n' \left( \frac{t}{x} \right) dt $$
(since $|f_n(t)| = \frac{|tf_n(t)|}{|t|} \leq \frac{C_{1,0}}{|t|} \to 0$ as $|t| \to \infty$). The integrand goes to $0$ uniformly by hypothesis, but the domain of integration is unbounded. My hope was to use the hypotheses to prove that the integration over the "ends" can be made arbitrarily small in a way independent of $|x|$, leaving only the bounded domain. However, the most straightforward approaches seem to be rigged to leave a factor of $|x|$ in the bound. For example,
$$ \left| \int_N^{\infty} f_n' \left( \frac{t}{x} \right) dt \right| \leq \int_N^\infty \left| f_n' \left( \frac{t}{x} \right) \right|dt \leq \int_N^\infty \frac{C}{1+(\frac{t}{x})^2} dt =C|x| \left( \frac{\pi}{2}-\tan^{-1} \left( \frac{N}{|x|} \right) \right)$$ for some $C>0$, and I cannot choose $N$ large enough such that this is less than some fixed $\epsilon$ for all $|x|>1$. Any hints, counterexamples, or pointers indicating that I'm missing something obvious would be much appreciated.
Hint: Take a sequence $a_n $ for which $a_n \to \infty$ and $a_n\|f_n\|_\infty \to 0$ as $n\to\infty$. Then, for $x \le a_n$, $$ |xf_n(x)| \le a_n \|f_n\|_\infty =o(1)$$ and for $|x|> a_n$, $$ |xf_n(x)| \le C_{2,0}|x|^{-1} = O(a_n^{-1}). $$ Can you fill in the gaps and continue from here? Hope this helps.