TITLE: $ |f''(x)+2xf'(x)+(x^2+1)f(x)|\leq1 $ for all $x$. Prove $ \lim_{x\rightarrow\infty }f(x)=0$
QUESTION [5 upvotes]: Let $f(x):(0, \infty)\rightarrow \mathbb{R} $ be a twice continuously  differentiable function 
such that
| $ f''(x)+2xf'(x)+(x^2+1)f(x) |\leq1 $ for all $x$. Prove $  \lim_{x\rightarrow\infty }f(x)=0$
I think this can be solved by applying  L'Hospital  rule twice on $\frac{e^{x^2}f(x)}{e^{x^2}}$. My problem is, Is it possible to apply the rule without knowing about numerator . Also I like to see different types of proofs

REPLY [2 votes]: Let $g(x)=e^{x^2/2}f(x)$. Then we have:
$$ \left| g''(x)\right| \leq e^{x^2/2} \tag{1}$$
$$ |g'(y)|\leq |g'(0)|+\int_{0}^{|y|}e^{t^2/2}\,dt \tag{2}$$
$$\begin{eqnarray*} |g(x)|&\leq& |g(0)|+|g'(0)||x|+\int_{0}^{|x|}\int_{0}^{|y|}e^{t^2/2}\,dt\,dy\\&\leq&C\sqrt{1+x^2}+\int_{0}^{1}\frac{-1+e^{x^2 t^2/2}}{t^2}\,dt \tag{3}\end{eqnarray*}$$
and now it is possible to exploit convexity or some variant of the Laplace's method to prove that the last quantity, when multiplied by $e^{-x^2/2}$, goes to zero as long as $|x|\to +\infty$.