micromass said:
9. For any infinitely differentiable monotonic function ##f:\mathbb{R}\rightarrow \mathbb{R}## such that ##\lim_{x\rightarrow +\infty} f(x) = 0##, also holds that ##\lim_{x\rightarrow +\infty} f^\prime(x) = 0##.
My old friend the
bump function will help us out again.
Let ##h_a^b## be the bump function with support ##a,b## and maximum value 1. Let ##B=\int_0^1 h_0^1(x)dx##. Then ##0<\int_a^b h_a^b(x)dx<(b-a)##.
Define ##g:\mathbb R\to\mathbb R## that is zero except when ##x\in[n,n+2^{-n}]## for some ##n\in\mathbb N## (here I take the natural numbers to start at 1, not 0), in which case ##g(x)=h_n^{n+2^{-n}}(x)##.
Note that
$$B\equiv\int_{-\infty}^\infty g(x)dx=\sum_1^\infty \int_n^{n+2^{-n}} h_n^{n+2^{-n}}(t)dt
<\sum_{k=1}^\infty 2^{-n}=1
$$Now define ##f:\mathbb R\to\mathbb R## such that $$f(x)=-B-e^{-x}+\int_{-\infty}^x g(t)dt$$
This is infinitely differentiable and monotonic increasing. The latter follows from the fact that ##g## is non-negative.
$$\lim_{x\to +\infty}f(x)=-B-0+\int_{-\infty}^\infty g(x)dx
=-B-0+B=0$$
But for every integer ##n>0## there exists ##x\in [n,n+2^{-n}]## where the bump function makes ##g## attain its maximum of 1, so that
$$f'(x)=e^{-x}+\frac d{dx}\int_{-\infty}^x g(t)dt
=e^{-x}+g(x)
=e^{-x}+1>1
$$
So we cannot have ##\lim_{x\to+\infty}f'(x)=0##. In fact, since outside those little intervals ##f'(x)## goes to zero, there is no limit of ##f'(x)## as ##x\to\infty##.