Let us assume that function $f:\mathbb{R}^{+}\rightarrow \mathbb{R}^{+}$ satisfies $$ \left|\frac{d^k f}{dx^k}\right|<\frac{C}{1+x^k} $$ for $k=0,1,2$ and some $C>0$.
Let $g(x)=f(x)-v(x)$, where $v(x)$ is an unknown function.
I wonder about condition for $v(x)$ that guarantees that $$ \left|\frac{d^k g}{dx^k}\right|<\frac{D}{1+x^k} $$ for $k=0,1,2$ and some $D>0$.
I know that I can use the triangle inequality, then: $$ \left|\frac{d^k g}{dx^k}\right| < \left|\frac{d^k f}{dx^k}\right| + \left|\frac{d^k v}{dx^k}\right| $$
and now I can say that $\left|\frac{d^k v}{dx^k}\right|<\frac{D}{1+x^k}$ is enough condition that I need.
My question is: Can I generalize this condition to receive only conditions on $v(x)$ function.
I suppose it can be the following one:
$$ \lim_{x\rightarrow \infty} v(x) = const. $$
We have $v=f-g$. Tren $$ v^{(k)}(x)=f^{(k)}(x)-g^{(k)}(x). $$ This shows that $v$ satisfies the same condition as $f$ and $g$ with constant $C+D$.