Order of remainder term in Taylor series approximation I'm having trouble verifying a bound on the remainder term of a Taylor series approximation. I have a $C^\infty$ function $f$ of compact support. Using the two-term Taylor series for $f$ centered at $x$, we have
$$f(x+y)=f(x)+f'(x)y+\frac{f''(x)y^2}{2!}+R(y)$$
where $R(y)=\frac{f^{(3)}(z)y^3}{3!}$ for some $z$ between $x$ and $y$. This paper claims (in equation 4) that for every $\epsilon >0$ there exists $\delta>0$ and $C<\infty$ such that for any $x,y \in \mathbb{R}$, $|R(y)| \leq \epsilon y^2$ if $|y| \leq \delta$, and $|R(y)| \leq C y^2$ otherwise. 
I see where the first bound comes from: $f^{(3)}$ is uniformly continuous as $f$ has compact support, so $|f^{(3)}|$ is bounded above by some constant $K$. Setting $\delta = (\epsilon \cdot 3!/K)$, if $|y| \leq \delta$ then $|R(y)|=|f^{(3)}(z)y^3/3!|\leq K/3! \cdot (\epsilon \cdot 3!/K) \cdot y^2 = \epsilon y^2$. My question is how to obtain the other bound, for when $|y|> \delta$. Taylor's theorem gives a bound on the order of $|y|^3$, not $y^2$.
 A: An integral representation of $R(y)$ is always more accurate.
Write $\displaystyle R(y)=\int_x^{x+y} \frac{f'''(t)}{2} (x+y-t)^2 dt=\int_0^y \frac{f'''(x+u)}{2}(y-u)^2du$
Note that $\displaystyle \frac{R(y)}{y^2}=\int_0^y \frac{f'''(x+u)}{2}(1-\frac uy)^2du=y\int_0^1\frac{f'''(x+vy)}{2}(1-v)^2 dv $
Since all the derivatives of $f$ have compact support, they all are bounded, so there's some $M>0$ such that $\forall x, |f'''(x)|\leq M$
That implies $\displaystyle\left| \frac{R(y)}{y^2}\right|\leq |y|\int_0^1\frac{M}{2}(1-v)^2=|y|\frac M3 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (1)$
Since $\displaystyle \frac{R(y)}{y^2} = \int_0^y \frac{f'''(x+u)}{2}(1-\frac uy)^2du$, the function $\displaystyle y\to  \frac{R(y)}{y^2}$ is continuous. Since $f'''$ has compact support, there is some $K_x>0$ such that $|y|\geq K_x \implies \displaystyle \frac{R(y)}{y^2} = \int_0^{K_x} \frac{f'''(x+u)}{2}(1-\frac uy)^2du$
Then $\displaystyle |y|\geq K_x \implies \left| \frac{R(y)}{y^2}\right|\leq \frac{MK_x}{2}$
The function $\displaystyle y\to  \frac{R(y)}{y^2}$ is therefore bounded over $\mathbb R$ by some $C_x >0 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (2)$
The bounds $(1)$ and $(2)$ almost answer your question. The problem is that $C_x$ depends on $x$...
