Suppose $f$ is a $C^2$ function with compact support. I.e. $f$ is $0$ outside a closed interval. Then $f,f',f''$ are uniformly continuous and bounded on $\mathbb{R}$. My textbook then claims that the following is true:
if $x, y \in \mathbb{R}$, then
$$f(y)-f(x) = (y-x)f'(x) +1/2(y-x)^2 f''(x) + r(x,y)$$
where $|r(x,y)| \leq (y-x)^2 h(x,y)$ and $h: \mathbb{R}^2 \to [0, \infty[$ is a uniformly continuous function with $\Vert h \Vert_\infty < \infty$ and $h(x,x)= 0$ for all $x \in \mathbb{R}$.
I figured this should be a consequence of Taylor's theorem:
By Taylor's theorem (expand around $x$), we can write
$$f(y) = f(x) + f'(x)(y-x) + 1/2f''(x)(y-x)^2 + r_x(y)(y-x)^2$$
How to proceed? Which function to take for $h$? I think the boundedness of $h$ will follow from the boundedness of $f,f',f"$. Thanks in advance.
Using integration by parts we see that $$f(y)-f(x)-(y-x)f’(x)=(y-x)^2\int_0^1(1-u)f’’(x+u(y-x))du$$ So, if $$\delta(x,y)=\dfrac{f(y)-f(x)-(y-x)f’(x)-\frac12(y-x)^2f’’(x)}{(y-x)^2}$$ for $x\ne y$ and $\delta(x,x)=0$, then $$\delta(x,y)=\int_0^1(1-u)\big(f’’(x+u(y-x))-f’’(x)\big)\,du$$ Now, we only need to set $$h(x,y)=\int_0^1(1-u)\big|f’’(x+u(y-x))-f’’(x)\big|\,du.$$ Clearly, $h(x,x)=0$ and $$\sup_{(x,y)\in\mathbb{R}^2}h(x,y) \le \sup_{x\in\mathbb{R}}|f’’(x)|$$ Further, the triangle inequality shows that the uniform continuity of $f’’$ implies the uniform continuity of $h$.