Prove that $\Vert f(b)-f(a)-f'(a)(b-a) \Vert\leq \sup_{x\in [a,b]} \Vert f''(x)\Vert\Vert b-a\Vert^2$ Let $\Omega$ be an open set in $\Bbb{R}^n$ and $f:\Omega\to \Bbb{R}^m $ be of class $\Bbb{C}^2.$ Let $a,b$ be in $\Omega$ such that $[a,b]\subset \Omega.$ Prove that 
\begin{align}\Vert f(b)-f(a)-f'(a)(b-a) \Vert\leq \sup_{x\in [a,b]} \Vert f''(x)\Vert\Vert b-a\Vert^2.\end{align}
where
$$f'(a)(b-a)=\big<\nabla f(x),(b-a)\big>=\sum^{n}_{k=1}\frac{\partial f_i}{\partial x_k}(a)(b_k-a_k).$$
I  think the idea of Mean Value Theorem for $C'$ functions in $\Bbb{R}^n$ (Reference for Mean Value Theorem in several variables) will help me in showing it but I don't know how to! Please, can anyone show me a proof or reference?
 A: Let
$$g(x)=f(b)-f(x)-f'(x)(b-x).$$
Note that $f'(x)$ here is defined to be the $m\times n$ matrix with entries $\partial f_i(x)/\partial x_j$, where $f(x)=(f_1(x),\ldots, f_m(x))$. Therefore, the product $f'(x)(b-x)$ is understood as the matrix product of $f'(x)$ and $b-x$. The result is a vector in $\mathbb{R}^m$ with makes $g(x)$ a meaningful function from $\Omega$ into $\mathbb{R}^m$. The derivative of $g(x)$ is again an $m\times n$ matrix whose $ij$-th entry is given by 
$$-\sum_{k}\dfrac{\partial f_i(x)}{\partial x_j\partial x_k}(b_k-x_k).$$
Then
$$\|g'(x)\|^2=\sum_{i,j} \|\sum_{k}\dfrac{\partial f_i(x)}{\partial x_j\partial x_k}(b_k-x_k) \|^2 \leq \left (\sum_{i,j,k} \left \|\dfrac{\partial f_i(x)}{\partial x_j\partial x_k} \right \|^2 \right ) \|b-x\|^2.$$
It follows from this and substitution in the inequality
$$\|g(b)-g(a)\| \leq \sup_{c\in [a,b]}\|g'(x)\|\|b-a\|,$$
that 
$$\|f(b)-f(a)-f'(a)(b-a)\| \leq \sup_{x\in [a,b]} \left (\sum_{i,j,k} \left \|\dfrac{\partial f_i(x)}{\partial x_j\partial x_k} \right \|^2 \right )^{1/2} \|b-a\|^2.$$
A reasonable definition for $\|f''(x)\|$ would be the triple sum given above and I assumed that's what it is meant by it. 
