Second order Taylor expansion with Lagrangian form of the remainder I have a vector-valued smooth function $\boldsymbol{f}:\mathbb{R}^m \rightarrow \mathbb{R}^n$ and I want to use the second order Taylor expansion to bound
$$\sum_{i=1}^{n}(f_i(\boldsymbol{v} +\boldsymbol{h})-f_i(\boldsymbol{v})-\boldsymbol{h}^{T} \nabla f_i(\boldsymbol{v}))^2.$$
(It's squared euclidean norm.) Taylor's theorem for each coordinate implies 
$$\sum_{i=1}^{n}(f_i(\boldsymbol{v} +\boldsymbol{h})-f_i(\boldsymbol{v})-\boldsymbol{h}^{T} \nabla f_i(\boldsymbol{v}))^2 \le \frac14 \sum_{i=1}^{n} \sup_{\xi_i \in [v,v+h]}|\boldsymbol{h}^T\nabla^2 f_i(\boldsymbol{\xi_i})\boldsymbol{h}|^2.$$
But can we change the order of sum and supremum? In other words, can we write that
$$\sum_{i=1}^{n}(f_i(\boldsymbol{v} +\boldsymbol{h})-f_i(\boldsymbol{v})-\boldsymbol{h}^{T} \nabla f_i(\boldsymbol{v}))^2 \le \frac14 \sup_{\xi \in [v,v+h]} \sum_{i=1}^{n} |\boldsymbol{h}^T\nabla^2 f_i(\boldsymbol{\xi})\boldsymbol{h}|^2.$$
Thanks! 
 A: Using the integral formulation of the remainder term (assume $f(0)=0$ just for the ease of typing) 
$$
f(h)=Df(0)h+\int_0^1 D^2f(sh)(h,h)(1-s)\, ds,
$$
one gets the estimate you want:
$$
\left\lVert \int_0^1 D^2f(sh)(h,h)(1-s)\, ds\right\rVert_{\mathbb{R}^n}\le \sup_{s\in[0,1]}\lVert D^2f(sh)(h,h) \rVert_{\mathbb{R}^n}\int_0^1(1-s)\,ds$$
and $\int_0^1(1-s)\,ds=\frac{1}{2}.$
A comment on notation: $$Df(0)h=\sum_i\left(\sum_j \frac{\partial f^i}{\partial x^j}(0)\, h_j\right)e_i$$
$$
D^2f(p)(h,h)=\sum_k\left(\sum_{i,j}\frac{\partial^2 f}{\partial x_i, \partial x_j}(p)\,h_ih_j\right) e_k, $$
where $e_k=(0,0,\ldots 1,\ldots, 0,0)$, the $1$ being in the $k$-th place.
EDIT Some more commentaries. If $f\colon \mathbb{R}^n\to \mathbb{R}$ is real valued (and $f(0)=0$ for simplicity), then one can have the Lagrangian form of the remainder, i.e. 
$$\tag{Lagr}\exists \xi \in (0,1)\quad \mathrm{s.t.}\quad f(h)=Df(0)h +  \frac{1}{2}D^2f(\xi h)(h,h).$$
However, if $f\colon \mathbb{R}^n\to \mathbb{R}^m$ is vector-valued, (Lagr) is not true. Take for example $m=2$, so that $f$ can be considered complex valued, and let $f(x)=e^{ix}-1$ for real $x$. Then (Lagr) becomes the identity
$$
\tag{!!} e^{ih}+\frac{1}{2}e^{i\xi h}h^2=1+ih, \qquad \text{some }\xi\in(0,1)$$
which is false. EDIT Explanation. By "(!!) is false" I mean in this case that there exists a neighborhood $(-\delta, \delta)$ of $0$ (which is the center of our Taylor expansion) such that for any $h\in(-\delta, \delta)$ there does not exist any $\xi\in(0,1)$ such that (Lagr) holds. 
Proof: one simply expands exponentials to second order in $h$, using the Peano formulation of the remainder (which is true for vector-valued functions also). One gets 
$$
1+ih+(1+i)\frac{h^2}{2}+O(h^3)=1+ih, $$
which would imply $(1+i)h^2=O(h^3)$, which is false in a small enough neighborhood of $0$, because the right hand side is asymptotically smaller than the left hand side.
