Correct Notation of Mean Value Theorem for Vector-Valued Function Let ${\bf f}({\bf x})$ be a function ${\bf f}: \mathbb{R}^n \to \mathbb{R}^m$ with continuous derivatives ${\bf H}({\bf x})$. We wish to approximate ${\bf f}({\bf x}_0)$ by ${\bf f}({\bf x})$.

It is well known that for $m &gt; 1$, we cannot guarantee the existence of a vector $\bf \tilde{x}$ between ${\bf x}_0$ and ${\bf x}$ such that
$${\bf f}({\bf x}) = {\bf f}({\bf x}_0) + {\bf H}({\bf \tilde{x}})({\bf x}-{\bf x}_0) $$

Hence, the MVT cannot be directly applied to vector-valued function. However, a straightforward modification to the MVT can yield fruitful results, but seems ignored by many sources (including this very site).
We can apply the MVT to each of the $m$ components of ${\bf f}$ separately, and thus we can write
$${f_k}({\bf x}) = f_k({\bf x}_0) + {\bf h}_k({\bf \tilde{x}}_k)({\bf x}-{\bf x}_0) , k = 1,...,m$$
I have seen some authors combine the $m$ results and write

$${\bf f}({\bf x}) = {\bf f}({\bf x}_0) + {\bf H}({\bf {x}_*})({\bf x}-{\bf x}_0) $$
where ${\bf {x}_*} =[{\bf \tilde{x}}_1, ...,{\bf \tilde{x}}_m]^T$ is now a $m \times n$ matrix, and ${\bf H}({\bf {x}_*}) = [{\bf h}_1({\bf \tilde{x}}_1),...,{\bf h}_m({\bf \tilde{x}}_m)]^T$.

Although I remember seeing this notation multiple times, I can't remember how exactly how it was expressed, or in which article I viewed it.
Could someone kindly reminds me how to properly use this notation for vector-valued functions?
The following might be what you are looking for: 

Given such an ${\bf f}$ with open domain $\Omega\subset{\mathbb R}^n$, let $\tilde\Omega:=\bigl\{({\bf x},{\bf y})\bigm| [{\bf x},{\bf y}]\subset\Omega\bigr\}$, whereby $[{\bf x},{\bf y}]$ denotes the segment connecting ${\bf x}$ with ${\bf y}$. Then there is a continuous function $L:\&gt;\tilde\Omega\to{\cal L}({\mathbb R}^n,{\mathbb R}^m)$, the linear maps from ${\mathbb R}^n$ to ${\mathbb R}^m$, such that
$${\bf f}({\bf y})-{\bf f}({\bf x})=L({\bf x},{\bf y}).({\bf y}-{\bf x})\qquad\forall\&gt;({\bf x},{\bf y})\in\tilde\Omega\ .$$
Furthermore $$L({\bf x},{\bf x})=d{\bf f}({\bf x})\qquad\forall\,{\bf x}\in\Omega\ .$$
This $L({\bf x},{\bf y})$ is exactly what you denote by ${\bf H}({\bf x_*})$. But neither your ${\bf x}_*$ is a point, nor ${\bf H}$ is a function of one vector variable.

Proof. Fix $({\bf x},{\bf y})\in\tilde\Omega$, and consider the auxiliary function
$$\phi(t):={\bf f}\bigl((1-t){\bf x}+t{\bf y}\bigr)\qquad(0\leq t\leq1)$$
with derivative $\phi'(t)=d{\bf f}\bigl((1-t){\bf x}+t{\bf y}\bigr).({\bf y}-{\bf x})$. One then has
$${\bf f}({\bf y})-{\bf f}({\bf x})=\phi(1)-\phi(0)=\int_0^1\phi'(t)\&gt;dt =\int_0^1 d{\bf f}\bigl((1-t){\bf x}+t{\bf y}\bigr).({\bf y}-{\bf x})\&gt;dt=\left(\int_0^1 d{\bf f}\bigl((1-t){\bf x}+t{\bf y}\bigr)\&gt;dt\right).({\bf y}-{\bf x})\ .$$
It follows that
$$L({\bf x},{\bf y}):=\int_0^1 d{\bf f}\bigl((1-t){\bf x}+t{\bf y}\bigr)\&gt;dt$$
has the stated properties.
The following might be what you are looking for: 

Given such an ${\bf f}$ with open domain $\Omega\subset{\mathbb R}^n$, let $\tilde\Omega:=\bigl\{({\bf x},{\bf y})\bigm| [{\bf x},{\bf y}]\subset\Omega\bigr\}$, whereby $[{\bf x},{\bf y}]$ denotes the segment connecting ${\bf x}$ with ${\bf y}$. Then there is a continuous function $L:\&gt;\tilde\Omega\to{\cal L}({\mathbb R}^n,{\mathbb R}^m)$, the linear maps from ${\mathbb R}^n$ to ${\mathbb R}^m$, such that
$${\bf f}({\bf y})-{\bf f}({\bf x})=L({\bf x},{\bf y}).({\bf y}-{\bf x})\qquad\forall\&gt;({\bf x},{\bf y})\in\tilde\Omega\ .$$
Furthermore $$L({\bf x},{\bf x})=d{\bf f}({\bf x})\qquad\forall\,{\bf x}\in\Omega\ .$$
This $L({\bf x},{\bf y})$ is exactly what you denote by ${\bf H}({\bf x_*})$. But neither your ${\bf x}_*$ is a point, nor ${\bf H}$ is a function of one vector variable.

Proof. Fix $({\bf x},{\bf y})\in\tilde\Omega$, and consider the auxiliary function
$$\phi(t):={\bf f}\bigl((1-t){\bf x}+t{\bf y}\bigr)\qquad(0\leq t\leq1)$$
with derivative $\phi'(t)=d{\bf f}\bigl((1-t){\bf x}+t{\bf y}\bigr).({\bf y}-{\bf x})$. One then has
$${\bf f}({\bf y})-{\bf f}({\bf x})=\phi(1)-\phi(0)=\int_0^1\phi'(t)\&gt;dt =\int_0^1 d{\bf f}\bigl((1-t){\bf x}+t{\bf y}\bigr).({\bf y}-{\bf x})\&gt;dt=\left(\int_0^1 d{\bf f}\bigl((1-t){\bf x}+t{\bf y}\bigr)\&gt;dt\right).({\bf y}-{\bf x})\ .$$
It follows that
$$L({\bf x},{\bf y}):=\int_0^1 d{\bf f}\bigl((1-t){\bf x}+t{\bf y}\bigr)\&gt;dt$$
has the stated properties.