Name this Mulltivariable Calculus Theorem In Robert Wald's  book General Relativity  a multivariable calculus theorem is cited on page 16, which states:
If $F:\mathbb{R}^n\mapsto \mathbb{R}$ is $C^{\infty}$ then for each $a=(a^1,...,a^n) \in \mathbb{R}^n$ there exist $C^{\infty}$ functions $H_{\mu}$ such that for all $x \in \mathbb{R}^n$ we have
$$F(x)=F(a)+\sum_{\mu=1}^n (x^{\mu}-a^{\mu})H_{\mu}(x)$$
(1) What is the name of this theorem?
(2) Where can I find a proof?
 A: The result is sometimes called Flanders' lemma. 
The remarkable point is that it does not need that $f$ is analytic, but just that it is $C^\infty$. So it does not relies  upon the Taylor series as it could seem at first glance, since that series may not converge.
It works in any open star-shaped neighborhood of points in $\mathbb R^n$. A set $A\subset \mathbb R^n$ is said to be star-shaped with respect to $p \in A$ if, when $q\in A$ the segment joining $p$ and $q$ is completely included in $A$. For example a convex set is star-shaped with respect to each point belonging to it.  
Theorem. Let $A\subset \mathbb R^n$ be open and starshaped with respect to $p\in A$. Consider a $C^\infty$ function $f: A \to \mathbb R$. 
Then there are $n$ functions $H_k=H_k(q)$ with $H_k \in C^\infty(A)$ such that:
$$f(q) = f(p) + \sum_{k=1}^n (q_k-p_k) H_k(q) \:,$$
and 
$$H_k(p) = \left.\frac{\partial f}{\partial x_k}\right|_p\:.$$
PROOF.
Keep $q\in A$ fixed and consider the smooth function $$[0,1]\ni t \mapsto g(t) := f(p+ t(q-p))\:.$$ Notice that $g(0)=f(p)$ and $g(1)=q$ so that
 we can write, in view of the second fundamental theorem of calculus:
$$f(q)= f(p) + (f(q) - f(p)) = f(p) + \int_0^1 \frac{d}{dt} g(t)  \:dt\:.$$ 
In other words:
$$f(q)=f(p) + \int_0^1 \frac{d}{dt} f(p+ t(q-p))  \:dt\:.$$ 
We  exploit the fact that $A$ is star-shaped when computing $f(p+ t(q-p))$ for $t\in [0,1]$, since $[0,1] \ni t \mapsto p+ t(q-p)$ is just the segment joining $p$ and $x$ and it must belong to the domain $A$ of $f$.
The derivative in the last integral can be computed as a derivative of a composite function, obtaining:
$$f(q) = f(p) + \int_0^1 \sum_{k=1}^n (q_k-p_k) \left.\frac{\partial f}{\partial x^k}\right|_{x=p+ t(q-p)}  \:dt\:.$$
In other words:
$$f(q) = f(p) + \sum_{k=1}^n (q_k-p_k) H_k(q) \:.$$
where:
$$H_k(q):=\int_0^1 \left.\frac{\partial f}{\partial x^k}\right|_{x=p+ t(q-p)}  \:dt\:.$$
Next observe that the $n$ functions $H_k= H_k(q)$ defined on $A$ are $C^\infty$ since, using standard theorems (Lebesgue's dominated convergence theorem and Lagrange's theorem), as the integrated function is jointly smooth in $(t,q)$ and the $t$ integration is performed over a compact set $[0,1]$, we can pass the symbol of $q$-derivatives of any type and order under the symbol of integration. 
Finally, just by the definition of $H_k$ one immediately finds:
$$H_k(p) = \left.\frac{\partial f}{\partial x_k}\right|_p\:.$$
QED
As a final remark I notice that the proof holds true also if $f \in C^k(A)$ with $1\leq k <+\infty$. In this case the functions $H_k$ turn out to be $C^{k-1}$, but verifying the remaining properties.
A: It is the Taylor expansion around a point $a$ of a multivariable function where $H_{\mu}$ is similar to the Hessian matrix but for order 1. You can find it in Vector Calculus of Marsden, Tomba, or just google it. This representation is written with the Einstein summation
A: Let us star with the fundamental theorem of calculus for $n=1$:
$F(x) - F(a) = \int_{a}^{x} F'(s)ds$
Now use the substitution $s=t(x-a)+a$. This is because $s$ rescales the interval $[a,x]$ to $[0,1]$. Then $ds = dt(x-a)$, then we get:
$F(x) - F(a) = (x-a)\int_{0}^{1} F'(t(x-a)+a)dt$,
and this proofs $n =1$. 
Now define a vector
$y^{\mu} = t(x^{\mu} -a^{\mu})$
Then by the chain rule:
$\frac{dF}{dt} = \sum_{\mu} \frac{\partial F}{\partial y^{\mu}} \frac{d y^{\mu}}{dt} = \sum_{\mu}F'(x^{\mu} -a^{\mu})$
and
$F(x) - F(a) = \int_{0}^{1}\frac{dF}{dt}dt = \int_{0}^{1} \sum_{\mu}F'(x^{\mu} -a^{\mu})dt =
\sum_{\mu}(x^{\mu} -a^{\mu}) \int_{0}^{1}F'(t) = \sum_{\mu}(x^{\mu} -a^{\mu})H_{\mu}$
where $H_{\mu} = \int_{0}^{1}F'(t)$
which gives the desired result 
