Slight confusion in proof of Hadamard's Lemma

["Don't panic!"]

I've been reading Wald's book on General Relativity and in chapter 3 he introduces and uses the so-called Hadamard's Lemma:
For any smooth (i.e. C^{\infty}) function F: \mathbb{R}^{n}\rightarrow\mathbb{R} and any a=(a^{1},\ldots,a^{n})\in\mathbb{R}^{n} there exist C^{\infty} functions H_{\mu} such that \forall \; x=(x^{1},\ldots,x^{n})\in\mathbb{R}^{n} we have F(x)=F(a)+\sum_{\mu =1}^{n}\left(x^{\mu}-a^{\mu}\right)H_{\mu}(x)
Furthermore, we have that H_{\mu}(a)=\frac{\partial F}{\partial x^{\mu}}\Bigg\vert_{x=a}

Now, I see how this can work for n=1 as it follows directly from the fundamental theorem of calculus that for F:\mathbb{R}\rightarrow\mathbb{R} we have F(x)-F(a)=\int_{a}^{x}\frac{dF(s)}{ds}ds
and so, upon making the substitution s=a+t(x-a), it follows that F(x)-F(a)=(x-a)\int_{0}^{1}\frac{dF(a+t(x-a))}{dt}dt and so we can choose H_{1}(x)=\int_{0}^{1}\frac{dF(a+t(x-a))}{dt}dt such that F(x)-F(a)=(x-a)H_{1}(x)
However, I'm unsure how to show this for general n?! I get that one could define a function h:[0,1]\rightarrow\mathbb{R} such that h(t)=F(a+t(x-a)), but I don't quite see how it follows that \frac{dh}{dt}=\sum_{i=1}^{n}(x^{i}-a^{i})\frac{\partial F}{\partial x^{i}} which is the form I've seen in some proofs. My confusion arises in how do you get \frac{\partial F}{\partial x^{i}}? Surely one would have to introduce a change of variables such that y=y(t)=a+t(x-a), and then \frac{dF(y)}{dt}= \sum_{i=1}^{n}\frac{dy^{i}}{dt}\frac{\partial F}{\partial y^{i}}=\sum_{i=1}^{n}(x^{i}-a^{i})\frac{\partial F}{\partial y^{i}} Or is it just that we consider F to define a one parameter family of functions (dependent on x), parametrised by t?!

 

[micromass]

Well, you could just apply the multivariable Taylor's theorem. But I'll give a hint for a direct proof:

1) Reduce to the case $F(0) = 0$.
2) Define for each $x$, the function $h_x(t) = F(tx)$. Then $F(x) = \int_0^1 h^\prime_x(t)dt$. I'll let you take it from here.

 

["Don't panic!"]

I'm not sure I quite understand the second step of the process that you've given? In a proof I've seen they define h: [0,1]\rightarrow\mathbb{R} such that t\mapsto F(a+t(x-a)) where a\in\mathbb{R}^{n} is fixed. It then states that clearly h'(t)=\sum_{i=1}^{n}\left(x^{i}-a^{i}\right)\frac{\partial F}{\partial x^{i}}

I mean, I see intuitively that h maps to a function of x=(x^{1},\ldots,x^{n}) with a fixed value of t (a kind of scaling factor?!), but I'm struggling to rationalise it in a mathematical sense in my head.
I may be being a little stupid, but I just don't see how the above relation follows (unless by the chain rule that I put in my first post)?! (Sorry, I seem to having a mental block over it).

Could one just define a function x':=g(t)=a+t(x-a) such that x'^{i}=a^{i}+t(x^{i}-a^{i}). Then, \frac{d(F \circ g) (t)}{dt}=\frac{dF(x'(t))}{dt} \\ \qquad\qquad\quad=\sum_{i=1}^{n}\frac{\partial F}{\partial x'^{i}}\frac{dx'^{i}}{dt} =\sum_{i=1}^{n}\frac{\partial F}{\partial x'^{i}}\left(x^{i}-a^{i}\right)

 

[micromass]

It should just be the multivariable chain rule.