I've been reading Wald's book on General Relativity and in chapter 3 he introduces and uses the so-called Hadamard's Lemma:(adsbygoogle = window.adsbygoogle || []).push({});

For any smooth (i.e. [itex]C^{\infty}[/itex]) function [itex]F: \mathbb{R}^{n}\rightarrow\mathbb{R}[/itex] and any [itex]a=(a^{1},\ldots,a^{n})\in\mathbb{R}^{n}[/itex] there exist [itex]C^{\infty}[/itex] functions [itex]H_{\mu}[/itex] such that [itex]\forall \; x=(x^{1},\ldots,x^{n})\in\mathbb{R}^{n}[/itex] we have [tex]F(x)=F(a)+\sum_{\mu =1}^{n}\left(x^{\mu}-a^{\mu}\right)H_{\mu}(x)[/tex]

Furthermore, we have that [tex]H_{\mu}(a)=\frac{\partial F}{\partial x^{\mu}}\Bigg\vert_{x=a}[/tex]

Now, I see how this can work for [itex]n=1[/itex] as it follows directly from the fundamental theorem of calculus that for [itex]F:\mathbb{R}\rightarrow\mathbb{R}[/itex] we have [tex]F(x)-F(a)=\int_{a}^{x}\frac{dF(s)}{ds}ds[/tex]

and so, upon making the substitution [itex]s=a+t(x-a)[/itex], it follows that [tex]F(x)-F(a)=(x-a)\int_{0}^{1}\frac{dF(a+t(x-a))}{dt}dt[/tex] and so we can choose [tex]H_{1}(x)=\int_{0}^{1}\frac{dF(a+t(x-a))}{dt}dt[/tex] such that [tex]F(x)-F(a)=(x-a)H_{1}(x)[/tex]

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

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# Slight confusion in proof of Hadamard's Lemma

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