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Slight confusion in proof of Hadamard's Lemma

  1. Feb 27, 2015 #1
    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. [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]?!
     
    Last edited: Feb 27, 2015
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  3. Feb 27, 2015 #2

    micromass

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    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.
     
  4. Feb 27, 2015 #3
    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 [itex]h: [0,1]\rightarrow\mathbb{R}[/itex] such that [tex]t\mapsto F(a+t(x-a))[/tex] where [itex]a\in\mathbb{R}^{n}[/itex] is fixed. It then states that clearly [tex]h'(t)=\sum_{i=1}^{n}\left(x^{i}-a^{i}\right)\frac{\partial F}{\partial x^{i}}[/tex]

    I mean, I see intuitively that [itex]h[/itex] maps to a function of [itex]x=(x^{1},\ldots,x^{n})[/itex] with a fixed value of [itex]t[/itex] (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 [itex]x':=g(t)=a+t(x-a)[/itex] such that [itex]x'^{i}=a^{i}+t(x^{i}-a^{i})[/itex]. Then, [tex]\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)[/tex]
     
    Last edited: Feb 27, 2015
  5. Feb 27, 2015 #4

    micromass

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    It should just be the multivariable chain rule.
     
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