Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
  2. jcsd
  3. Feb 27, 2015 #2
    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
    It should just be the multivariable chain rule.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook