# Slight confusion in proof of Hadamard's Lemma

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1. Feb 27, 2015

### "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 $t$?!

Last edited: Feb 27, 2015
2. Feb 27, 2015

### 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.

3. Feb 27, 2015

### "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, $$\frac{d(F \circ g) (t)}{dt}=\frac{dF(x'(t))}{dt} \\ \qquad\qquad\quad=\sum_{i=1}^{n}\frac{\partial F}{\partial x'^{i}}\frac{dx'^{i}}{dt} =\sum_{i=1}^{n}\frac{\partial F}{\partial x'^{i}}\left(x^{i}-a^{i}\right)$$

Last edited: Feb 27, 2015
4. Feb 27, 2015

### micromass

It should just be the multivariable chain rule.