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In Robert Wald's book General Relativity a multivariable calculus theorem is cited on page 16, which states:

If $F:\mathbb{R}^n\mapsto \mathbb{R}$ is $C^{\infty}$ then for each $a=(a^1,...,a^n) \in \mathbb{R}^n$ there exist $C^{\infty}$ functions $H_{\mu}$ such that for all $x \in \mathbb{R}^n$ we have

$$F(x)=F(a)+\sum_{\mu=1}^n (x^{\mu}-a^{\mu})H_{\mu}(x)$$

(1) What is the name of this theorem?

(2) Where can I find a proof?

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Would Mathematics be a better home for this question? –  Qmechanic Apr 10 '14 at 5:44

4 Answers 4

up vote 7 down vote accepted

The result is sometimes called Flanders' lemma.

The remarkable point is that it does not need that $f$ is analytic, but just that it is $C^\infty$. So it does not relies upon the Taylor series as it could seem at first glance, since that series may not converge.

It works in any open star-shaped neighborhood of points in $\mathbb R^n$. A set $A\subset \mathbb R^n$ is said to be star-shaped with respect to $p \in A$ if, when $q\in A$ the segment joining $p$ and $q$ is completely included in $A$. For example a convex set is star-shaped with respect to each point belonging to it.

Theorem. Let $A\subset \mathbb R^n$ be open and starshaped with respect to $p\in A$. Consider a $C^\infty$ function $f: A \to \mathbb R$. Then there are $n$ functions $H_k=H_k(q)$ with $H_k \in C^\infty(A)$ such that:

$$f(q) = f(p) + \sum_{k=1}^n (q_k-p_k) H_k(q) \:,$$


$$H_k(p) = \left.\frac{\partial f}{\partial x_k}\right|_p\:.$$

PROOF. Keep $q\in A$ fixed and consider the smooth function $$[0,1]\ni t \mapsto g(t) := f(p+ t(q-p))\:.$$ Notice that $g(0)=f(p)$ and $g(1)=q$ so that we can write, in view of the second fundamental theorem of calculus: $$f(q)= f(p) + (f(q) - f(p)) = f(p) + \int_0^1 \frac{d}{dt} g(t) \:dt\:.$$ In other words: $$f(q)=f(p) + \int_0^1 \frac{d}{dt} f(p+ t(q-p)) \:dt\:.$$ We exploit the fact that $A$ is star-shaped when computing $f(p+ t(q-p))$ for $t\in [0,1]$, since $[0,1] \ni t \mapsto p+ t(q-p)$ is just the segment joining $p$ and $x$ and it must belong to the domain $A$ of $f$.

The derivative in the last integral can be computed as a derivative of a composite function, obtaining:

$$f(q) = f(p) + \int_0^1 \sum_{k=1}^n (q_k-p_k) \left.\frac{\partial f}{\partial x^k}\right|_{x=p+ t(q-p)} \:dt\:.$$ In other words: $$f(q) = f(p) + \sum_{k=1}^n (q_k-p_k) H_k(q) \:.$$ where: $$H_k(q):=\int_0^1 \left.\frac{\partial f}{\partial x^k}\right|_{x=p+ t(q-p)} \:dt\:.$$ Next observe that the $n$ functions $H_k= H_k(q)$ defined on $A$ are $C^\infty$ since, using standard theorems (Lebesgue's dominated convergence theorem and Lagrange's theorem), as the integrated function is jointly smooth in $(t,q)$ and the $t$ integration is performed over a compact set $[0,1]$, we can pass the symbol of $q$-derivatives of any type and order under the symbol of integration.

Finally, just by the definition of $H_k$ one immediately finds:

$$H_k(p) = \left.\frac{\partial f}{\partial x_k}\right|_p\:.$$


As a final remark I notice that the proof holds true also if $f \in C^k(A)$ with $1\leq k <+\infty$. In this case the functions $H_k$ turn out to be $C^{k-1}$, but verifying the remaining properties.

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It is the Taylor expansion around a point $a$ of a multivariable function where $H_{\mu}$ is similar to the Hessian matrix but for order 1. You can find it in Vector Calculus of Marsden, Tomba, or just google it. This representation is written with the Einstein summation

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The Hessian matrix involves second derivatives but immediately after this theorem the author writes "Furthermore, we have $H_{\mu}(a)=\frac{\partial{F}}{\partial x^{\mu}}|_{x=a}$" Also, the equation uses "=" rather than $\approx$, so I am not convinced that this is an infinite series expansion. –  user37222 Apr 10 '14 at 0:10
-1: you're looking for the Jacobian matrix, which is basically the differential; however, $H_\mu$ is not given by the differential - it just agrees with it at point $a$ –  Christoph Apr 10 '14 at 8:37

Let us star with the fundamental theorem of calculus for $n=1$:

$F(x) - F(a) = \int_{a}^{x} F'(s)ds$

Now use the substitution $s=t(x-a)+a$. This is because $s$ rescales the interval $[a,x]$ to $[0,1]$. Then $ds = dt(x-a)$, then we get:

$F(x) - F(a) = (x-a)\int_{0}^{1} F'(t(x-a)+a)dt$,

and this proofs $n =1$.

Now define a vector

$y^{\mu} = t(x^{\mu} -a^{\mu})$

Then by the chain rule:

$\frac{dF}{dt} = \sum_{\mu} \frac{\partial F}{\partial y^{\mu}} \frac{d y^{\mu}}{dt} = \sum_{\mu}F'(x^{\mu} -a^{\mu})$


$F(x) - F(a) = \int_{0}^{1}\frac{dF}{dt}dt = \int_{0}^{1} \sum_{\mu}F'(x^{\mu} -a^{\mu})dt = \sum_{\mu}(x^{\mu} -a^{\mu}) \int_{0}^{1}F'(t) = \sum_{\mu}(x^{\mu} -a^{\mu})H_{\mu}$

where $H_{\mu} = \int_{0}^{1}F'(t)$

which gives the desired result

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I'll share a one dimensional example and see if this helps to shed any light on the matter.

Suppose $f(x)$ is an analytic function (i.e. the taylor series of $f$ converges to $f$). Then we can write,

$$ f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2 + \cdots $$

$$ = f(a) + (x-a) \left[ f'(a) + f''(a)(x-a)/2 + \cdots \right] $$

$$ = f(a) + (x-a) H(x) $$

Where by inspection we see that,

$$ H(x) \equiv f'(a) + f''(a)(x-a)/2 + \cdots $$

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No, you are assuming that the function is analytic. Here just $C^\infty$ is enough. –  Valter Moretti Apr 10 '14 at 6:42

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