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Assume that $M$ is a smooth oriented compact manifold with boundary and assume that $\omega$ is a Lipschitz $(n-1)$-form on $M$.

Question Is there a published simple proof of the Stokes theorem $$ \int_M d\omega=\int_{\partial M} \omega\ \ ? $$

This result is well known in geometric measure theory. However, geometric measure theory is not easy and I don't like using strong results for elementary claims. I am sure many people would benefit from such a reference.

Edit. Since integration by parts holds for Lipschitz functions, the standard proof (see [1,Theorem 16.11]) applies verbatim to the case of Lipschitz forms. When I asked the question I didn't realize that it was that simple.

[1] Lee, John M., Introduction to smooth manifolds, Graduate Texts in Mathematics. 218. New York, NY: Springer. xvii, 628 p. (2002). ZBL1030.53001.

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  • $\begingroup$ Are you specifically looking for a "direct", self-contained proof? I guess you can show this by following the proof of Stokes with a bit more attention, but in practice taking the smooth Stokes for granted and applying a quick mollifier-argument would probably be far more efficient. Sadly I have no good reference for either. $\endgroup$
    – mlk
    Commented Oct 11 at 7:33
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    $\begingroup$ Maybe Federer, geometric measure theory. $\endgroup$
    – coudy
    Commented Oct 11 at 8:14
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    $\begingroup$ @mlk Taking a mollification works, but in the Lipschitz case it is not easy at all and not written anywhere. $\endgroup$
    – Piotr Hajlasz
    Commented Oct 11 at 12:31
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    $\begingroup$ @PiotrHajlasz, if I understand correctly, the crucial point is that the product of two Lipschitz functions is Lipschitz. I prefer to work with Sobolev spaces, but Holder spaces do have this advantage. $\endgroup$
    – Deane Yang
    Commented Oct 14 at 14:59
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    $\begingroup$ @DeaneYang The proof is fairly easy. With a partition of unity argument it is reduced to the half-space. You do not need to multiply Lipschitz functions except for multiplication by smooth partition of unity. The proof extends also to the case of Sobolev forms in $W^{1,n}$, but then the restriction of $\omega$ to $\partial M$ is in the trace space $W^{1-\frac{1}{n},n}$. $\endgroup$
    – Piotr Hajlasz
    Commented Oct 14 at 16:08

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I am not aware of a specific result for the case of Lipschitz (i.e. $C^{0,1}$) forms $\omega$ and compact smooth manifolds $M$: nevertheless I am aware that a generalized Stokes formula for continuous (class-zero, $C^0$) forms on compact smooth manifolds was first proved by Paul Pierre Gillis in 1943 (see [1]) entirely by using classical methods. His result was proved again in considerably simplified way somewhat 14 years later by Beniamino Segre in [2] and this paper is 9 pages long so it could suit your needs. Even Segre uses classical methods, notably the theory of $C^1$ forms and approximation theory: (addendum) interestingly, he does not use a mollification procedure but instead he approximates the continuous coefficients of $\omega$ by rational functions constructed from Stieltjes polynomials and proves their uniform convergence on $M$.

References

Paul Gillis, Sur les formes différentielles et la formule de Stokes (in French), Académie royale de Belgique, Classe des Sciences, Mémoires Vol. XX, No. 3, pp. 95 (1943), MR0016175, Zbl 0061.20403.

Beniamino Segre, "Sul differenziale delle forme esterne di classe zero" (in Italian), Rendiconti del seminario matematico di Messina , Vol. I p. 1-9 (1955), MR0090081.

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    $\begingroup$ What does it mean that a smooth manifold is bounded? $\endgroup$
    – Ben McKay
    Commented Oct 12 at 20:31
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    $\begingroup$ @BenMcKay I should have written "compact" (and I'll correct the post). Segre uses the locution "$n$-field" (precisely $n$-campo) which is a old fashioned Italian word for a compact manifold: however the slip is mine as the first time when I looked at this papers I was working on compact manifolds embedded in $\Bbb R^n$, which can be thought as bounded sets. $\endgroup$
    – Daniele Tampieri
    Commented Oct 12 at 21:03
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    $\begingroup$ @BenMcKay note that Gillis uses a somewhat more modern terminology, calling $M$ a "variété fermé". $\endgroup$
    – Daniele Tampieri
    Commented Oct 12 at 21:10
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    $\begingroup$ Thanks. I wasn't thinking about how many languages and terminological traditions you are straddling. $\endgroup$
    – Ben McKay
    Commented Oct 13 at 12:24

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