Poincaré lemma for star shaped domain I would like to know if someone can help me out to prove the Poincaré lemma for a star shaped domain without using the Stokes theorem.
 A: Because the lost link of the accepted answer, I will post a proof here.
Lemma: Let be $E\subset{\Bbb R}^n$ star-shaped respect to $0$ and ${\bf F}:E\longrightarrow{\Bbb R}^n$ a $C^1$ vector field. We define:
$$
h:{\Bbb R}^n\longrightarrow{\Bbb R}^n,\qquad  h(x) = tx,\qquad t\in{\Bbb R};
$$
$$
g:[0,1]\longrightarrow{\Bbb R}^n,\qquad g(t) = t{\bf F}(tx),\qquad x\in{\Bbb R}^n;
$$
$$
\phi:E\longrightarrow{\Bbb R},\qquad \phi(x) = {\bf F}(tx)\cdot x = \sum_{i=1}^n F_i(tx) x_i,\qquad t\in{\Bbb R}.
$$
Then, we have the following equalities ($D =$ differential, superscript $t =$ transpose):
$$Dh(x)  = tI;$$
$$g'(t)  = {\bf F}(tx) + t D{\bf F}(tx)x;$$
$$D\phi(x) = t x^t D{\bf F}(tx) + {\bf F}(tx)^t.$$
Proof:
The first two are obvious and $D\phi(x)$ is a row vector with components:
$$
\partial_j \phi(x) = \sum_{i=1}^n\partial_j(F_i\circ h)(x) + F_j(tx) = $$
$$ \sum_{i=1}^n x_i\sum_{k=1}^n(\partial_k F_i)(tx)\partial_j h(x)+ F_j(tx) = $$
$$\sum_{i=1}^n x_i(\partial_j F_i)(tx) t+ F_j(tx),$$
so
$$D\phi(x) = (\partial_1\phi(x),\cdots,\partial_n\phi(x)) =$$
$$
t(x_1,\cdots,x_n)
\begin{pmatrix}
\partial_1 F_1(tx)&\dots&\partial_n F_1(tx)\\\
\vdots&\ddots&\vdots\\\
\partial_1 F_n(tx)&\dots&\partial_n F_n(tx)
\end{pmatrix}
+ (F_1(tx),\cdots,F_n(tx))
= t x^t D{\bf F}(tx) + {\bf F}(tx)^t.\Box
$$
Theorem (Poincaré Lemma): Let be $E\subset{\Bbb R}^n$ star-shaped and ${\bf F}:E\longrightarrow{\Bbb R}^n$ a $C^1$ vector field s.t. for $i,j = 1,\cdots,n: \partial_i F_j = \partial_j F_i$. Then, exists a scalar field (potential of ${\bf F}$) $f:E\longrightarrow{\Bbb R}$ with ${\bf F} = \nabla f$.
Proof: WLOG, we can suppose $E$ star-shaped respect to $0$. Let be
$$f(x)=\int_0^1 {\bf F}(tx)\cdot x\,dt.$$
By derivation under the integral sign an the previous lemma,
$$
D f(x) = \int_0^1 D\phi(x)\,dt =
\int_0^1(t x^t D{\bf F}(tx) + {\bf F}(tx)^t)\,dt.
$$
Gradient is simply the transpose of differential and by hypothesis $D{\bf F}(tx)$ is symmetric:
$$
\nabla f(x) = D f(x)^t = \int_0^1(t D{\bf F}(tx)^t x + {\bf F}(tx))\,dt = \int_0^1(t D{\bf F}(tx) x + {\bf F}(tx))\,dt.$$
By the previous lemma again:
$$
\nabla f(x) = \int_0^1 g'(t)\, dt =
g(1) - g(0) = 1{\bf F}(1x) - 0{\bf F}(0x) = {\bf F}(x).\Box
$$
A: B. Dacorogna has a proof that at least does not invoke the Stokes theorem explicitly.
