# A closed $1$-form on a convex open set is exact

Baby Rudin Exercise 10.24:

Let $\omega = \sum a_i(\mathbf x) \, dx_i$ be a $1$-form of class $\mathscr{C}''$ in a convex open set $E \subset \mathbb{R}^n$. Assume $d \omega = 0$ and show that $\omega$ is exact in $E$, by completing the following outline:

Fix $p \in E$, Define $$f(\mathbf{x}) = \int_{\bf [p,x]} \omega$$ and apply Stokes' theorem to affine-oriented 2-simplexes $\bf[p, x, y]$ in E. Deduce that

$$f(\mathbf y) - f(\mathbf x) = \sum_{i=1}^n (y_i - x_i) \int_0^1 a_i((1-t) \mathbf x + t \mathbf y)\,dt$$

for $\mathbf x \in E, \mathbf y \in E$. Hence $(D_i f)(\mathbf x) = a_i(\mathbf x)$.

The boundary of $\bf[p, x, y]$ is $\bf [x, y] - [p, y] + [p, x]$, from which

$$f(\mathbf y) - f(\mathbf x) = \int_{\bf [x,y]} \omega= \int_{\bf [x,y]} \sum_{i=1}^n a_i(\mathbf x) \, dx_i$$

follows. Next take $\gamma(t) = (1-t) \mathbf x + t \mathbf y$ for $0 \leq t \leq 1$. This is a $1$-surface. By differentiation of $1$-forms, we have

$$\int_\gamma \omega = \int_{\bf [x, y]} \omega = \int_0^1 \sum_{i=1}^n a_i((1-t) \mathbf x + t \mathbf y)(y_i - x_i) \, dt$$

which is the same as what Rudin gives. At this point I'm not sure what to do. I suppose I should partially differentiate both sides with respect to one of the basis vectors of $\mathbb{R}^n$, but I'm not sure how to go forth with that.

Edit. I have followed JohnMa's advice and here is the rest of the solution (for verification and archival purposes):

Fix any $\mathbf x$ and and take $\mathbf y = \mathbf x + v \mathbf e_k$. Then $$f(\mathbf x + v \mathbf e_k) - f(\mathbf x) = v \int_0^1 a_k ((1-t) \mathbf x + t (\mathbf x + v \mathbf e_k))\, dt = v \int_0^1 a_k(\mathbf x + tv \mathbf e_k) \, dt$$ Dividing by $v$ and taking $v \to 0$ gives $$(D_k f)(\mathbf x) = a_k (\mathbf x)$$ since $a_k$ is continuous. Since $\mathbf{x}$ was arbitrary, we have $D_k f = a_k$ for $1 \leq k \leq n$, hence $$\omega = \sum_{i=1}^n (D_i f)(\mathbf x) \, dx_i = d f$$

Since $f$ is a $0$-form.

• Just ilke what you said, try to put $y = x + te_i$, then make up the term $\frac{f(x + te_i) - f(x)}{t}$ and take limit $t\to 0$.
– user99914
Commented Nov 15, 2015 at 23:36
• @JohnMa Ah, I see. So we have a general form for any $\mathbf x, \mathbf y$, so use it with specific $\mathbf y$ now. Got it.
– MT_
Commented Nov 15, 2015 at 23:48