# Show that $\omega = dF$ when $\omega = \sum_{i=1}^n f_i(x)dx_i$ and $F = \frac{1}{p+1} \sum_{i=1}^n x_if_i(x)$

Let $$f_i : \mathbb{R}^n \setminus \{0\} \to \mathbb{R}$$ be a homogeneous function of degree $$p \neq -1$$ for every $$1 \le i \le n$$. Namely, $$f_i(tx) = t^pf_i(x)$$ for every $$x \in\mathbb{R}^n \setminus \{0\}$$ and $$t >0$$.

We define $$\omega = \sum_{i=1}^n f_i(x)dx_i$$ and assume it is closed.

I want to prove that $$\omega = dF$$ when $$F = \frac{1}{p+1} \sum_{i=1}^n x_if_i(x)$$

Basically want we want to show is that $$\frac{\partial F}{\partial x_i} = f_i(x)$$.

By straightforward deriving I get: $$\frac{\partial F}{\partial x_i} = \frac{1}{p+1} (\sum_{j=1}^n x_j\frac{\partial f_j(x)}{\partial x_i} + f_i(x))$$

Here I got stuck. I don't know how to continue. I also don't see how the information that $$\omega$$ is closed (namely, $$\frac{\partial f_i(x)}{\partial x_j} = \frac{\partial f_j(x)}{\partial x_i}$$ for $$i \neq j$$) helps me here. I also don't know how to use the fact thar $$f_i$$ are homogeneous because it is not given that $$x_i > 0$$.

Help would be appreciated.

• If I understand correctly, you should sum over, say, $j$ in your final formula: for instance $\partial F/\partial x_i=(p+1)^{-1}\left(\sum_j x_j \partial f_j/\partial x_i+f_i\right)$ – Brightsun Jul 2 '19 at 11:50
• Thanks, I corrected that. – Gabi G Jul 2 '19 at 11:51

Hint: $${d\over{dt}}_{t=1}f_i(tx)={d\over{dt}}_{t=1}t^pf_i(x)=pf_i(x)$$.

You can also compute it as follows

$${d\over{dt}}_{t=1}f_i(tx)=(\sum_{j=1}^{j=n}x_j{{\partial f_i}\over{\partial x_j}})_{t=1}$$

$$=\sum_{j=1}^{j=n}x_j{{\partial f_i}\over{\partial x_j}}$$.

If you replace the second formula by the first in your computations, you obtain the result.

• Could you explain the first line in your answer? Why $t=1$ becomes $t=0$ and where does the $pf_i(x)$ come from? Moreover, I am not sure to which computations in my question you are referring, so if you can clear it up I would be glad. – Gabi G Jul 2 '19 at 13:17
• @Gabi G For any positive $t$, we have $f_i(tx)=t^pf_i(x)$. Now if you take the ordinary derivative w.r. to $t$ on both sides you get $d/dt\, f_i(tx)=p t^{p-1}f_i(x)$, which is what Tsemo wrote in the first line substituting $t=1$. Then, you can also calculate the derivative using the chain rule: $d/dt\, f_i(tx)=\sum_j \partial f_i(tx)/\partial x_j\, d(tx_j)/dt=\sum_j x_j \partial f_i/\partial x_j\,$; substituting again $t=1$ yields the last formula by Tsemo. – Brightsun Jul 2 '19 at 13:41
• @Tsemo Aristide, I edited out a $t$ from your next-to-last line, guessing it was a typo. – Brightsun Jul 2 '19 at 13:43
• @GabiG After you establish $p f_i = \sum_j x_j \partial f_i/\partial x_j$ you can use the closedness hypothesis to conclude that $\sum_j x_j \partial f_j/\partial x_i=p f_i$ which is precisely what you need in order to conclude your argument. – Brightsun Jul 2 '19 at 13:45
• Thanks to both of you, I get it now – Gabi G Jul 2 '19 at 13:52

Assume $$x_1$$, $$x_2$$, $$\ldots$$, $$x_n$$ are all positive. Then, for fixed $$i$$, we can write $$f_i(x_1,\ldots, x_i, \ldots, x_n) = x_i^{p} f_i\left(\tfrac{x_1}{x_i},\ldots,\tfrac{x_{i-1}}{x_i}, 1,\tfrac{x_{i+1}}{x_i}, \ldots,\tfrac{x_n}{x_i}\right)\,.$$ Now, $$F(x)=\sum_{i=1}^n x_i f_i(x)=\sum_{i=1}^n x_i^{p+1}f_i\left(\tfrac{x_1}{x_i},\ldots,\tfrac{x_{i-1}}{x_i}, 1,\tfrac{x_{i+1}}{x_i}, \ldots,\tfrac{x_n}{x_i}\right)\,.$$ Consider, for fixed $$j$$, $$\frac{\partial}{\partial x_j}F= (p+1)x_j^{p+1}f_j\left(\tfrac{x_1}{x_j},\ldots,\tfrac{x_{j-1}}{x_j}, 1,\tfrac{x_{j+1}}{x_j}, \ldots,\tfrac{x_n}{x_j}\right)\\ +\sum_{i=1}^n x_i^{p+1}\frac{\partial}{\partial x_j}f_i\left(\tfrac{x_1}{x_i},\ldots,\tfrac{x_{i-1}}{x_i}, 1,\tfrac{x_{i+1}}{x_i}, \ldots,\tfrac{x_n}{x_i}\right)\,.$$ We recognize the first term on the right-hand side as $$(p+1)f_j(x)$$. The second term, appropriately separating the terms in the sum, gives $$\sum_{i\neq j} x_i^{p+1}\frac{\partial }{\partial x_j} f_i\left(\tfrac{x_1}{x_i},\ldots,\tfrac{x_{i-1}}{x_i}, 1,\tfrac{x_{i+1}}{x_i}, \ldots,\tfrac{x_n}{x_i}\right)\\ + x_j^{p+1} \frac{\partial }{\partial x_j}f_j\left(\tfrac{x_1}{x_j},\ldots,\tfrac{x_{j-1}}{x_j}, 1,\tfrac{x_{j+1}}{x_j}, \ldots,\tfrac{x_n}{x_j}\right)\\ = \sum_{i\neq j} \frac{x_i^{p+1}}{x_i}\frac{\partial f_i}{\partial x_j}\left(\tfrac{x_1}{x_i},\ldots,\tfrac{x_{i-1}}{x_i}, 1,\tfrac{x_{i+1}}{x_i}, \ldots,\tfrac{x_n}{x_i}\right)\\ +\sum_{k\neq j} \frac{-x_j^{p+1}x_k}{x_j^2} \frac{\partial f_j}{\partial x_k}\left(\tfrac{x_1}{x_j},\ldots,\tfrac{x_{j-1}}{x_j}, 1,\tfrac{x_{j+1}}{x_j}, \ldots,\tfrac{x_n}{x_j}\right)\\ =\sum_{k\neq j} x_k\left(\frac{\partial }{\partial x_j}f_k-\frac{\partial}{\partial x_k}f_j\right)=0\,.$$ In the last step we used the closedness property; note that, for $$k\neq j$$, $$\frac{\partial}{\partial x_k}f_j(x)=x_j^{p-1} \frac{\partial f_j}{\partial x_k}\left(\tfrac{x_1}{x_j},\ldots,\tfrac{x_{j-1}}{x_j}, 1,\tfrac{x_{j+1}}{x_j}, \ldots,\tfrac{x_n}{x_j}\right)\,.$$ When some $$x_i$$ is allowed to be negative, the argument can be repeated by replacing the $$1$$ with $$-1$$ appropriately.