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I was reading a solution when I came across this statement.

So $\frac{\partial f_i}{x_j}=\frac{\partial f_j}{x_i}.$ Then there exists a differentiable function $g$ on $\mathbb{R}^n$ such that $\frac{\partial g}{\partial x_i}=f_i$.

Why is this true?

1 Answers 1

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The following proofs assume 2 variables.

Proof of necessary condition:

If $(f_i, f_j)$ is the gradient of a function $F$, it means that:

$ \frac{\partial{F}}{\partial{x_i}} = f_i \\ \frac{\partial{F}}{\partial{x_j}} = f_j $

Now, if $F$ has continuous second partial derivatives, then according to Clairaut's theorem:

$ \frac{\partial^2{F}}{\partial{x_i}\partial{x_j}} = \frac{\partial^2{F}}{\partial{x_j}\partial{x_i}} $

Therefore:

$ \frac{\partial{f_i}}{\partial{x_j}} = \frac{\partial{f_j}}{\partial{x_i}} $

Proof of sufficient condition:

The function $F$, if it exists, has the property:

$ \frac{\partial{F}}{\partial{x_i}} = f_i $

By integrating with $x_j$ constant:

$ F = \int_{x_{i_0}}^{x_i} f_i \, dx_i + R(x_j) \tag{1} $

Now take partial derivatives of both sides with respect to $x_j$:

$ \frac{\partial{F}}{\partial{x_j}} = \frac{\partial}{\partial{x_j}}\int_{x_{i_0}}^{x_i} f_i \, dx_i + R'(x_j) = f_j $

Using differentiation under integral sign:

$ \frac{\partial{F}}{\partial{x_j}} = \int_{x_{i_0}}^{x_i} \frac{\partial{f_i}}{\partial{x_j}} \, dx_i + R'(x_j) = f_j $

Using the assumption that $\displaystyle \dfrac{\partial f_i}{\partial x_j} = \dfrac{\partial f_j}{\partial x_i}$:

$ \frac{\partial{F}}{\partial{x_j}} = \int_{x_{i_0}}^{x_i} \frac{\partial{f_j}}{\partial{x_i}} \, dx_i + R'(x_j) = f_j $

Which we can write as:

$ \left. f_j \right|_{x_{i_0}}^{x_i} + R'(x_j) = f_j $

Therefore:

$ R'(x_j) = f_j(x_{i_0}, x_j) $

And:

$ R(x_j) = \int_{x_{j_0}}^{x_j} f_j \, dx_j $

Plug in back into (1): $ F = \int_{x_{i_0}}^{x_i} f_i \, dx_i + \int_{x_{j_0}}^{x_j} f_j \, dx_j $

Therefore, we have shown that $F$ exists.

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    Nothing automatic. You have to type \tag{1}, \tag{2}, \tag{3} etc. all manually :/2012-06-04