$\frac{\partial f_i}{x_j}=\frac{\partial f_j}{x_i}\implies(f_1,\ldots,f_n)$ is a gradient

Solution 1:

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.