If $\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y}$ then $f(x,y) = g(x + y)$

Question

This exercise is from my textbook:

Let $f : \mathbb{R}^2 \rightarrow \mathbb{R}$ such that $\frac{\partial f}{\partial x}(x,y) = \frac{\partial f}{\partial y}(x,y)$ for all $(x,y)\in\mathbb{R}^2$. Show that there exists $g : \mathbb{R} \rightarrow \mathbb{R}$ such that $g$ is differentiable and $f(x,y) = g(x + y)$ for all $(x,y)\in\mathbb{R}$.

I think that the key is to define $g(x) := f(x,0)$, but I'm not sure how to show that $f$ satisfies $g(x + y) = f(x + y, 0) = f(x,y)$, Can you give me some hint? I found the same question here but I'm not allowed to use the chain rule for the composition of two functions $f\circ g$ where $f:\mathbb{R}^n \rightarrow \mathbb{R}$ and $g : \mathbb{R}^m \rightarrow \mathbb{R}^n$ that is used in the answer.

Answer 1

This is pushing the boundaries of semantic trickery, here is a solution that is identical to the answer you posted that does not explicitly use the chain rule. This is usually solved using a change of variables; if we let:

$$ \begin{cases} u = x + y\\ v = x - y \end{cases} $$ we will get:

$$ \begin{cases} x = \frac{1}{2}(u + v)\\ y = \frac{1}{2}(u - v) \end{cases} $$

Therefore $$ f(x, y) = f\left(\frac{1}{2}(u + v), \frac{1}{2}(u - v)\right) $$

So if we let $\varphi:\mathbb{R}^2\to\mathbb{R^2}, (u, v)\mapsto \left(\frac{1}{2}(u + v), \frac{1}{2}(u - v)\right)$, we get $f(x, y) = (f\circ\varphi)(u, v)$.

Let us then define the auxiliary function $F = f\circ\varphi$, the jacobian $J$ of $F$ is the product of that of $f$ and that of $\varphi$. It is easy then to verify that this implies

$$ \begin{cases} \frac{\partial F}{\partial u} = \frac{1}{2}\left(\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\right)\\ \frac{\partial F}{\partial v} = \frac{1}{2}\left(\frac{\partial f}{\partial x} - \frac{\partial f}{\partial y}\right) \end{cases} $$

By using the PDE, we get $\frac{\partial F}{\partial v} = \frac{1}{2}\left(\frac{\partial f}{\partial x} - \frac{\partial f}{\partial y}\right) = 0$, and hence $f(x, y) = F(u, v) = g(u) = g(x+y)$ for some differentiable function g.