How to prove that a 2D function f(x,y) is always equal to a 1D function g(x) if f(x,y) always has zero gradient in the y direction

Question

Prove that if $f(x,y)$ is continuously differentiable in the whole plane, and if $f_y\equiv0$, then $f(x,y) = g(x)$, where g is a function of one variable. [Hint: let $g(x) = f(x,0)$ and apply the Mean Value Theorem to $f(x,y) - f(x,0)$]

Here's the statement of the Mean Value Theorem which I am required to work with:

Let $f(x,y)$ be continuously differentiable in the domian D, and let $(x_1, y_1)$ and $(x_2,y_2)$ be points of D such that the entire line segment L from $(x_1, y_1)$ to $(x_2, y_2)$ lies in D. Then there is a point $(x_0, y_0)$ on L such that:

$f(x_2,y_2) - f(x_1, y_1) = (x_2 - x_1) f_x(x_0, y_0) + (y_2 - y_1) f_y(x_0, y_0)$

The result is intuitively obvious to me because the 2D function always has zero gradient in the y direction and so its value is always only determined by x and so it is always equal to the value of a function of only x, but I can't see how to prove this via the above hint.

So, if I just use the Mean Value Theorem and apply it to the function $f(x,y) - f(x,0)$ I get (remembering that $f_y \equiv 0$):

$f(x_2,y_2) - f(x_2, 0) - f(x_1, y_1) + f(x_1, 0) = (x_2 - x_1) f_x(x_0, y_0)$

but I am not sure how to proceed. Thanks for any help.

Answer 1

To look at the problem another way, you want to show that $f(x,y)$ is constant with respect to $y$. With this in mind, try selecting a $x_2$ that is related to $x_1$ to reduce the amount of variables.

Set $x_2 = x_1$. This effectively reduces it to a single variable problem, at which point you can use an idea reminiscent of single variable calculus.

To finish,

By MVT, $f(x,y) - f(x,0) = 0$ for any $y$ so $f(x,y) = f(x,0) = g(x)$ for any $y$