I am currently reading through several multivariable calculus books to understand the proofs better (most of which go back to introduce functions in $\mathbb{R}$ for which the results are already evident). Ultimately I stumbled across:
Mean Value Theorem: Let $f: \Omega \to \mathbb{R}$ be a differentiable function where $\Omega \subset \mathbb{R}^n$ is open.
Let $x,y \in \Omega$ such that $[x,y] \subset \Omega$ then there $\exists \xi \in [x,y]$ such that $$ f(y)-f(x)= \frac{\partial f}{\partial(y-x)}(\xi)$$
I understand the proof of this theorem (it's mainly a clever trick to reduce it to a problem in $\mathbb{R}$ and the chain rule for curves) but then the Author (Pöschel) gives 2 comments:
1) Similar as in $\mathbb{R}$ if all the partial derivatives are equal to $0$ then $f$ is constant.
Nothing to say here, this is evidently true by the formula above. What bothers me is the 2nd comment:
2) Consider the case $f: \mathbb{R}^2 \to \mathbb{R}$ differentiable with $\frac{\partial f}{\partial x}(x,y)=0, \forall (x,y) \in \mathbb{R}^2$ then $f$ does not depend on $x$
When I first read that comment it was 'intuitively clear' to me. I thought that I should try to prove it and this is where I fail.
My Approach: I want to show that $f(x,y)=h(y)$ where $h: \mathbb{R} \to \mathbb{R}$ and by obvious reasons I want to do so by applying the mean value theorem.
I often find it easier to just remind myself that $x,y \in \mathbb{R}^2$ rather than to write $x=(x_1,x_2), y=(y_1,y_2)$ but maybe in this situation it's necessarly.
I only know that $\partial_x f(x,y)=0, \ \forall(x,y) \in \mathbb{R}^2 $ the only way I see is to establish this situation in the theorem is by letting $y=0$ and considering $f(-x)$ rather than $f(x)$ then I'd have $\exists \xi \in [x,0]$ such that $$f(0)-f(-x)=\frac{\partial f}{\partial x}(\xi)=0 \implies f(0)=f(-x) $$
But the equation $f(0)=f(-x)$ doesn't help me at all to understand that comment. Is there something very obvious that I am missing out? Or do I mainly need to focus on $\partial_yf(x,y)$ ?
Let's suppose that there exist $x_0$, $x_1$, $y_0$ such that $$f(x_0,y_0)\neq f(x_1,y_0)$$
Define $h_{y_0}(x)=f(x,y_0)$. Our hypothesis is that $h_{y_0}'\equiv 0$. But MVT tells us that there exists $\xi\in(x_0,x_1)$ (this is an interval, not a point!) such that $$h_{y_0}'(\xi)=\frac{h_{y_0}(x_1)-h_{y_0}(x_0)}{x_1-x_0}$$ and since $h_{y_0}'(\xi)=0$ we have that $h_{y_0}(x_1)=h_{y_0}(x_0)$, a contradiction.
We conclude that $h_{y_0}$ is comstant for every $y_0\in \Bbb R$.