Is there something wrong with this multivariable calculus proof?

Question

There are $2$ parts to the question:

1. Let $f:\mathbb{R} ^{2}\rightarrow \mathbb{R}$ be differentiable in $\mathbb{R} ^{2}$ such that $\nabla f\equiv 0$ for every point $\left( x,y\right) \in \mathbb{R} ^{2}$. proof that f is constant.

2. Let $f:\mathbb{R} ^{2}\rightarrow \mathbb{R}$ be a function that satisfies $\left| f\left( x_{1},y_{1}\right) -f\left( x_{2},y_{2}\right) \right| \leq \left( x_{1}-x_{2}\right) ^{2}+\left( y_{1}-y_{2}\right) ^{2}$ for every $\left( x_{1},y_{1}\right) ,\left( x_{1},y_{1}\right) \in \mathbb{R} ^{2}$. proof that $f$ is constant

Before I go on with the proofs, I would even argue that in the first question, the word "differentiable" is superfluous because if we are given that $\nabla f\equiv 0$, then it's clear that $f$ is differentiable because the partial derivatives are continuous (because they are $0$ everywhere, and what more continuous than the constant function $0$?), is this correct?

I came up with the following proofs:

1. Given $\nabla f\equiv 0$ this means that $f_{x}\equiv f_{y}\equiv 0$ , if we "freeze" $y$ to be $y_{0}$ and let $x$ vary then $f(x,y_{0})$ is constant as a result of lagrange's theorem because $f_{x}\equiv 0$, if we did something similar on $x$ we get that $f(x_{0},y)$ is constant for every $y$. thus for every $\left( x,y\right) \in \mathbb{R} ^{2}$ we get that $f\left( x,y\right) =f\left( 0,y\right) =f\left( 0,0\right) $ (since it is constant if we move parallel to the axis) meaning $f$ is constant.

2. We first prove this for the 1-dimensional case: meaning if f satisfies $\left| f\left( x\right) -f\left( y\right) \right| \leq \left| x-y\right| ^{2}$ for every $x$ and $y$ then (if $x \neq y$) $$\left| \dfrac{f\left( x\right) -f\left( y\right) }{x-y}\right| \leq \left| x-y\right| $$ Therefore $$\left| f^{'}\left( x\right) \right| = \lim _{y\rightarrow x}\left| \dfrac{f\left( x\right) -f\left( y\right) }{x-y}\right| \leq \lim _{y\rightarrow x}\left| x-y\right| = 0$$ meaning $f$ is differentiable and constant. back to the $2$ dimensional case: if we "freeze" $x_{1}$ to be $x_{2}$ to be $x_{0}$ then we get that $$\left| f\left( x_{0},y_{1}\right) -f\left( x_{0},y_{2}\right) \right| \leq \left( y_{1}-y_{2}\right) ^{2}$$ and from the lemma at the begging we proved we conclude that $f_{y}\equiv 0$ the same argument goes for $x$, and we get that $f_{x}\equiv f_{y}\equiv 0$; this means that (because the partial derivatives are continuous because they're $0$) $f$ is differentiable in $2$ variables and this coupled with the conclusion from $1$ (the first question) we get that $f$ is constant.

These are the proofs I had in mind, but I don't see what's wrong with them.

Can someone provide me with insight?

Answer 1

This looks like 1st semester exercises. I was an examiner for 1st semester courses and a main focus was rigor. I heard some people say that they corrected solutions like this:

If I have to think (about what you mean) while reading your solution, you will get no points.

Sure thats too harsh, but the students should definitely learn to write their ideas rigorously and precisely, especailly for 1st semesters. Sure in later semesters its fine to sometimes 'describe' a proof or idea instead of doing it in full detail.

So I will assume that rigor was the thing bothering your examinor and I can see why. A good basis is to always define everything you are talking about -- dont have variables flying around without stating from which set they are.

For instance you write

if we "freeze" $y$ to be $y_0$ and let $x$ vary then $f(x,y_0)$ is constant

or

... if we did something similar on $x$ ...

which is not rigorous. Its more a description than logical reasoning. Again -- this is fine in most cases but maybe the examiner was expecting full detail. Here is a proof of Exercise 1 with which I would have been fine with:

proof of 1: Lets consider the 1-Dimensional case first.

Lemma 1: Let $f:\mathbb R \to \mathbb R$ be differentiable with $f'$ being constant $0$. Then $f$ is constant.

proof of Lemma 1: Assume by contradiction that $f$ is not constant. Then there exist $a,b\in\mathbb R$ such that $f(a)\neq f(b)$. Assume wlog $a\lt b$. The restriction $f\vert_{[a,b]}$ is differentiable on the open interval $(a,b)$ since $f$ is differentiable. Now by the Lagranges Theorem (Mean Value Theorem) there exists some $c\in [a,b]$ such that $$ f'(c)=\frac{f(a)-f(b)}{a-b}. $$ But this is a contradiction since $f(a)-f(b)\neq 0$ so $f'(c)\neq 0$ but $f'$ was assumed to be constant $0$. This finishes the proof of the lemma.

proof of Exercise 1: Let $x_0\in\mathbb R$ and consider the map $f_{x_0}:\mathbb R \to \mathbb R,x\mapsto f(x_0,x)$. Then $f_{x_0}'$ is constant $0$ since $\nabla f=0$ (this could be written in greater detail. How exactly are $f_{x_0}'$ and $\nabla f$ related?). So by Lemma 1 we get that $f_{x_0}$ is constant. Now let $y_0\in\mathbb R$. Then analogously we get that the map $f^{y_0}:\mathbb R\to \mathbb R, x\mapsto (x,y_0)$ is constant. Hence we have $f(x,y)=f_x(y)=f_x(y_0)=f^{y_0}(x)=f^{y_0}(x_0)=f(x_0,y_0)$ for every $x,y\in\mathbb R^2$ thus proving that $f$ is constant.

This is exactly your idea only in greater detail. For part 2 you again write undefined things

if we "freeze" $x_1$ to be $x_2$ to be $x_0$ then we get that ...

which seems to be a central thing for your argmuent.