Let $f\colon\Bbb{R}^2\to \Bbb{R}$ such that $|f(x)-f(y)|\leq \Vert x-y\Vert^2.$ Prove that $f$ is a constant

Question

Edit: Several questions of this type have been asked here before but not on the same domain $\Bbb{R}^2.$ Please, how do I deal with a function of this type or could anyone show me a reference or a similar question with the same domain?

Let $f\colon\Bbb{R}^2\to \Bbb{R}$ such that $|f(x)-f(y)|\leq \Vert x-y\Vert^2.$ Prove that $f$ is a constant.

What if we assume that $f$ is differentiable. Is there another way of showing that $f$ is a constant?

Answer 1

Here is a couple of definitions of differentiability for functions in $\mathbb{R}^2$ which can easily be extended into $\mathbb{R}^n$.

Definition 1: Function $f:\mathbb{R}^2\rightarrow\mathbb{R}$ is differentiable at point $x$ if there are two constants $c_1$ and $c_2$ such that $$\lim_{h\to0}\frac{f(x+h)-f(x)-c_1h_1-c_2h_2}{|h|}=0$$

Definition 2: Function $f:\mathbb{R}^2\rightarrow\mathbb{R}$ is differentiable at point $x$ if there are two constants $c_1$ and $c_2$ and function $\phi:\mathbb{R}^2\rightarrow\mathbb{R}$ such that $$f(x+h)-f(x)=c_1h_1+c_2h_2+\phi(h)$$ and $$\lim_{h\to0}\frac{\phi(h)}{|h|}=0$$

For both of these definitions $c_1$ and $c_2$ are constants with respect to $h$. i.e. if change $x$ constants and $\phi$ are allowed to change.

So in order to establish differentiability we need to present or show existence of $c_1$, $c_2$, and functions $\phi$ that would satisfy required properties.

If we choose definition 1. We claim that $c_1=c_2=0$ will satisfy the conditions: $$\left|\lim_{h\to0}\frac{f(x+h)-f(x)-c_1h_1-c_2h_2}{|h|}\right|=\left|\lim_{h\to0}\frac{f(x+h)-f(x)}{|h|}\right|=$$ $$=\lim_{h\to0}\frac{|f(x+h)-f(x)|}{|h|}\leq\lim_{h\to0}\frac{|h|^2}{|h|}=0$$

Here we must mention that absolute value is a continuous function so if $\lim|F|$ exists so must $|\lim F|$ and they must be the same.

In order to verify definition 2 we must present both constants $c_1$ and $c_2$ and functions $\phi$ that satisfy the necessary condition. We claim that the constants can be chosen as $c_1=c_2=0$ and $$\phi(h)=f(x+h)-f(x).$$ The first identiy is trivial since $\phi$ was chosen this way. The limit identity follows from the condition provided by the theorem: $|f(y-x)|\leq|y-x|^2$, set $y=x+h$:

$$\left|\lim_{h\to0}\frac{\phi(h)}{|h|}\right|=\lim_{h\to0}\frac{|\phi(h)|}{|h|}=\lim_{h\to0}\frac{|f(x+h)-f(x)|}{|h|}\leq\lim_{h\to0}\frac{|(x+h)-x)|}{|h|}\leq\lim_{h\to0}\frac{|h|^2}{|h|}=0$$

Here we must perform the standard calculus dance with existing limits: since the last limit in the chain exists and equals $0$ and all qualities are non-negative, the first limit in the chain within the absolute value exists as well and it is $0$ ("Two policemen" or "Sandwich" theorem)

As a conclusion, to establish something in math, we must present certain objects, or prove their existence by other means. In this case, we presented $c_1$, $c_2$, and $\phi$ and verified that these objects satisfy all required conditions. Hence the functions is differentiable and $c_1=c_2=0$ are partial derivatives existing at any point $x$ and they are $0$: $$f'(x)=[0,0]\Longrightarrow\frac{\partial f(x,y)}{\partial x}=\frac{\partial f(x,y)}{\partial y}=0\Longrightarrow f(x,y)=\text{const}(x)=\text{const}(y)=\text{const}$$ Q.E.D.

Answer 2

HINT: $$ \lim_{h\to0}\frac{|f(x+h)-f(x)|}{\|h\|}\leq\lim_{h\to0}\|h\|=0 $$ It means that the derivative is equal to 0 for all $x\in\mathbb{R}^2$. But, as the set is $\mathbb{R}^2$, it means that $f$ is constant on every interval, connecting two points.

Answer 3

Lemma. Suppose $k>0$ and $|f(x)-f(y)|\leq k \|x-y\|^2$ for all $x,y\in \Bbb R^2.$ Then for all $x,y\in \Bbb R^2$ we have $|f(x)-f(y)|\leq \frac {k}{2}\|x-y\|^2.$

Proof: Let $z=\frac {x+y}{2}.$ We have $\|x-z\|=\|z-y\|=\frac {1}{2}\|x-y\|.$ So we have $$|f(x)-f(y)|\leq |f(x)-f(z)|+|f(z)-f(y)|\leq$$ $$\leq k\|x-z\|^2+k\|z-y\|^2=\frac {k}{2}\|x-y\|^2.$$

Corollary. By induction on $n\in Z^+,$ if $|f(x)-f(y)|\leq \|x-y\|^2$ for all $x,y\in \Bbb R^2$ then $|f(x)-f(y)|\leq 2^{1-n}\|x-y\|^2$ for all $n\in \Bbb N,$ for all $x,y\in \Bbb R^2.$

Clearly we cannot have $0<|f(x)-f(y)|\leq 2^{1-n}\|x-y\|^2$ for all $n\in \Bbb N.$

Answer 4

Divide the line segment from $x$ to $y$ into $n$ segments of length $\| x- y\|/n$, with endpoints

$$x = x_0 , x_1, \ldots, x_n = y$$

Apply the inequality to each pair $(x_i, x_{i+1})$ and use the triangle inequality to get $$|f(x) - f(y)| \leq n \cdot \frac{\|x-y\|^2}{n^2}$$

Now let $n \to \infty$ for fixed $x,y$.