Vtyjkyuk

Edit: Several questions of this type have been asked here before but not on the same domain $BbbR^2.$ 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 $fcolonBbbR^2to BbbR$ such that $|f(x)-f(y)|leq Vert x-yVert^2.$ 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?

Here is a couple of definitions of differentiability for functions in $mathbbR^2$ which can easily be extended into $mathbbR^n$.

Definition 1: Function $f:mathbbR^2rightarrowmathbbR$ is differentiable at point $x$ if there are two constants $c_1$ and $c_2$ such that $$lim_hto0fracf(x+h)-f(x)-c_1h_1-c_2h_2=0$$

Definition 2: Function $f:mathbbR^2rightarrowmathbbR$ is differentiable at point $x$ if there are two constants $c_1$ and $c_2$ and function $phi:mathbbR^2rightarrowmathbbR$ such that $$f(x+h)-f(x)=c_1h_1+c_2h_2+phi(h)$$ and $$lim_hto0fracphi(h)=0$$

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:
$$left|lim_hto0fracf(x+h)-f(x)-c_1h_1-c_2h_2right|=left|lim_hto0fracf(x+h)-f(x)right|=$$ $$=lim_hto0fracleqlim_hto0frach=0$$

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 $y=x+h$:

$$left|lim_hto0fracphi(h)right|=lim_hto0frac=lim_hto0fracleqlim_hto0fracleqlim_hto0frach=0$$

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 $0$: $$f'(x)=[0,0]Longrightarrowfracpartial f(x,y)partial x=fracpartial f(x,y)partial y=0Longrightarrow f(x,y)=textconst(x)=textconst(y)=textconst$$ Q.E.D.

HINT:
$$
lim_hto0fracleqlim_hto0|h|=0
$$
It means that the derivative is equal to 0 for all $xinmathbbR^2$. But, as the set is $mathbbR^2$, it means that $f$ is constant on every interval, connecting two points.

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
$$|f(x) - f(y)| leq n cdot frac^2n^2$$

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

Lemma. Suppose $k>0$ and $|f(x)-f(y)|leq k |x-y|^2$ for all $x,yin Bbb R^2.$ Then for all $x,yin Bbb R^2$ we have $|f(x)-f(y)|leq frac k2|x-y|^2.$

Proof: Let $z=frac x+y2.$ We have $|x-z|=|z-y|=frac 12|x-y|.$ So we have $$|f(x)-f(y)|leq |f(x)-f(z)|+|f(z)-f(y)|leq$$ $$leq k|x-z|^2+k|z-y|^2=frac k2|x-y|^2.$$

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

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