Suppose that $f(y)-f(x) \leq (y-x)^2 $ for all $x\in \mathbb{R}$ and $y\in \mathbb{R}$. Then how can i show $f$ is a constant function.
I encountered this problem reading Calculus book written by Michael Spivak. He said that this would imply that $|f(y)-f(x)| \leq (y-x)^2$ and I could derive this exchanging y and x. And his final hint was Divide the interval from $x$ to $y$ into $n$ equal pieces.
Let $a, b\in \Bbb R$. Pick $n\in \Bbb N$ and for $0\le i\le n$, let $x_i=a+i\frac{b-a}n$ (so $x_0=a$, $x_n=b$). Then $$\begin{align} f(b)-f(a)&=f(x_n)-f(x_{n-1})+f(x_{n-1})-f(x_{n-2})+\cdots +f(x_1)-f(x_0)\\ &\le (x_n-x_{n-1})^2+(x_{n-1}-x_{n-2})^2+\cdot +(x_1-x_0)^2\\ &=n\cdot\frac{(b-a)^2}{n^2}\\&=\frac {(b-a)^2}{n}.\end{align}$$ As $n$ was arbitrary, conclude that $f(b)\le f(a)$. By symmetry, also $f(a)\le f(b)$.
Let $x \neq y$ and w.l.o.g $x<y$ and $n \in \Bbb{N}$
Then take the partition $P$ of $[x,y]$ such that $P=\{x+\frac{k(y-x)}{n}: k=0,1,2,...,n\}$
Then $$|f(x)-f(y)| \leq \sum_{k=0}^{n-1} |f(x+\frac{(k+1)(y-x)}{n})-f(x+\frac{k(y-x)}{n})|$$ $$ \leq \frac{(y-x)^2}{n}$$
from hypothesis. Can you continue from here?
Instead of dealing with sums: this approach seems to me a bit more elegant. We claim: If $f(y)-f(x)\leq c\cdot(y-x)^2$ for a non-negative constant $c$, the already $f(y)-f(x)\leq \frac12c\cdot(y-x)^2$.
Prove similar as above: $$\begin{align} f(y)-f(x)&=f(y)-f\bigl(\frac12(x+y)\bigr)+f\bigl(\frac12(x+y)\bigr)-f(x)\\ &\leq c\cdot \left(y-\frac12(x+y)\right)^2+c\cdot \left(\frac12(x+y)-x\right)^2\\ &=c\cdot \left(\frac12(y-x)\right)^2+c\cdot \left(\frac12(x-y)\right)^2\\ &=\frac12 c\cdot(y-x)^2. \end{align} $$
From the given,
$$\lim_{y\to x}\left|\frac{f(y)-f(x)}{y-x}\right|\le \lim_{y\to x}|y-x|=0$$ and the derivative is everywhere zero.