If $x,y\in\mathbb{R}$ and $|f(x)-f(y)|\leq (x-y)^2$. Then find value of $f(x)$ is , it is given that $f(1)=2$
Try: $$\lim_{h\rightarrow 0}|f(y+h)-f(y)|\leq \lim_{h\rightarrow 0}(y+h-y)^2$$
$$\lim_{h\rightarrow 0}\bigg|\frac{f(y+h)-f(y)}{h}\bigg|\leq \lim_{h\rightarrow 0}h =0$$
So $|f'(y)|\leq 0\Rightarrow |f'(y)|=0$
So $f(y)=\mathcal{C}$ with using condition we have $f(x)=2$
My Question is How can i solve without using First derivative.
Could some help me to explain it , thanks
Start with $|f(y+h)-f(y)| \le h^2$ (basically your fist line without the limit). Substituting $y+h$ for $y$ gives $|f(y+2h)-f(y+h)| \le h^2$. Via the triangle inequality this leads to $|f(y+2h)-f(y)| \le 2h^2$. Continue this via mathematical induction and you get $|f(y+nh)-f(y)| \le nh^2$.
If you set $y=1$ and $h=\frac{1}{n}$, you get $|f(2)-f(1)| \le n(\frac{1}{n})^2 =\frac{1}{n}$. Since you can make $n$ arbitrarily large, this proves $f(1)=f(2)=2$.