$f:D \rightarrow R, D = (0,1), f(x) = x^2 $
Let $ \epsilon > 0 $ choose $\delta = \frac{\epsilon}{2} $ Then $\forall x,y \in (0,1)$ with $|x-y|< \delta$ we have: $$ |x^2 - y^2|< \epsilon $$ $$ |x^2-y^2| = |x-y||x+y| < 2|x-y| < 2\delta = \epsilon $$ after that I tried to prove if this funtion lipschitz
if the function is lipschitz there exists $L \geq 0 $ such that $$ |x^2−y^2| \leq L |x−y| \text{ . } \forall x,y \in [0,1] $$ $$ |x-y||x+y| \leq L |x-y| $$ $$ |x+y| \leq L $$ $$ |x| + |y| \leq L $$ $$ 2 \leq L $$
Is that prove right?
The ideas are correct, but you are not expressing yourself clearly. For instance, you write “we have: $|x^2-y^2|<\varepsilon$”. Well, you only have that after having proved it (which you did).
And you proof of the fact that $f$ is Lipschitz consists in several lines of mathematical expressions without test or any other connection between them. You can do it simply as follows: if $x,y\in(0,1)$, then$$|x^2-y^2|=|x+y||x-y|\leqslant\bigl(|x|+|y|\bigr)|x-y|\leqslant2|x-y|.$$