Statement 1   Let $f(x)=x^2$ . Then $f$ satisfies the Lipschitz condition on $[a,b]\subset\R$ for finite real numbers $a<b$ .

Proof. We want to show that for some real constant $L$ , and for all $x,y\in[a,b]$ ,$$ \modulus{x^2-y^2}\leq L\modulus{x-y}.$$ Let $x,y\in[a,b]$ . Clearly if $x=y$ , the above inequality holds, so assume $x\neq y$ . Since $x$ and $y$ are interchangable in the above equation, it can be assumed without loss of generality that $x<y$ .

Since $f$ is differentiable on $(a,b)$ , by the mean-value theorem, there is a $z\in(x,y)$ such that$$ \frac{f(x)-f(y)}{x-y}=f'(z),$$ that is,$$ \frac{x^2 - y^2}{x-y} = 2z.$$ Taking the modulus of both sides gives$$ \frac{\modulus{x^2-y^2}}{\modulus{x-y}} = 2\modulus{z}.$$ Finally, to find $L$ it is necessary to consider all possible values of $z$ :

Thus, for all $x,y\in[a,b]$ ,$$ \modulus{f(x) - f(y)} \leq 2\max\{\modulus{a},\modulus{b}\}\modulus{x-y}$$ as required.

Statement 2   Additionally, $L=2\max\{\modulus{a},\modulus{b}\}$ is the Lipschitz constant of $f$ .

Proof. Assume $\modulus{b}\geq\modulus{a}$ , since if $\modulus{b}<\modulus{a}$ , it is possible to consider $-f$ instead of $f$ . This also implies that $b>0$ . Let $\ve>0$ be sufficiently small that $a<b-\ve$ and that higher powers of $\ve$ can be ignored. Now,
By the assumption above, $b=\max\{\modulus{a},\modulus{b}\}$ . Thus, since $b,b-\ve\in[a,b]$ and by the definition of the Lipschitz condition,$$ L\geq \frac{\modulus{f(b)-f(b-\ve)}}{\modulus{b-(b-\ve)}} = 2\max\{\modulus{a},\modulus{b}\}.$$ However, the result from the previous proof gives$$ \frac{\modulus{f(b) - f(b-\ve)}}{\modulus{b-(b-\epsilon)}} \leq L \leq 2\max\{\modulus{a},\modulus{b}\}.$$ Combining these inequalities provides$$ 2\max\{\modulus{a},\modulus{b}\}\leq L\leq 2\max\{\modulus{a},\modulus{b}\},$$ and the result follows by trichotomy.