TITLE: How to solve for Lipschitz constant?
QUESTION [1 upvotes]: Please, how do I show that the Lipschitz constant is 8 Given the function $f(x) =x^2$ on $[-1, 4]$. It was a given assignment which I have been trying to solve.  I just need a hint on how to solve it

REPLY [2 votes]: For $x,y\in [-1,4]$ we have $|x^2-y^2|= |x-y|\cdot |x+y|\leq |x-y|(|x|+|y|)\leq |x-y|(4+4)=8|x-y|$. This is the sharpest possible, because if $x=4$ and $y=4(1-2^{-n})$ then $|x^2-y^2|/|x-y|=8-4^{-n}.$ Another way to do this is to observe that $d(x^2)/dx=2x.$ So for $-1\leq x< y\leq 4$ we have $|x^2-y^2|\leq |x-y|\cdot \max \{|2z| :z\in [x,y]\}\leq 8|x-y|.$
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