What is the easiest way to check whether the function is Globally Lipschitz continuous or Locally? Say,for example the function, $sin(x^2)$? Many thanks in advance.
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For differentiable functions like this, this is easy to find out:
A differentiable function $f$ is (locally) Lipschitz if and only if the derivative $f'$ is (locally) bounded.
To see this, first note that if $f$ is Lipschitz with constant $L$ on $(a,b)$, this implies
$$ |f'(x)| = |\lim_{y \to x} \frac{f(y)-f(x)}{y-x}| \leq L, $$
because each term in the limit has absolute value $\leq L$ (if $x \in (a,b)$, then $y \in (a,b)$ for $y$ close to $x$.)
For the converse, assume that $|f'(x)| \leq L$ for all $x \in (a,b)$. Let $x,y \in (a,b)$ and use the mean value theorem to get $\xi \in (a,b)$ with
$$ |f(x) -f(y)| = |f'(\xi)| \cdot |x-y| \leq L |x-y|. $$
PhoemueX
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