A function $f: \mathbb{R} \rightarrow \mathbb{R}$ is defined to be Lipschitz if there is a constant $K>0$ such that for all $a,b\in \mathbb{R}$
$$
|f(a) – f(b)| \le K|a – b|
$$
$(a)$ Use the mean value theorem to show that the function $f(x) = 2\sin(x)$ is Lipschitz.$(b)$ Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ is Lipschitz. Prove that $f$ is continuous. (Hints: Use squeeze law and the epsilon-delta definition of limit.
This is how I did part (a):
$2 \sin(x)$ is continuous on $\mathbb{R}$, in particular on $[b,a]$ and differentiable on $(b,a)$.
By MVT, there exists $c \in (b,a)$ such that
$$f'(c) = \frac{f(a) – f(b)}{a – b}$$
But $f'(x) = 2\cos(x)$,
therefore $f'(c) = 2\cos(c)$,
therefore
$$
2\cos(c) = \frac{f(a) – f(b)}{a – b} \\
|2\cos(c)| = \left|\frac{f(a) – f(b)}{a – b}\right|
$$
But $|2\cos(c)| \le 2$,
therefore $\left|\frac{f(a) – f(b)}{a – b}\right| \le 2$
i.e. $|f(a) – f(b)| \le 2|a – b|$ and $f(x)$ is Lipschitz.
Can someone please verify for me that this solution is correct and if possible, give me some pointers on how to do part $(b)$? Thanks for any help!
Best Answer
For every $x_0,h\in\mathbb{R}$, $|f(x_0+h)-f(x_0)|\leq K|x_0+h-x_0|=K|h|\rightarrow 0$ as $h\rightarrow 0$