Recall that a function f A rightarrow R is Lipschitz on A if

Recall that a function f: A rightarrow R is Lipschitz on A if there exists M > 0 such that |f(x) - f(y) lessthanequalto M|x - y| for all x, y A. Show that if f is differentiable on a dosed interval [a, 6] and if f\' is continuous on [a,b], then f is Lipschitz on [a, b]. Prove that f: [0,2phi] rightarrow R given by f(x) = sin(z) is Lipschitz 011 [0,2phi).

Solution

/ Recall Lipschitz condition: a function f satisfies Lipschitz if there is a real number N such that |f(x)f(y)|N|xy|

2/ First, I plan to show this :

(*) Knowing f(x) is continuous at any point xf(x) is bounded at some neighborhood about x

Using definition of continuity on f(x), I say that for any >0, there is >0 such that for any y, we have |xy|<=>|f(x)f(y)|<.

After some works, I realize that f(y) is in the neighborhood of (f(x),f(x)+).

So if I let my N=max{f(x)+,f(x)+}, I reach the conclusion that f(x) is bounded (at least below)

3/ Then, I plan to use (*) to say this :

If function f has a derivative f such that f is bounded by some number K f satisfies Lipschitz condition on any interval [a,b]

I plan to use the Mean Value Theorem, provided that by (*), there is a derivative f(x)<K where x is in between some x1 and x2 in [a,b]

Would someone please check if my ideas are correct?

I feel very shaky about part 2 of my work. If the derivative is bounded, then I think the proof will be way easier. But to conclude that continuous bounded, I\'m not sure if I can claim such thing .