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### GR 9367 Question 25

Posted: **Mon Apr 05, 2010 11:57 am**

by **mathQ**

Ques. Let F be real valued function with domain [0,1]. if there is some K > 0 such that

f(x) - f(y) <= K |x-y| for all x and y in [0,1], which of the following will be true ?

A) f is discontinuous at each point of (0,1)

B) f is not continuous on (0,1)but it is discontinuous at only countably many points of (0,1)

C) f is continuous on (0,1)but it is differentiable at only countably many points of (0,1)

D) f is continuous on (0,1)but may not be differentiable on (0,1)

E) f is differentiable on (0,1)

Please explain your answer. Thanks

### Re: GR 9367 Question 25

Posted: **Mon Apr 05, 2010 1:47 pm**

by **mrb**

Well, pick any point x. Now pick a sequence y_n approaching x. Then:

f(x) - f(y_n) <= K| x - y_n|

The right side obviously approaches 0, so the left side must too.

Actually it's slightly more complicated than that; you have to see that we must also have

f(y_n) - f(x) <= ...

as well, but you get the idea. This is pretty much exactly the definition of continuity.

On the other hand, there's just not any reason to expect the function to be differentiable. I'm not going to think of a counterexample right now, but you can see that the relation:

f(x) - f(y_n)

--------- <= K

| x - y_n |

doesn't prevent the left side from oscillating all over the place, as long as it stays less than K.

So it's D.

### Re: GR 9367 Question 25

Posted: **Tue Apr 06, 2010 11:43 pm**

by **enork**

by the way, this condition on F is called Lipschitz continuity if you're interested in finding out more.

### Re: GR 9367 Question 25

Posted: **Wed Apr 07, 2010 1:37 am**

by **EugeneKudashev**

let me correct you man, this is not exactly Lipschitz condition. Lipschitz continuity bounds the distance between f(x) and f(y), that is, |f(x)-f(y)|. in this very example, function is not bounded from below - only from above, thus, the equivalence fails.

### Re: GR 9367 Question 25

Posted: **Wed Apr 07, 2010 1:58 am**

by **enork**

I misread it before, but I think what's written is still equivalent to Lipschitz condition.

f(x) - f(y) <= K|x-y| and f(y) - f(x) <= K|y-x| implies that |f(x) - f(y)| <= K|x-y|.

### Re: GR 9367 Question 25

Posted: **Wed Apr 07, 2010 11:57 am**

by **obpizdon**

I guess the counterexample is |x-1/2|. It's continious, but not differentiable in 1/2