Ques. Let F be real valued function with domain [0,1]. if there is some K > 0 such that
f(x)  f(y) <= K xy 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
GR 9367 Question 25
Re: GR 9367 Question 25
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.
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
by the way, this condition on F is called Lipschitz continuity if you're interested in finding out more.

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Re: GR 9367 Question 25
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
I misread it before, but I think what's written is still equivalent to Lipschitz condition.
f(x)  f(y) <= Kxy and f(y)  f(x) <= Kyx implies that f(x)  f(y) <= Kxy.
f(x)  f(y) <= Kxy and f(y)  f(x) <= Kyx implies that f(x)  f(y) <= Kxy.
Re: GR 9367 Question 25
I guess the counterexample is x1/2. It's continious, but not differentiable in 1/2