Lipschitz constant, am I correct?

[Meital]

I am reading ODE ( Ordinary DE) notes, and there is a statement that says " A function that is not uniform, even a continuous function that is not uniform, cannot have a lipschitz constant. As an example is the function 1/x on the open interval (0,1).

 

I want to see if I understood this correctly.

 

I will assume that we can find a number k >= 0 such that

| f(x) - f(y) | =< K*|x - y|

we have x and y in (0,1)

so 0 < x < 1

0 < y < 1 .....(1)

Now,

| 1/x - 1/y | =< k*|x - y|

Find common denominator

 

| y - x|/|xy| =< k*|x-y|

| y - x | = |x - y| then we cancel the term from both sides of inequality, so we get

 

1/|xy| =< k ...($)

 

k >= 0, and from (1) 1/xy > 1 so is 1/|xy| > 1

 

but then the equation ( $ ) becomes

1 < 1/|xy| =< k , but k >= 0 so this is contradiction? I am not sure, can someone tell me why 1/x doesn't have a lipschitz constant? I mean since we are in (0,1) then there is no upper bound for 1/|xy| so we can't have an upper bound for it (lipschitz constant)

[Dave]

That's the idea. I think the argument I saw a long time ago said that basically no matter which k we pick, we can always find x or y big enough so that 1/|xy| > k, so there could never exist a k big enough.