# Coercive function

1. Sep 8, 2009

### hsong9

1. The problem statement, all variables and given/known data

Find a function f(x,y) on R2 such that for each real number t, we have
lim x->∞ f(x,tx) = lim y->∞ f(ty,y) = ∞, but such that f(x,y) is not coercive.

2. Relevant equations

3. The attempt at a solution
I know that f(x,y) = x^2 -2xy + y^2 = (x-y)^2 is not coercive
but I am not sure this function can be used in above question.

2. Sep 8, 2009

### Dick

What makes you think there is something wrong with your example? The limits of f(x,tx) and f(ty,y) are infinity, aren't they?

3. Sep 8, 2009

### hsong9

I want to know f(x,y) = x^2 -2xy + y^2 can be the answer for my question.

4. Sep 8, 2009

### Dick

Why NOT?? If the limits of f(x,tx) and f(ty,y) are infinity then you have solved the problem. Can you show those limits are both infinity?

5. Sep 8, 2009

### hsong9

Thanks,
Actually, I know how both infinity are, but
My question is the function is "not coercive"
I don't know how it works lim f(x,y) != infinite. --> this means not coercive.. right?

6. Sep 8, 2009

### Dick

I think so, if I understand what you are saying. If x and y go to infinity, and x=y then the limit is zero, right? Not infinity. So not coercive.

7. Sep 8, 2009

### hsong9

hmm, yes you understand exactly.
it does not matter lim x->∞ f(x,tx) = lim y->∞ f(ty,y) = ∞ ?
If I consider about it, x - tx is not equal to zero..

8. Sep 9, 2009

### Dick

Wait a minute. I was wrong. You can't have shown f(x,tx) and f(ty,y)->infinity for all t. They don't. If t=1 then f(x,x)=f(y,y)=0. Your condition only says f(x,y) approaches infinity along straight lines approaching infinity. Your function IS zero along a line approaching infinity. You need to find a function that approaches infinity along lines approaching infinity but doesn't approach infinity along a curve that goes to infinity. Can you think of one?

Last edited: Sep 9, 2009