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