"Finding a Non-Coercive Function f(x,y) on R2

[hsong9]

Homework Statement

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.

Homework Equations

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.

 

[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?

 

[hsong9]

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

 

[Dick]

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

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?

 

[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?

 

[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.

 

[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..

 

[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?