Showing a function in R2 is unbounded (no least upper bound)

[cantidosan]

Homework Statement

Show that this function has no absolute max by showing that it is unbounded

Homework Equations

f(x,y) = (x-1)^2 + (y+2)^2 -4

The Attempt at a Solution

my initial idea is to construct a sequence of points {(xk, yk)} so that the sequence {f(xk, yk)} becomes unbounded.

to show that : Let M=f(x,y)
∨M>0 ∃xk, yk s.t xk,yk∉ B(M,(1,-2)). This issue i have is determining an adequate sequence of values to use.

 

[pasmith]

Observe that f(x,y) \geq (x - 1)^2 -4. Is that bounded above?

 

[cantidosan]

Observe that f(x,y) \geq (x - 1)^2 -4. Is that bounded above?

I'm assuming by f(x,y) you're referring to the original equation and no it isn't bounded above. I'm sorry, i have a feeling I'm supposed to make some intellectual leap with that example,but I am still somewhat lost.

 

[cantidosan]

Observe that f(x,y) \geq (x - 1)^2 -4. Is that bounded above?

Actually, i noticed that you reduced it to a single variable? To what end?

 

[pasmith]

If (x - 1)^2 - 4 is not bounded above, and f(x,y) \geq (x-1)^2 -4, can it be the case that f(x,y) is bounded above?

 

[PeroK]

I'm assuming by f(x,y) you're referring to the original equation and no it isn't bounded above. I'm sorry, i have a feeling I'm supposed to make some intellectual leap with that example,but I am still somewhat lost.

Perhaps this will help, but I'm not sure:

Can you think of any function (even of a single variable) that is not bounded above?

 

[cantidosan]

Perhaps this will help, but I'm not sure:

Can you think of any function (even of a single variable) that is not bounded above?

f(x,y) = X^2 for instance, For the sake of a concrete proof. Would it be sufficient to say that if f(x) is unbounded and f(x,y) >f(x) then f(x,y) is also unbounded. Seems incomplete, or have we just skimmed the surface in terms of reasoning?

 

[PeroK]

f(x,y) = X^2 for instance, For the sake of a concrete proof. Would it be sufficient to say that if f(x) is unbounded and f(x,y) >f(x) then f(x,y) is also unbounded. Seems incomplete, or have we just skimmed the surface in terms of reasoning?

Ok, so you know that $f(x) = x^2$ is not bounded above. What about $f(x) = (x-1)^2$? Not bounded above?

 

[HallsofIvy]

The sequence or points (1, 1), (2, 2), ..., (n, n) for any integer n leaps out at you.