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

1. Nov 5, 2014

### cantidosan

1. The problem statement, all variables and given/known data
Show that this function has no absolute max by showing that it is unbounded

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

3. 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.

Last edited: Nov 5, 2014
2. Nov 5, 2014

### pasmith

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

3. Nov 5, 2014

### cantidosan

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 im still somewhat lost.

4. Nov 5, 2014

### cantidosan

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

5. Nov 5, 2014

### 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?

6. Nov 5, 2014

### PeroK

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?

7. Nov 5, 2014

### cantidosan

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?

8. Nov 5, 2014

### PeroK

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

9. Nov 6, 2014

### HallsofIvy

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