# Existence of a minima/maxima of function on a non-compact set

I have a set $$S=\{(x,y,z) \in \mathbb{R}^3 \mid g(x,y,z)=x^2+y^2-z^2 = 1\}$$ on which we have a function $$f:S \rightarrow R$$ defined by $$f(x,y,z)=x+y+z^2$$ I have to find all global minima and maxima. By Lagrange, I have found the global minima to be at $$(x=-\sqrt{\frac{1}{2}}, y=-\sqrt{\frac{1}{2}}, z=0)$$. If $$z \neq 0,$$ I get a complex solution, which is not what we are looking for.

Before solving I have to argue that such a minimum/maximum exist or does not exist. I know that a continious function on a compact set assume its minima and maxima, but $$S$$ is unbounded in our case and therefore not compact. Talking about maxima I have shown that a maximum cannot exist by contradiction. Assuming that there exists a maximum with some value, and then choosing another point based on this value, that satisfies $$g=1,$$ which turns out to have a bigger value according to $$f.$$

How do I argue that the minima does exist and is there an easier way to show that there are no global maximum? I guess I have to show that $$f$$ is closed and bounded from below, subject to the condition, because then a minima will exist, but I cant figure out how. I know from the Hessian that $$f$$ is convex while $$g$$ is not convex, can I use that in any way?

For existence of minima, you have to show that $$f(x,y,z)\to\infty$$ if $$\|(x,y,z)\|\to\infty$$. Then sets of the type $$\{(x,y,z):f(x,y,z)\le C\}$$ are bounded.
Let $$(x,y,z)$$ satisfy the constraint. Then $$z^2=x^2+y^2-1$$ and $$f(x,y,z)= x+y+z^2 = x+y+ \frac12 z^2 +\frac12 x^2 + \frac 12 y^2-\frac12 =\frac12(x+1)^2 + \frac12(y+1)^2 +\frac12 z^2 - \frac32.$$ The right hand-side tends to $$+\infty$$ for $$\|(x,y,z)\|\to\infty$$.
To see that this is enough, take the feasible point $$(1,0,0)$$. Any minimum has to have a smaller value of $$f$$ than $$1$$. So we can restrict our search to all feasible points with $$f\le 1$$. Due to the above inequality, such $$(x,y,z)$$ satisfy $$\frac12(x+1)^2 + \frac12(y+1)^2 +\frac12 z^2 - \frac32 \le 1$$ or $$(x+1)^2 + (y+1)^2 + z^2 \le 5,$$ that is, they are in a circle around $$(-1,-1,0)$$ with radius $$\sqrt 5$$. The set of such feasible points is compact, so we get existence of minimum.
• Convexity of $f$ is of no use as the feasible set is not convex. Also $\exp(x)$ has no minimum. and yes that property implies non-existence of maximum. (This could be achieved much easier using the feasible points $(1,n,n)$).