Consider the following condition ($\star$) for the real-valued continuous function $f$ on $\mathbb{R}^2$
($\star$) For any positive real number $R$, the following set is bounded.
$\{(x,y) \in \mathbb{R}^2 |\, |f(x,y)| \leq R \}$
Answer the following questions.
- Give an example of a continuous function $f$ that satisfies the condition ($\star$) and show that it satisfies ($\star$)
- When the continuous function $f$ satisfies the condition ($\star$) show that one of the following holds
- $f$ has a maximum value but no minimum value
- $f$ has a minimum value but no maximum value
Answer: My idea is the following but I don't feel that it is correct or enough.
The example of $f$
$f(x,y) = \frac{1}{1+x^2+y^2} \qquad \forall (x,y)\in \mathbb{R}^2$
$\bigg|\frac{1}{1+x^2+y^2}\bigg| \leq 1 \qquad\qquad\qquad\quad \text{since}\,\, \,\,x^2+y^2 \geq 0 \,\,\,\forall (x,y) \in \mathbb{R}^2$
So the example is satisfied the condition ($\star$) for all $R\geq 1$
We can see that if and only if $x=0$ and $y=0$, $f(x,y)$ attains its absolute maximum and for all other $(x,y) \in \mathbb{R}^2 \setminus (0,0)$ the value of the function is less than 1. The range of the function is $(0,1]$. Therefore there is no global minimum and the global maximum value is 1.
Any idea would be appreciated.