Does $g: \mathbb{R^3} \rightarrow \mathbb{R}: (x,y,z) \rightarrow \frac{f}{1+x^2+y^2+z^2}$ reaches its max/min value

Question

Does $g: \mathbb{R^3} \rightarrow \mathbb{R}: (x,y,z) \rightarrow \frac{f}{1+x^2+y^2+z^2}$ reaches her max/min value when we know that the function $f:\mathbb{R^3}\rightarrow \mathbb{R}$ a bounded function that is also continuous.

I wanted to aproach this as the product of 2 functions $g=fh$ with $h=\frac{1}{1+x^2+y^2+z^2}$. Because $h$ is continuous and bounded by $0<h\leq 1$: both functions will reach their minimum and maximum value. (According to the supremum theorem)

Is this a correct method?

Answer 1

The method is not correct.

Consider the following one-dimensional counterexample (which can be extended to functions $\mathbb{R}^3\to \mathbb{R}$ easily). Let $f\colon \mathbb{R}\to\mathbb{R}$ be the constant function $f(x)=1$ and let $g\colon \mathbb{R}\to\mathbb{R}:x\mapsto \frac{f(x)}{1+x^2}=\frac{1}{1+x^2}$. Then $g$ does not have a minimum.

Answer 2

For some $f$ the maximum is attained, for others is is not attained. $f \equiv 0$ is an obvious example where the maximum is attained. Now let $f(x)=\frac {x^{2}} {1+x^{2}}$. Then the supremum is $1$ but it is never attained.

[ $\frac {x^{2}} {1+x^{2}} \frac 1{1+x^{2}+y^{2}}=1$ implies that each of the factors equals $1$ but $\frac {x^{2}} {1+x^{2}}$ cannot be equal to $1$].