Hayley Bernard

2022-07-18

I am preparing for my exam and need help with the following task:
Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous function with the estimation ${x}^{2}\le f\left(x\right)$$\mathrm{\forall }x\in \mathbb{R}$. Show that f takes on its absolute minima.
If a function is continuous, then $\underset{x\to a}{lim}f\left(x\right)$=$f\left(a\right)$.
Our function has an absolute minima in ${x}_{0}$$\in \mathbb{R}$, if $f\left(x\right)\ge f\left({x}_{0}\right)$ for $x\in \mathbb{R}$.
At first I thought the task is pretty easy. We learned how to prove that if $f:\left[a,b\right]\to \mathbb{R}$ is continuous, then f has an absolute maxima and an absolute Minima in [a,b]. The Problem here is, that the domain of our function here is unbounded. Thats why I don't have any idea what I could and should use for the proof. We should probably use the estimation ${x}^{2}\le f\left(x\right)$ $\mathrm{\forall }x\in \mathbb{R}$. This gives us the information, that f is bounded from below with $f\left(x\right)\ge 0$ $\mathrm{\forall }x\in \mathbb{R}$. But how does this help me? And what else do we have?

dasse9

Expert

Let $a=f\left(0\right)\ge 0$. Then, if $|x|\ge \sqrt{a}$, we have:
$\begin{array}{}\text{(1)}& f\left(x\right)\ge {x}^{2}\ge a=f\left(0\right)\end{array}$
As $\left[-\sqrt{a},\sqrt{a}\right]$ is compact and $f$ is continuous on it, there is some ${x}_{0}\in \left[-\sqrt{a},\sqrt{a}\right]$, such that:
$f\left({x}_{0}\right)=\underset{\left[-\sqrt{a},\sqrt{a}\right]}{min}f$
Clearly, $f\left({x}_{0}\right)\le f\left(0\right)$. Therefore by (1), we have:
$f\left({x}_{0}\right)=\underset{\mathbb{R}}{min}f$

jlo2ni5x

Expert