$f:\mathbb{R}\rightarrow\mathbb{R}$ continuous function with $x^2\leq f(x)$ $\forall x\in \mathbb{R}$. Show that f takes on its absolute minima. I am preparing for my exam and need help with the following task:
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function with the estimation $x^2\leq f(x)$ $\forall x\in \mathbb{R}$. Show that f takes on its absolute minima.
If a function is continuous, then $\lim\limits_{x \rightarrow a}{f(x)}$=$f(a)$.
Our function has an absolute minima in $x_0$ $\in \mathbb{R}$, if $f(x)\geq f(x_0)$ for $x\in \mathbb{R}$.
At first I thought the task is pretty easy. We learned how to prove that if $f:[a,b]\to\mathbb{R}$ 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 $x^2\leq f(x)$ $\forall x\in\mathbb{R}$. This gives us the information, that $f$ is bounded from below with $f(x)\geq 0$ $\forall x\in \mathbb{R}$. But how does this help me? And what else do we have?
Is there anyone who could give an advice? I would be very grateful.
 A: Let $a =f(0) \geq 0$. Then, if $|x|\geq \sqrt{a}$, we have :
$$f(x) \ge x^2 \geq a = f(0) \tag 1$$
As $[-\sqrt a, \sqrt a]$ is compact and $f$ is continuous on it, there is some $x_0 \in [-\sqrt a, \sqrt a]$, such that :
$$f(x_0) = \min_{[-\sqrt a, \sqrt a]} f$$
Clearly,  $f(x_0)\leq f(0)$. Therefore by $(1)$, we have :
$$f(x_0) = \min_{\mathbb R}f$$
A: Since $\lim_{x\to\infty}x^2=\infty$, there is some $M>0$ such that $x\geqslant M\implies x^2>f(0)$, and therefore $x\geqslant M\implies f(x)>f(0)$. And, for the same reason, there is some $N<0$ such that $x\leqslant N\implies f(x)>f(0)$. The restriction of $f$ to $[N,M]$ has a minimum, which is attained at some $x_0\in[N,M]$ and, since $0\in[N,M]$, $f(0)\geqslant f(x_0)$. So, $f$ attains its absolute minimum at $x_0$. In fact, if $x\in\Bbb R$, then either $x$ belongs to $[N,M]$ or it doesn't. If it doesnt, then $f(x)\geqslant f(0)\geqslant f(x_0)$; if it does, then $f(x)\geqslant f(x_0)$.
A: You can prove the claim by contradiction invoking the Bolzano-Weierstrass theorem.
Because of the condition $x^2\leq f(x)$ we know that it exists
$$m := \inf_{x\in\mathbb{R}}f(x) \geq 0$$
Now, assume that the infimum is not a minimum. Then, for each $n\in\mathbb{N}$ you find an $x_n\in\mathbb{R}$ such that
$$m < f(x_n) < m+\frac 1n$$
Because of Bolzano-Weierstrass, the sequence $(x_n)$ must be unbounded. Otherwise this sequence would have a convergent subsequence and the continuity of $f$ would enforce the infimum to be attained.
So, there is an $N \in\mathbb{N}$ such that
$$\color{blue}{m+2 <} x_N^2 \leq f(x_N) < m+ \frac 1N \leq \color{blue}{m + 1 } \color{blue}{\text{  Contradiction!}}$$
