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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.

Analysis_Mark
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  • A continuous coercive function always have a minimum https://math.stackexchange.com/q/2240269/399263 – zwim Mar 15 '22 at 20:40

3 Answers3

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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 : $$f(x_0) = \min_{\mathbb R}f$$

SolubleFish
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  • I don't understand why there must be a $x_0$ $\in[-\sqrt{a},\sqrt{a}]$ where f reaches it's minimum on [-a,a]. First..why (-a)? $f(x)\geq 0$. If we have a positive a, - a doesnt make sense...or am I wrong.

    Second, we could have a function where $f(x_0)$>a $\forall x_0 \in (-\sqrt{a},\sqrt{a})$ or am I wrong. I mean, only because $x^2$<a, it doesn't make f(x) necessiraly smaller then a.

     – Analysis_Mark Mar 15 '22 at 20:57
  • There was a typo (the $[-a,a]$ should have been $[-\sqrt a, \sqrt a]$). I corrected this, and edited my answer to, hopefully, make it clearer. – SolubleFish Mar 15 '22 at 21:03
  • Thanks but I still don't understand why clearly $f(x_0)$ $\leq f(0)$. I mean if $|x|<\sqrt{a}$ we have: $x^2<a=f(0)$. But since $x^2$ $\leq$ $f(x)$, we still could have $x^2<f(0)\leq f(x_0)$. I am probably missing something though... – Analysis_Mark Mar 15 '22 at 21:26
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    Because $0 \in [-\sqrt a,\sqrt a]$ and $f$ reaches a minimum at $x_0$ – SolubleFish Mar 16 '22 at 08:17
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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 $f(x)\geqslant f(x_0)$.

José Carlos Santos
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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

$$\color{blue}{m+2 <} x_N^2 \leq f(x_N) < m+ \frac 1N \leq \color{blue}{m + 1 } \color{blue}{\text{ Contradiction!}}$$

trancelocation
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  • Thanks for the answer....but where does m+2 come from? – Analysis_Mark Mar 15 '22 at 19:42
  • You can choose any number greater than $m+1$ to produce the contradiction. The point here is, that $f(x_n) < m+\frac 1n \leq m+1$ per construction of the sequence $(x_n)$ holds, while the sequence $(x_n)$ must be unbounded. – trancelocation Mar 15 '22 at 19:45
  • So we don't need m+2 for the contradiction. I mean we have m<$f(x_n)$<m+$\frac{1}{n}$$\leq m+1$. So $f(x_n)$ is upper and lower bounded...and therefore bounded. – Analysis_Mark Mar 15 '22 at 19:59
  • You can argue like that but you still need to lead this to a contradiction. For example by mentioning that $(x_n)$ is unbounded and because of the condition $x^2\leq f(x)$ the function $f$ cannot be bounded on an unbounded set. – trancelocation Mar 15 '22 at 20:18