Prove that $f$ has a minimum

Question

Let $f$ be a positive and continuous function in $[0,\infty)$, such that $\lim\limits_{x\to \infty} f(x)=2$.

Prove that if $f(0)<2$, $f$ has a minimum in $[0,\infty)$.

I am stuck in the final step, in my opinion.

From the limit's definition, there is $M \in \Bbb R$ such that for every $x>M$, $|f(x)-2| < \varepsilon$

In the closed interval, $[0,M]$, the function has a minimum, according to Weierstrass theorem.

I have to prove that it is the global minimum. It won't be a global minimum if there is some $x_0>M$ which is a minimum. I think that's where I have to reach a contradiction which I can't formalize.

Thanks for you help.

score 1 · Answer 1 · answered Feb 12 '16 at 22:20

1

HINT: You need to choose $\epsilon$ small enough to guarantee that the minimum on $[0,M]$ is in fact the global minimum. I.e., you need to choose $\epsilon$ small enough so that you know $f(x)\le 2-\epsilon$ for some $x\in [0,M]$.

answered Feb 12 '16 at 22:20

Ted Shifrin

125,228

score 1 · Answer 2 · answered Feb 12 '16 at 22:20

Assume $f(0) = 2-2\epsilon$ for some positive $\epsilon$, then there is an $M>0$ such that

$$x\geq M \Rightarrow |x-2|\leq \epsilon$$

According to Weierstrass, there is a minimum in $[0,M]$. This is also the global minimum, since

$$f([M,\infty))\subseteq [2-\epsilon,2+\epsilon],\ 2-2\epsilon\in f([0,M]).$$

Balloon · Accepted Answer · 2016-02-12T22:28:56.197

1

Note that in your formula, for $\varepsilon=2-f(0)>0,$ it exists $M\in\mathbb{R}$ such that $$\forall x\in[0,+\infty),x\geq M\implies |f(x)-2|\leq 2-f(0),$$ which implies $f(0)-2\leq f(x)-2\leq2-f(0)$ and so $f(0)\leq f(x).$ Now note $m$ the minimum of $f$ on $[0,M]$ which exists according to Weierstrass theorem, and as $m\leq f(0)\leq f(x)$ for $x\geq M,$ then $m$ is your global minimum.

edited Feb 12 '16 at 22:28

answered Feb 12 '16 at 22:20

Balloon

8,624

Thank you for your help! I understand it now. Gotta pay attention to the fact the you can set an epsilon according to $f(0)$. – Alan Feb 12 '16 at 22:52
You're welcome ! – Balloon Feb 12 '16 at 22:59
Your logic implies that the minimum of $f$ must be smaller or equal than $1$, which is not the case – Gregor de Cillia Feb 12 '16 at 23:28
@GregordeCillia : I don't think so, $m$ can be $2-\frac{1}{2}$ for example. – Balloon Feb 13 '16 at 11:34

score 1 · Answer 4 · answered Feb 12 '16 at 22:23

1

hint

Let $t\in ]f(0),2[$. Since $\lim_{x\to \infty }f(x)=2$, there is $M$ s.t. $f(x)>t$ for all $x>M$. The rest follow...

answered Feb 12 '16 at 22:23

user301068

778

score 1 · Answer 5 · answered Feb 12 '16 at 22:26

Hint

Since $\lim\limits_{x\to \infty} f(x)=2$ there exists $M>0$ such that

$$x>M \implies |f(x)-2|<\min\{\frac{2-f(0)}{2},1\}(>0).$$ Now, on $[0,M]$ the function $f$ has a global minimum $x_0.$ This is the global minimum, because if $x>M$ it is

$$f(x)>f(0)\ge f(x_0).$$

Prove that $f$ has a minimum

5 Answers5

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