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I want to show that a continuous function $f:[0, \infty) \to \mathbb{R}$ satisfying $f(x) \ge 2+x$ must has a minimum.

I want to use the extreme value theorem but i don't know how to rewrite the interval into a closed and bounded one in order to satisfy all the requirements of the Extreme Value Theorem.

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    This is very hard to follow. You should try and make the question clearer. – 2017-02-26
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    @Fimpellizieri i want to know how i can apply the extreme value theorem for this question in order to show that the inequality function has a minimum. – 2017-02-26
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    Is $f(x) \geq 2 +x$? – 2017-02-26
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    Yes it is f(x)>=2+x – 2017-02-26
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    Let $x \in [0, +\infty)$ with $x > f(0)$, then $x = f(0) + \epsilon$ for some $\epsilon > 0$. Now $f(x) = f(f(0) + \epsilon) >= 2 + f(0) + \epsilon > f(0)$. Do you know which closed bounded interval to proceed with? – 2017-02-26
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    Where is the 2 coming from? F(0)=2 and why is x > f(0)? – 2017-02-26
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    @Mark The "$2$" comes from the assumption in the question $f(x) >= x + 2$ $\forall x \in [0, +\infty)$. I say $x > f(0)$ because I want to show for large enough $x$, $f(x) > f(0)$. Do you understand the idea? – 2017-02-26
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    Yes i do now, but my question is what are we looking for exactly?? And how do i ever knew that i should approach this like this? Also the closed interval is not that clear.. – 2017-02-26
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    @Mark Hmm... The idea is that when $x > f(0)$, $f(x) > f(0)$ would imply the infimum of the function cannot be in $(f(0), +\infty)$. Do you know why $\inf f$ always exists in $\mathbb{R}$? – 2017-02-26
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    No i dont know that inf f always exists in R. – 2017-02-26
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    @Mark Well $\forall x \in [0, +\infty)$ $f(x) >= x + 2 >= 2$ implies $f$ is bounded below by 2. Since $f$ is bounded below, by the completeness axiom of $\mathbb{R}$, $\inf f$ exists in $\mathbb{R}$. – 2017-02-26
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    @Mark Do you want me to give a complete answer or think about it yourself? – 2017-02-26
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    First i dont know why you said that if f(x)>f(0) would imply that infimum would not exist in (f(0),infinity). If we use f(0)>=2 why is 2 not the infimum it is the lowest bound right? Because the lim x-> infinity will be infinity? Now if f(0)>=2 this statement says that we mean all the values from 2 till infinity right? – 2017-02-26
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    @Mark No, I mean $\inf f$ **exists** because $2$ is a lower bound of $f$. – 2017-02-27
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    Ok can you give the full answer and proof now? Because that is still missing – 2017-02-27
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    I know i can proceed with the close bounded interval [0,f(0)]? But can you still show it complete in proof form – 2017-02-27
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    You can show the complete answer – 2017-02-27
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/54417/discussion-between-mark-and-alex-vong). – 2017-02-27

2 Answers 2

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You said:

I want to use the extreme value theorem but i don't know how to rewrite the interval into a closed and bounded one in order to satisfy all the requirements of the Extreme Value Theorem.

The idea is to use the extreme value theorem on a "suitable" closed and bounded interval, and then show that the minimum of $f$ on this interval is in fact the overall minimum of $f$.

$f(0) \ge 2 + 0 > 0$, therefore we can define the interval $I$ as $$ I = [0, f(0)] \, . $$ $I$ is a closed, bounded interval and $f$ is continuous, therefore $f$ has a minimum on $I$, i.e. there is a $x_0 \in I$ such that $$ \tag{*} f(x_0) \le f(x) \text{ for all } x \in I \, . $$ In particular, $f(x_0) \le f(0)$.

Now for arbitrary $x \ge 0$, we have either $$ 0 \le x \le f(0) \Longrightarrow f(x) \ge f(x_0) $$ because of $(*)$, or $$ x > f(0) \Longrightarrow f(x) \ge 2 + x > 2 + f(0) > f(0) \ge f(x_0) \, . $$ Therefore $f(x_0)$ is the overall minimum of $f$.

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    What does the statement a:=f(0) means? What is a:?? – 2017-02-26
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    @Mark: It just means that $a$ is *defined* as $f(0)$. You can also write "Let $a = f(0)$" if you prefer. – 2017-02-26
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    So will our bound be [0,2]? Because f(0)=>=2 and 0<=x<=2=a ??? – 2017-02-26
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    @Mark: Sorry, I don't get you, which bound? You asked for a *proof* that $f$ has a *minimum.* – 2017-02-26
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    The expression f(0)=a implies a>=2? So a will be at least 2?? – 2017-02-26
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    @Mark: Yes. $f(x) \ge x+2$ for all $x$, $ \Longrightarrow a = f(0) \ge 2 + 0$. – 2017-02-26
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    Yes then if you make an interval 0<=x<=a implies 0<=x<=2?? – 2017-02-26
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    @Mark: No. Why should it? Can you explain which step in the proof you don't understand? – 2017-02-26
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    You said a=f(0) which is >=2 so if we set x in the interval 0<=x<=a than a=<2? Because a=>2? – 2017-02-26
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    @Mark: I have rewritten the proof. Is it more clear now? – 2017-02-26
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    The last part of your new proof you say 2+f(0)>f(0) where does 2+f(0) come from? – 2017-02-26
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/54372/discussion-between-martin-r-and-mark). – 2017-02-26
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Let $m=\inf_{x\in[0,\infty)}f(x)$. Of course, $m\geq 2$. If $x>m$, then $f(x)> m+2$, so any minimizing sequence must eventually lie in $[0,m]$.

But $[0,m]$ is compact, and hence the infimum is attained (ie, there is a minimum) at some point inside it.

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    I dont fully understand this approach why areyou bringing sequences into the mix? – 2017-02-26
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    That's just one approach which I believe makes it very clear. Take a sequence $x_n$ such that $f(x_n)\to m$. Such a sequence exists by the infimum property. The inequality then assures us that at most finitely many $x_n$ lie outside $[0,m]$. Then by compacity there is a convergent subsequence $x_{n_k}\to x$, and by continuity we must have that $f(x_{n_k})\to m$. In other words, the infimum is attained, so it is a minimum. Does this longer explanation help? – 2017-02-26
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    Not really what does f(x_n) means? And why bring subsequences into the proof?? – 2017-02-26
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    $f$ is the function. $f(x_n)$ is $f$ applied to the $n$-th element of the sequence. Because compacity involves subsequences and compacity is key here. – 2017-02-26
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    There are to much technical terms in your explenation. F(x_n) what does that imply? Also i am confused is f(I) the same as saying f(x_n)? Where I is interval – 2017-02-26
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    If $D$ is the domain of $f$ and $x\in D$ then $f(x)$ is the image of $x$ under $f$. If $S$ is a *subset* of the domain of $f$, then $f(S)$ is the set $$\{f(x)\,|\,x \in S\}$$ – 2017-02-26
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    What is the minimizing sequence in your proof? What does it mean? – 2017-02-27
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    Minimizing sequence is a sequence $x_n$ such that $f(x_n)\to m$ as $n\to\infty$. – 2017-02-27
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    Ok but is the subsequence in the interval [0,m]?? And is the sequence x_n all the x values? Like x_1, x_2 etc? – 2017-02-27
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    One may choose the subsequence so that it lies entirely in $[0,m]$. I don't understand your second question, but a sequence $(x_n)$ means $x_1,x_2,\dots,x_n,\dots$. It goes on infinitely. – 2017-02-27
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    Yes but it is not very common to bring sequences in connection with functions. That was very confusing for me to comprehend. You should have made it more concrete as how we can bring in sequences for this problem. Not everyone in here has a phd in mathematics from MIT. It costs me many hours to understand it all. – 2017-02-27
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    It is very common actually. It is very usually shown in *entry level* mathematics that $\epsilon$-$\delta$ continuity is the same as continuity in terms of sequences. The answer I gave is appropriate for an introductory course to real analysis -- very, *very* far from a PhD. – 2017-02-27
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    Yes it might be common in proving a limit. This question was regarding proving a minimum for a function. And you brought sequences and subsequences. You should have give an explanation as why we bring those and give an example as we can say that all the x values of the function can be treated as a sequences like x_1, x_2 etc and then went on and explain the whole proof that would have spared me many many hours and maybe others as well. And your not the only one who does this here. It should be taken into consideration in answering the questions in here as why you do things and the assumptions – 2017-02-27
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    You should either study harder, or provide more context in your questions. – 2017-02-27