So I need help with this exercise. If $f$ is a positive and continuous function with $$\lim_{x \to -\infty} f(x) = \lim_{x \to +\infty} f(x) = 0 $$ Prove that $f(x)$ has a maximum. Thanks in advance :)
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what did you try? – imranfat Nov 12 '13 at 21:43
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I reeally dont know what to try. It seems logical to me that a max exists,but I can't really prove it.... – george Nov 12 '13 at 21:43
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As the question is posed $f: \mathbb{R}\backslash {0} \rightarrow \mathbb{R}_+$, $f(x)=1/|x|$ would be a counterexample. – Listing Nov 12 '13 at 22:11
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In my proof, I assumed the function was continuous on $\mathbb{R}$. – Christopher K Nov 12 '13 at 22:13
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@Listing, I think he means on $\mathbb{R}$? – Christopher K Nov 12 '13 at 22:15
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I do,it is continuous on all of R. – george Nov 12 '13 at 22:16
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I just wanted to point out that the question should probably be adjusted. – Listing Nov 12 '13 at 22:20
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By the way, you don't actually need the 'positive' part in the statement "if $f$ is a positive and continuous function". – AJMansfield Nov 13 '13 at 00:17
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And, like @Listing said, this depends hugely on exactly how you define "$f$ is a continuous function". It can mean that $f$ is continuous $\forall x \in \Bbb R$ (in which case it is correct), or just that $f$ is continuous $\forall x$ for which $f(x)$ is defined (in which case it is false), or something else, depending on what author you have. – AJMansfield Nov 13 '13 at 00:20
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Well, go back to the definition of a limit that $\forall\; \varepsilon>0$, there exists $b$ such that $|f(x)-L|<\varepsilon\; \forall\; x>b$. Similarly, there exists $a$ such that $|f(x)-L|<\varepsilon\; \forall\; x<a$. So fix $\varepsilon > 0$. Now, $f(x)$ is bounded by let's say $M>0$ on $[a, b]$ by compactness and the continuity of $f$. But it's also bounded by $\varepsilon$ everywhere else. Hence $f$ is bounded above by $\text{max}(M, \varepsilon)$. Now, continuity demands that $f$ reaches its supremum (it exists by boundedness). That is $f$ has a maximum.
EDIT: Note that I assumed $f$ was continuous on all of $\mathbb{R}$. Of course, counterexamples exist otherwise.
Christopher K
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Thanks man,you've been really helpful. Just one thing,can you elaborate further on why the function is bounded by M? – george Nov 12 '13 at 22:17
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Every continuous function is bounded on a closed interval in $\mathbb{R}$. Hence, there is some $M$ such that $f$ is bounded on a given closed interval. For more info, check Spivak. I believe he calls it one of "Three Hard Theorems". – Christopher K Nov 12 '13 at 22:20
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Choose an $B$ such that $$x > B \implies |f(x) - 0| < 1$$ and an $A$ such that $$x < A \implies |f(x)- 0| < 1$$
Now consider $f$ which is continuous on $[A, B]$.
I will show the existence of the real number $x_1 \in [A,B]$ such that $f(x)\le f(x_1)$ for every $x\in [A,B]$.
Note the set $S = \{f(x) : x\in [A,B]\}$ is bounded above. So let $M = \sup S$. Then $f(x) \le M$ for every $x \in [A,B]$. Suppose on the contrary that there does not exist $x_1 \in [A,B]$ such that $f(x_1) = M$. Then $f(x) < M$ for every $x\in [A,B]$.
Now it then follows that the function $$g(x) = \dfrac{1}{M-f(x)}$$ is continuous in the closed inverval $[A,B]$, and is therefore bounded above on $[A.B]$. Suppose that $g(x) \le K$ in the closed interval, where $K>0$ since $g(x)>0$ for every $x \in [A,B]$.
This gives $f(x) \le M-\dfrac1K$ contradicting the assumption that $f(x) < M$ for every $x\in [A,B]$.
Hence there exists $x_1\in [A,B]$ such that $f(x_1)=M$.
You can do the same and show that $f(x)$ has maximum in $(-\infty,A]$ and $[B,\infty)$ and therefore $f(x)$ has maximum in the whole $\mathbb{R}$.
MathsMy
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