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Let $f$ and $g$ be non-negative continuous functions on $[0,1]$ such that $f(x)>g(x)$ for all $x$ in $[0,1]$. Show that there exists a constant $c>1$ such that for all $x$ in $[0,1]$ we have $f(x)\ge c g(x)$.

I tried using the fact that since $[0,1]$ is compact and $f$ and $g$ are continuous $f$ and $g$ each attain a maximum and a minimum on $[0,1]$, but it didn't get me anywhere. Anyone know how to prove this?

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    How can an account be deleted 2 minutes after asking a question?2012-12-20
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    The question asked $f(x) \geq Mg(x)$.2012-12-20
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    @N.S. Ah, made a mistake while converting to $\LaTeX$. Fixed now.2012-12-20
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    @MattN. Good question. I thought only moderators could delete users.2012-12-20
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    Maybe his IP was flagged or something....2012-12-20
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    @MattN. This person appears to be posting exactly once then deleting their account, which [this (unclearly phrased) meta.SO post](http://meta.stackexchange.com/a/7979) indicates is possible. There have already been two "answers" posted here, intended as comments, and most likely by the same person as who posted the question, which are now deleted because they immediately deleted those accounts as well.2012-12-20
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    @ZevChonoles Thank you for resolving the mystery. That's weird behaviour -- one can but shrug.2012-12-20
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    http://math.stackexchange.com/questions/262750/a-question-regarding-uniform-continuity/262751#2627512012-12-20
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    Perhaps we can merge these two threads instead of closing? (I suppose we should merge the other one into this one?)2012-12-20

5 Answers 5

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Take $h(x)=\frac{g(x)}{f(x)}$. This function is continuous and well-defined because $f(x)>g(x)\geq 0$.

But the above says that $h(x)<1$ for all $x$. Then $h$ has a maximum, $m$ and it is less than $1$. We have that $\frac{g(x)}{f(x)}\leq m$ and therefore $m\cdot f(x)\geq g(x)$. Take $c=\frac1m$ then $c>1$ and we have $f(x)\geq c\cdot g(x)$.

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    elegant, thanks!2012-12-20
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    @Eric: Thanks!!2012-12-20
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    This the same, as N.S.'s solution [here](http://math.stackexchange.com/questions/262750/a-question-regarding-uniform-continuity). Still, very nice!2012-12-20
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    @dtldarek: Well it is a reasonably obvious solution if you think about it... :-)2012-12-20
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    @dtldarek: And now it's N.S.'s solution *here*. :-)2012-12-21
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    @AsafKaragila Still, very nice! ^^2012-12-21
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Since $f(x) >g(x) \geq 0$ you know that $f(x) \neq 0$.

Then the function $\frac{g(x)}{f(x)}$ is continuous on $[0,1]$ and thus It attains it's maximum.

Let $N$ be its maximum, and let $x_0$ be the point where it is attained. Then

$$N=\frac{g(x_0)}{f(x_o)} <1$$

and since $N$ is the maximum

$$f(x)\geq \frac{1}{N} g(x) \,.$$

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    Is there an example where this fails when the domain is no longer [0,1] but the whole real line R?2012-12-20
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    yes. $f(x)=1+\frac{1}{x^2+1}$ and $g(x)=1$.2012-12-20
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Edit: The function $f(x)-g(x)$ is continuous on $[0,1]$ and strictly positive, so it attains a minimum $k_1>0$. The function $g(x)$ is also continuous on this interval, and attains a maximum $k_2\geq 0$. If $k_2=0$, then $g(x)=0$ and the problem is trivial, so assume that $k_2>0$. With this maximum and minimum in mind, consider the function

$$h(x)=f(x)-\left(1+\frac{k_1}{2k_2}\right)g(x).$$

What can we say about it's minimum on $[0,1]$?

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Set $\epsilon>0$ and $$ C=\left(1+\epsilon\right)>0 $$ Note that $\epsilon\cdot g(x)>0$ and \begin{align} f(x)-C\cdot g(x)= & f(x)-\left(1+\epsilon\right)\cdot g(x) \\ = & f(x)-g(x)+\epsilon\cdot g(x) \\ > & \epsilon\cdot g(x) \\ > & 0 \end{align}

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Hint:

Let $A = \{x \mid g(x) \neq 0\}$. Set $c = \inf_{x \in A} \frac{f(x)}{g(x)}$.