# Prove existence of $c$ such that $f(x)\ge c g(x)$

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?

• How can an account be deleted 2 minutes after asking a question? Dec 20, 2012 at 16:43
• The question asked $f(x) \geq Mg(x)$. Dec 20, 2012 at 16:45
• @N.S. Ah, made a mistake while converting to $\LaTeX$. Fixed now. Dec 20, 2012 at 16:46
• @MattN. Good question. I thought only moderators could delete users. Dec 20, 2012 at 16:49
• @MattN. This person appears to be posting exactly once then deleting their account, which this (unclearly phrased) meta.SO post 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. Dec 20, 2012 at 17:08

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 $f(x)\geq c\cdot g(x)$.

• elegant, thanks! Dec 20, 2012 at 23:53
• @Eric: Thanks!! Dec 20, 2012 at 23:53
• This the same, as N.S.'s solution here. Still, very nice! Dec 20, 2012 at 23:54
• @dtldarek: Well it is a reasonably obvious solution if you think about it... :-) Dec 20, 2012 at 23:55
• @dtldarek: And now it's N.S.'s solution here. :-) Dec 21, 2012 at 0:27

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

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

• Is there an example where this fails when the domain is no longer [0,1] but the whole real line R?
– R.Q
Dec 20, 2012 at 17:04
• yes. $f(x)=1+\frac{1}{x^2+1}$ and $g(x)=1$. Dec 20, 2012 at 18:27

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 $[0,1]$?

Hint:

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

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}