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If $f: [1, \infty ) \rightarrow \mathbb{R}$ is uniformly continuous, prove that there exists a finite positive number M such that $|f(x)| \leq Mx$ for all $ x\geq 1$.

In my mind, I must prove that $M$ is positive, and then prove that $M$ is finite. Proving that M is nonnegative is obvious due to the definition of absolute value. I'm having a hard time proving that $M \neq 0$ though. If $M = 0$, then $|f(x)| \leq 0 \ \Rightarrow f(x) =0$ for all $ x\geq 1$. But the problem states nothing about $f$ being onto, so $f(x) = 0$ seems to be a legitimate function.

I'm clueless as to how to prove $M$ is finite. It seems to me that if $|f(x)| \leq Mx$, then $|f(x)| \leq \infty \times x = \infty$. So why can't $M$ be infinite?

Claycrusher
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    You're completely misinterpretting the problem. You need to prove that there EXISTS some $M$ with those properties. You're assuming...I'm not sure what. Hint, what does uniformly continuous mean? Write this out, and see what happens. –  Nov 18 '15 at 07:12
  • @avid19: I guess I'm assuming that if $M$ exists, then it is positive and finite. – Claycrusher Nov 18 '15 at 07:16
  • That is not the way to do this problem. :) See for example http://math.stackexchange.com/questions/305014/every-uniformly-continuous-real-function-has-at-most-linear-growth-at-infinity –  Nov 18 '15 at 07:16
  • One way you could interpret the problem is this: If $M$ exists, show that we can choose $M$ such that it is positive and finite. – Joey Zou Nov 18 '15 at 07:18
  • For example, if $f(x) = 0$, then you could choose $M=0$, and it would satisfy the bound (though it would not be positive). But you could choose $M=1$, and it would satisfy the bound, as well as being finite and positive. – Joey Zou Nov 18 '15 at 07:19
  • @JoeyZou: What you're saying makes sense. I guess the trick is proving that M exists. – Claycrusher Nov 18 '15 at 07:21
  • Apparently first we should prove the existance of $M$ and this is the logical form. I'm in doubt if it is in mathematial terminology that we suppose $M$ is exist!. – Nosrati Nov 18 '15 at 07:27

2 Answers2

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If we can prove that $S$={$\frac{|f(x)|}x$| $x\in[1,\infty)$} has finite supremum ,i.e. sup$S$ < $\infty$, then there exists M and M can be chosen exactly sup$S$.

So let's prove $S$ has finite sup.

By @avid19's comment , we can choose $\delta_1$ s.t. $|x-y|<\delta_1 \Rightarrow |f(x)-f(y)|<1$ and it implies that $|f(x)|<m_x+|f(1)|$ where $m_x :=$smallest natural number $k$ with $|x-1|=x-1<k\delta_1$

(Note that $x$ is in $[1,\infty)$ so $|x-1|=x-1$)

Then our second problem is done , since $x>1$ and $\frac{x-1}\delta_1<m_x<\frac{x-1}\delta_1+1$

$\frac{|f(x)|}x<(m_x+|f(1)|)\frac1x<(\frac1\delta_1(x-1)+1+|f(1)|)\frac1x=\frac1\delta_1\frac{x-1}x+\frac1x(1+|f(1)|)<M$

where $M=\frac1\delta_1+(1+|f(1)|)$

user128766
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Since $f$ is uniformly continuous on $[1,\infty)$, $\exists\delta>0$ such that $\forall x,y$ satisfying $|x-y|<\delta$, we have $|f(x)-f(y)|<1$. Now consider the interval $[1,1+\delta]$. Now, for any $x\in [1,\infty)$, $\exists n$ such that $x=n\delta+x'$ such that $x'\in [1,1+\delta]$. Now,

$|f(x)|=|f(x'+n\delta)|\le |f(x')|+|f(x'+\delta)-f(x')|+\cdots +|f(x'+n\delta)-f(x'+(n-1)\delta)|\le |f(x')|+n$. Hence , $|\frac{f(x)}{x}|=|\frac{f(x'+n\delta)}{x'+n\delta}|\le |\frac{f(x')}{x'+n\delta}|+|\frac{n}{x'+n\delta}|\le M+\frac{1}{\delta}$, where $M=\sup_{[1,1+\delta]}f(x)$. Hence the result is proven.

Ribhu
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