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Question. Let $f:\mathbb R\to \mathbb R$ be a uniformly continuous function. Show that there exists $a,b>0$ such that $|f(x)|\le a|x|+b,$ $\forall x\in\mathbb R$.

Proof. Since $f$ is uniformly continuous, $\forall\epsilon>0$ $\exists\delta>0$, $|x_1-x_2|<\delta$ we have $|f(x_1)-f(x_2)|<\epsilon$. Let $\epsilon$ and $\delta$ be fixed. Then $\forall x\in\mathbb R,$ $\exists n\in\mathbb Z,$ $x_0\in (-\delta,\delta)$ such that $x=n\delta + x_0.$ Then we have $$f(x)=\sum_{k=1}^n\{f(k\delta+x_0)-f((k-1)\delta)+x_0\}+f(x_0).$$ Note that $f$ is continuous on $[-\delta,\delta],$ then it is also bounded on $[-\delta,\delta].$ That is $\exists M>0, |f(x)|<M,$ $\forall x\in [-\delta,\delta].$ By the triangle inequality we have $$|f(x)|\le \sum_{k=1}^n|f(k\delta+x_0)-f((k-1)\delta)+x_0|+|f(x_0)|\le |n|\epsilon + M.$$ Since $x=n\delta + x_0, |n|=\frac{|x-x_0|}{\delta},$ we have $$|f(x)|\le\dfrac{|x-x_0|}{\delta}\epsilon+M\le \frac{\epsilon}{\delta}|x|+\left(\frac{x_0}{\delta}\epsilon+M\right).$$ Set $a=\frac{\epsilon}{\delta}, b=\frac{|x_0|}{\delta}\epsilon+M.$ We have $\forall x\in\mathbb R, |f(x)|\le a|x|+b$.

I need some help understanding the justification of the use of the first summation. I get the rest of this, but I would complete my understanding if I could figure out where the summation came from. Thanks

Brian Fitzpatrick
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homegrown
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  • One can "explain" this as follows. You want to show that $\vert f\vert$ cannot increase too much between $x_0$ and $x$. What you know is that it does not increase by more than $\varepsilon$ if you "jump" from some point to some other point which is within $\delta$. Now, to pass from $x_0$ to $x$, you need $n$ jumps of size $\delta$. – Etienne Feb 26 '14 at 20:50

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I agree with Etienne's comment for why we would want to do such a summation. The justification is that (after moving a parenthesis) it is a telescoping sum. i.e. I believe there is a typo in the proof. The first line should be:

$$f(x)=\sum_{k=1}^n\{f(k\delta+x_0)-f((k-1)\delta+x_0)\}+f(x_0)$$

And so all the middle terms in the sum cancel out. The extra $f(x_0)$ cancels when $k$ = 1. And we get the $f(x)$ term from $k = n$.

Andrew Kelley
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