# Is a uniformly continuous function vanishing at $0$ bounded by $a|x|+c$?

Let $g: \mathbb{R} \rightarrow \mathbb{R}$ be uniformly continuous with $g(0)=0,c\geq 0, c \in \mathbb{R}$. Show: $$\exists a\geq 0 \in \mathbb{R}: \forall x \in \mathbb{R}: |g(x)| \leq a \cdot |x|+c$$

I could also say $g(x) \in \mathcal{O}(x)$.

Notes: I could not make up any counterexample so I guess it could be true, all uniformly continuous functions I know grow too slowly.

My approach:

Given $\epsilon > 0$, we have that: $$\exists \delta(\epsilon): |x-y|<\delta=> |g(x)-g(y)|<\epsilon$$ because of the continuity of $g$. Now choose $n=\text{max}\{n \in \mathbb{N}: (n-1)\delta/2\leq|x|\}$. Obviously, such an $n$ exists, and $n > 0$. We also easily see that an upper bound for $n$ is $n \leq \frac{2}{\delta}|x|+1$.

Now we use this to separate $|x|$ into $n-1$ distinct parts of size $s<\delta/2$, and the last part which is smaller than $\delta$ : $$|x|=|x_1-x_0|+|x_2-x_1|+|x_3-x_2|+...+|x_n-x_{n-1}| < (n-1)\delta/2 + \delta = (n+1)\delta/2.$$

\begin{align} \Rightarrow |g(x)| & =|g(x_1)-g(x_0)+g(x_2)-g(x_1)+g(x_3)-g(x_2)+...+g(x_n)-g(x_{n-1})| \\ & \leq |g(x_1)-g(x_0)|+|g(x_2)-g(x_1)|+|g(x_3)-g(x_2)|+...+|g(x_n)-g(x_{n-1})| \\ & \lt n \cdot \epsilon \leq (\frac{2}{\delta}|x|+1) \cdot \epsilon = \frac{2\epsilon}{\delta} \cdot |x|+\epsilon \end{align}

So we can see that the constant $c$ we were given can be set as the $\epsilon := c$, and that was also the reason why generally speaking $c>0$. Then we can choose $a := \frac{2\epsilon}{\delta}$, as our $\delta$ only depends on the $\epsilon$, and we have that $|g(x)| \leq a \cdot |x| + c$ for $c > 0$. $\quad \square$

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This is not really functional analysis. –  Jonas Teuwen Dec 18 '10 at 0:56
Is this homework? I ask because I would like to help, but don't want to give a full solution if it is something you are supposed to solve. –  Jonas Meyer Dec 18 '10 at 1:02
Yea it is homework, I adjusted the tags also –  Listing Dec 18 '10 at 1:09
@GottfriedLeibniz: A uniformly continuous function need not have a derivative, and if a uniformly continuous function has a derivative, it need not be bounded. However, if a function is Lipschitz continuous and everywhere differentiable, then its derivative is bounded. From the stronger hypothesis of Lipschitz continuity would follow the stronger conclusion that $c$ can be taken to be $0$. –  Jonas Meyer Dec 18 '10 at 6:32
Yes a counterexample for $c=0$ is not hard. For example $x^{\frac{1}{3}}$ is some very nasty function. Its slope will go against infinity, and therefore you cannot "beat" it with a simple $a \cdot |x|$ function. And yea, its u.c. –  Listing Dec 19 '10 at 10:34

It is false if $c=0$. To see this, try to think of a continuous function that grows very rapidly near $0$.

It is true if $c\gt 0$. One way to show it is by taking a number of very small steps from $0$ to $x$, small enough to guarantee (using uniform continuity) that the function changes no more than a certain fixed amount at each step. Trying to write out the details should lead you to what this fixed amount is, and to what value of $a$ will work.

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Nice find with $c=0$, howevery. I don't get what you mean with small steps. I see that if you would devide $0$ to $x$ into e.g. $n$ steps, then the function should change less than $a/n$ with each step. –  Listing Dec 18 '10 at 1:26
"Small" will mean less than $\delta$ (corresponding to some $\epsilon$). Taking $n$ steps of less than $\delta$ results in your function changing by less than [fill in the blank]. The number $n$ of steps has a bound depending on $\delta$ and $|x|$, so you should end up with an inequality involving $|g(x)|$ and $|x|$. Then you should be able to see what adjustments are necessary (if any) to finish the job. –  Jonas Meyer Dec 18 '10 at 1:30
Thank you, I understand it now :-) –  Listing Dec 18 '10 at 1:34
I did it like this now, could you please check if its correct? (I edited the answer into the question because it is too long for a comment) –  Listing Dec 18 '10 at 16:48
@Jonas, sorry, I don't understand what goes wrong for $c = 0$. Doesn't the same proof work? Or have you missed the condition $g(0) = 0$? –  Soarer Dec 18 '10 at 17:08