# 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 $c > 0$. $\quad \square$

• It is easy to show that if a function is uniformly continuous then its derivative everywhere is bounded. Perhaps this result could be of use. – AnonymousCoward Dec 18 '10 at 5:40
• @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
• @user3123: Why is the $n$ you define an integer? – Jonas Teuwen Dec 18 '10 at 18:14
• @Jonas because I defined $\delta=\text{max}\{z \leq \hat\delta: \frac{|x|}{z} \in \mathbb{N}\}$, so $n$ has to be an integer according to the definition of $\delta$! – Listing Dec 18 '10 at 20:22
• 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.
• 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
• @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