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

  • $\begingroup$ 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. $\endgroup$ – AnonymousCoward Dec 18 '10 at 5:40
  • 1
    $\begingroup$ @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 $0$. $\endgroup$ – Jonas Meyer Dec 18 '10 at 6:32
  • $\begingroup$ @user3123: Why is the $n$ you define an integer? $\endgroup$ – Jonas Teuwen Dec 18 '10 at 18:14
  • $\begingroup$ @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 $\delta$! $\endgroup$ – Listing Dec 18 '10 at 20:22
  • 1
    $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ 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. $\endgroup$ – Listing Dec 18 '10 at 1:26
  • $\begingroup$ "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. $\endgroup$ – Jonas Meyer Dec 18 '10 at 1:30
  • $\begingroup$ Thank you, I understand it now :-) $\endgroup$ – Listing Dec 18 '10 at 1:34
  • $\begingroup$ 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) $\endgroup$ – Listing Dec 18 '10 at 16:48
  • $\begingroup$ @Jonas, sorry, I don't understand what goes wrong for $c = 0$. Doesn't the same proof work? Or have you missed the condition $g(0) = 0$? $\endgroup$ – Soarer Dec 18 '10 at 17:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.