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

t.b.
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Listing
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  • 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 05:40
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    @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 06:32
  • @user3123: Why is the $n$ you define an integer? – JT_NL 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
  • @user3123: $\delta$ depends on $|x|$, so $a$ is not a constant in your argument. As I pointed out, examples show that you will not be able to prove $|g(x)|\leq a|x|$. – Jonas Meyer Dec 18 '10 at 20:33
  • @user3123: As I recently commented on my answer, your proof for $c\gt0$ looks good. If you want it to be more formal, you can make explicit how it is a proof by induction. Do you yet have a counterexample for $c=0$? – Jonas Meyer Dec 19 '10 at 05:32
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    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
  • @user3123: Good. – Jonas Meyer Dec 19 '10 at 18:13
  • Related:https://math.stackexchange.com/questions/691143/show-that-there-are-a-b-geq-0-so-that-fx-leq-axb-forall-x-geq-0 – Watson Nov 18 '16 at 19:01

1 Answers1

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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.

Jonas Meyer
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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 01:26
  • "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 01:30
  • Thank you, I understand it now :-) – Listing Dec 18 '10 at 01:34
  • 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 $g(0) = 0$? –  Dec 18 '10 at 17:08
  • @Soarer: No, the same proof doesn't work for $c=0$. You need $|g(0)|\lt c$ for it to be true. To be more explicit about my hint for $c=0$, think of a continuous function $g$ such that $g(0)=0$ and $g'(x)\to\infty$ as $x\to 0$. If the example you think of isn't already uniformly continuous on $\mathbb{R}$, you can just make it constant outside say $[-1,1]$. – Jonas Meyer Dec 18 '10 at 20:18
  • @user3123: I already commented on your question pointing out the problem with your argument. A suggestion: Don't think that the interval $[0,|x|]$ needs to be divided exactly evenly. Just take steps of equal size (smaller than $\delta$, but independent of $x$) until you come within $\delta$ of $x$. For example, steps of size $\delta/2$, in which case you can take $n$ to be the largest integer such that $(n-1)\delta/2\leq|x|$. I probably won't be posting further hints for a while. – Jonas Meyer Dec 18 '10 at 20:39
  • I see, I tried to add this to the proof and also see why $c>0$ now. If its still incorrect, forget it then I'm probably a helpless case. If people are interested into the full proof I can also add it after we discussed this task. Thanks for your help in any case, I profited from it, even if the proof is still wrong :-) – Listing Dec 18 '10 at 22:27
  • @user3123: Your modified proof is good. Stylistically, I would recommend just setting $\epsilon=c$ in the beginning. – Jonas Meyer Dec 19 '10 at 05:25
  • (When I wrote "You need $|g(0)|<c$ for it to be true" in my comment to Soarer, I meant that without such a restriction there would exist a $g$ not satisfying the inequality in question. Of course, there are plenty of examples of functions $g$ with $|g(0)|=c$ that satisfy the inequality.) – Jonas Meyer Dec 19 '10 at 06:52
  • Ok thanks a lot then, without ur hints it would have been impossible for me to show this. Now that I am into the proof it was not hard to make up a counterexample for $c=0$ (check original post) – Listing Dec 19 '10 at 10:38
  • @user3123:You're welcome. – Jonas Meyer Dec 19 '10 at 23:55