Given a continuous function $f$ over an interval, must there exist a continuous, increasing function $g$ such that for all $x,y$

$$|f(x)-f(y)| \leq |g(x)-g(y)|$$

I've tried assuming the opposite, that all $g$'s fail for some pair $x,y$ to reach a contradiction, since for every pair $x,y$, there is some $g$ that satisfies the inequality. Similarly for a finite number of pairs $x,y$ there is always a $g$ to satisfy. But I haven't gotten far with that nor used the continuity of $f$.

So I tried using the $\varepsilon$-$\delta$ definition on $f$ around a point since I might be able to connect $g$ to the values of $\varepsilon$, but I haven't found a way to do that either.

In the end, if a $g$ does exist (which I'm somewhat convinced is true), I'd like to find the "minimal" $g_0$ such that for all satisfying $g$'s and $x,y$

$$|g_0(x)-g_0(y)| \leq |g(x)-g(y)|$$

For example, $f=x^2$ and $g_0=x|x|$. Of course if there's a $g_0$, any $g_0+C$ is another one.

Thank you.

  • $\begingroup$ Does "increasing" here mean "strictly increasing"? $\endgroup$ – Omnomnomnom Oct 13 '15 at 11:11
  • $\begingroup$ @Omnomnomnom No, since anywhere the function is flat/constant, $g_0$ must be also. $\endgroup$ – user137794 Oct 13 '15 at 11:12
  • $\begingroup$ Intuitively, draw the curve of $f$ progressively from left to right, and each time the curve goes down, you inverse the direction such that it goes up by the same scale. The the final curve you get represents the function $g$ $\endgroup$ – Petite Etincelle Oct 13 '15 at 11:14
  • 3
    $\begingroup$ I have a feeling that it will only be possible to find such a $g$ if $f$ is of bounded variation. So, for example, there might be no $g$ for $f(x) = x \sin(1/x)$ on an interval including $0$. $\endgroup$ – Omnomnomnom Oct 13 '15 at 11:16
  • $\begingroup$ @Omnomnomnom : You are 23 second ahead of me to answer the question... $\endgroup$ – user99914 Oct 13 '15 at 11:16

Let $f:[a, b]\to\mathbb R$ be a continuous function which is not of bounded variation, see here for an example. Then such a $g$ cannot be found. To see this, assume that $$|f(x) - f(y)| \le | g(x) - g(y)|$$ then by assumption, for any $M$, there is a partition $a =y_0< y_1 < y_2 < \cdots y_{n-1} < y_n = b$ so that

$$\sum_{k=1}^n |f(y_{k}) - f(y_{k-1})| \ge M.$$

In particular,

$$\sum_{k=1}^n |g(y_{k}) - g(y_{k-1})| \ge M.$$

But $g$ is inceasing, so $g(b) \ge M + g(a)$. But $M$ is arbitrary, so $g$ cannot be continuous.

  • $\begingroup$ It is notable also that if such a $g$ exists, then $$ g(x) = V_a^x f $$ is the "minimal" $g_0$ desired. $\endgroup$ – Omnomnomnom Oct 13 '15 at 11:22


Consider the function $f : [-1,1] \to \mathbb R$ defined by: $$\begin{cases}f(x) = x \sin \frac{1}{x} \text{ if } x \neq 0\\ f(0)=0\end{cases}$$

Considering $f$ on $[-1,0]$, what can be the value of $g(0)$?

$f$ is continuous, but it is not of bounded variation, which forces a potential $g$ to take infinite values.


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.