Creating an increasing function $g$ given a continuous function $f$, such that $|f(x)-f(y)| \leq |g(x)-g(y)|$

Question

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.

@Omnomnomnom No, since anywhere the function is flat/constant, $g_0$ must be also. — user137794, Oct 13 '15 at 11:12
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 $g$ — Petite Etincelle, Oct 13 '15 at 11:14
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 $0$. — Ben Grossmann, Oct 13 '15 at 11:16
@Omnomnomnom : You are 23 second ahead of me to answer the question... — , Oct 13 '15 at 11:16

score 12 · Accepted Answer · edited Apr 13 '17 at 12:20

12

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.

edited Apr 13 '17 at 12:20

Community

1

answered Oct 13 '15 at 11:16

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

score 1 · Answer 2 · answered Oct 13 '15 at 11:17

Hint.

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.

Creating an increasing function $g$ given a continuous function $f$, such that $|f(x)-f(y)| \leq |g(x)-g(y)|$

2 Answers2