If $|g(a)-g(b)| \leq |f(a)-f(b)|,$ for every $a,b$, and $f$ is a Darboux function, then $g$ is a Darboux function. Let $f,g : \mathbb R \to \mathbb R$ be two functions. In particular, $f$ is a Darboux function (that is, functions satisfying the intermediate value property).
If $|g(a)-g(b)| \leq |f(a)-f(b)|,$ for every real numbers $a,b$, then I need to prove that $g$ is also a Darboux function.
I feel like this is easier than most problems/exercises related to the properties of Darboux functions, however, I'm not sure how to approach this one.
Source: I just checked the source again, and this is how the problem is written (a Romanian book with no solutions)
 A: Let $D:=f(\mathbb{R})$, because $f$ is a Darboux function, $D$ is an interval.
Because "$f(x)=f(y)$" implies "$g(x)=g(y)$", there is a function $h : D \rightarrow \mathbb{R}$ such that $g(x)=h(f(x))$ for all $x$. Again, because $\text{Im}(f)=D$, the initial condition gives $|h(s)-h(t)| \le |s-t|$ for all $s,t \in D$.
So $h$ is a continuous function hence has Darboux property.
Now for any $x,y\in \mathbb{R}$ and value $c$ lying between $g(x),g(y)$, note that $g(x)=h(f(x)), g(y)=h(f(y))$ and that $h$ has Darboux property, there is a $u \in D$ lying between $f(x),f(y)$ such that $h(u)=c$ (1) .
But $f$ has Darboux property, then there is a $z$ lies between $x,y$ such that $f(z)=u$ (2).
So from (1) and (2), we imply that $g(z)=h(f(z))=h(u)=c$.
In summary, for any $x,y\in \mathbb{R}$ and value $c$ lies between $g(x),g(y)$, there is a number $z$ which lies between  $x,y$ such that $g(z)=c$.
Or, $g$ has Darboux property.
Side note: The result still holds if we can replace  $|f(x)-f(y)|$ by $H( f(x),f(y))$  for any nonegative continuous function $H$ such that $H(s,t)=0$ iff $s=t$
A: This question was recently reposted to QUORA (don't ask why).  Andy Bruckner answered it there.  I am reposting it here for Andy as I am sure some StackExchangers will find it interesting. Quora repost
As the question was taken from a Romanian textbook I would venture a guess it was written by Solomon Marcus who was a great fan of the Darboux property and well-known to me and Andy as a prolific poser of problems.  Marcus often posed such problems for which he himself did not know the answer.


A function that has the Intermediate Value Property (IVP) is called a Darboux function because in 1875 Jean Gaston Darboux (1842-1917) showed that derivatives had this property.
The poser’s question is an interesting one. Let me explain.
Some 19th century mathematicians thought this IVP was a good candidate for the notion of continuity. Darboux dispelled that belief with some of his examples. (Even today, one sees casual explanations of continuity as “a function one can draw without taking the hand off the paper”.  This is closer to suggesting a Darboux function than a continuous one.
If the answer to the poser’s question is “yes” it would give a rare example of a statement that is true for continuous functions is also true for Darboux functions.
(If one replaces “Darboux” with “continuous” the corresponding statement is clearly true).
Here are some well known useful facts about continuous functions that fail for Darboux functions:

*

*The sum of two continuous functions is continuous
(every function is the sum of two Darboux functions).


*A uniform limit of continuous functions is continuous
(not so for Darboux).


*A continuous function on $[a,b]$ achieves an absolute max and an absolute min.
( A Darboux function on $[a,b]$ can take every value in every interval- even in every nonempty perfect set.)


*A continuous function on $[a,b]$ is Riemann integrable.
(A Darboux function, even a bounded one, can fail to be integrable on any interval, relative to any notion of integral in use.)


*A continuous function on $[a,b]$ has a connected graph.
(Not so for Darboux.)


*A continuous function is determined by its values on a dense set.
(The only set that is determining for a Darboux function is the entire domain interval.)
This is just a sample list. The only property shared by continuous functions and Darboux functions that comes to mind is that both classes are closed under composition.
So I was curious whether the poser’s question had an affirmative answer. It does!
Let’s prove it.
Suppose $f$ and $g$ are defined on $R$, that $g(a) < c < g(b)$. We wish to show there exists $d$ in $(a,b)$ such that $g(d) = c$.
We will use a fact about Darboux functions that is easy to prove. If $f$ is such a function, then for every $x \in R$ there are sequences $\{x_n\}$ increasing to $x$, and $\{y_n\}$ decreasing to $x$, such that $\lim f(x_n) = f(x)$ and $\lim f(y_n) =f(x)$   as $n \to ∞$.    (1)
For the remainder of this discussion we restrict our work to the interval $(a,b)$.
Let $A =\{x:g(x)>c\} $, $B =\{x:g(x)<c\}$, and $D$ be the boundary of $A$. The set $D$ is closed  and every open interval containing a point of $D$ contains points of $A$ and of $B$. We will show there is a point $d$ in $D$ such that $g(d) = c$.
The set $D$ must contain at least one of three kinds of sets: closed intervals, Cantor sets, and isolated points . One can check the several cases and verify that there exists a point which is isolated from at least one side in $D$. For example, an endpoint of an interval complementary to a Cantor set is isolated form one side.  Thus, for such a point $d$ there will be an interval $(d,e)$ or $(e,d)$ contained in $A$ or in $B$.
Choose such a point $d$ that is isolated from at least one side in $D$. If $d$ is in $A$ then some interval with $d$ as an endpoint, say $(d,e)$, is contained in $B$ This means $g(d) > c$ while $g(x) <c$ for all $x$ in $(d,e)$. This shows
$|g(d) -g(x)| > |g(d) - c | $  for all $x$ in $(d,e)$. (2)
On the other hand, by (1) , there exists a sequence $\{d_n\}$, contained in $(d,e)$  approaching $d$ such that $f (d_n)$ approaches $f(d)$. Thus, for $n$ sufficiently large,
$|f(d) - f(x_n)| < |g(d) - c|$     (3)
Comparing (2) with (3) we see this is in violation of our basic assumption that
$|g(x) - g(y)|  \leq |f(x) -f(y)|$ for all $x, y \in R$.
We have shown $d$ cannot be in $A$. A similar argument shows $d$ cannot be in $B$.
The only remaining possibility is that $f(d) = c$.
This verifies that $g$ has The Darboux property.

Incidentally, if $f$ is in the first class of Baire, then so is $g$.  This follows from the fact that a function is a Baire 1 function iff every nonempty perfect set has a point of relative continuity. Such a point for $f$ will also be such a point for $g $ because of our hypotheses.
