Estimate the Lipschitz constant of the product $xf(x)$, where $f$ is Lipschitz It seems to me that this is such a simple question, but the answer has eluded me for a month or more. 
I will give two interpretations. The first is more general. The second is for those possessing an intimacy with iterated function systems or maybe even dynamical systems. For neither of these will I show my work thus far, because everyone I have shown this work to has trod the same route as I have.


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*For a contractive function $\left|\,f(x) - f(y) \right| \leq c \cdot d(x,y)$, where $c \in [0,1), \; x,y \in X=[0,1]$ what is the best bound we can give for $$\left|\,x\cdot f(x)-y\cdot f(y)\right|.$$
The best I have found thus far is $$\leq \left| \,f(x)-y \right| \cdot d(x,y)$$ which is far from the best after reviewing examples.

*I am  doing research in Iterated Function Systems with place-dependent probabilities. 
For a while now I have been struggling to prove the contractivity of the Markov Operator in order to prove the existence of an invariant measure. When using the Hutchinson metric, I come to a halt when trying to bound 
$$\displaystyle \sum_{i=1}^N \left| \, p_i(x) \cdot f(w_i(x)) - p_i(y) \cdot f(w_i(y))\right|$$
where 
$$\displaystyle \sum_{i=1}^N p_i(x) = 1 \qquad \forall x$$
and $$0\leq p_i(x) \leq 1 \qquad \forall i.$$
Additionally, $$f(w_i(x)) - f(w_i(y)) \leq c_i \cdot d(x,y) \qquad 0 \leq c_i \leq 1.$$
The best I have been able to find is,
$$< \displaystyle \sum_{i=1}^N \left| \, 2 \cdot f(w_i(x)) \right| - \max_i c_i\cdot d(x,y)$$
Which is bad for various reasons, mostly because I am looking for an expression like $$< "M" \cdot \, d(x,y)$$
 A: I answer Question 1. In the special case when $f$ is differentiable, the condition $\left|\,f(x) - f(y) \right| \leq c \cdot d(x,y)$ is equivalent to $|f'(x)|\le c$. By the product rule, 
$$(xf(x))' = f(x) + xf'(x)
$$
Here we can estimate $|xf'(x)|\le c$, but there is no quantitative control on $f$ itself. So one is left to say that 
$$|xf(x)|' \le c + \sup|f| \tag{1}
$$
hence
$$|xf(x)-yf(y)| \le \left(c + \sup|f|\right) |x-y| \tag{2}
$$
This bound is optimal: consider the constant function $f$, for example.
If the supremum looks bad, use the mean value theorem to estimate it by $|f(0)|+c$. Then
$$|xf(x)|' \le 2c + |f(0)|  \tag{3}
$$
General case
Without assuming differentiability, the conclusion (2) remains the same. One way to see this is  to use the fact that Lipschitz functions are differentiable almost everywhere and absolutely continuous. Then the estimate (1) holds at almost every point.  The absolute continuity of $xf(x)$ allows us to use the fundamental theorem of calculus: 
$$
yf(y)-xf(x) = \int_x^y (tf(t))'\,dt
$$
and conclude with (2).
