Suppose I have a decreasing Lipschitz function $f:[0, \infty)\to [0,\infty)$ and an increasing Lipschitz function $g:[0,\infty) \to [0,\infty).$ If the product $f\cdot g$ is decreasing, does it imply that $f\cdot g$ is Lipschitz? Intuitively, it seems to me that multiplying by an increasing function can only make $f$ decrease less slowly, but I'm unable to formalize this.
I've tried proving something like $$f(x)g(x)-f(y)g(y) \leq K(f(x)-g(y))$$ for $x<y$ and some constant $K>0.$ But I have not yet been able to.
If it helps to be more specific: I'm particularly interested in the case where $g(x) = x+c$ for some constant $c>0$ and it is known that $f(x)\left[g(x)\right]^\alpha$ is also decreasing, for some $\alpha>4.$
I might of course also be wrong with my intuition, in that case I would also really appreciate a counterexample.
Since $f\cdot g$ is supposed to be decreasing, there is not a lot that could happen (monotone bounded continuous functions are very nice). Intuitively, the only thing that could go wrong is to have at some $x$ a situation similar to $-\sqrt{x}$ around $0$, i.e. an vertical tangent. And that would be very weird... (I know, that is not the most valid mathematical argument.)
More formally, we know that $f$ and $g$ are differentiable almost everywhere, same for $f\cdot g$. Furthermore, $(fg)' = f'g + fg'$ almost everywhere, and $fg$ is Lipschitz if and only if $(fg)'$ has finite $L^\infty$-norm. Now, since $fg$ is decreasing, we would need to have $(fg)'$ to go to $-\infty$. This is not possible towards $+\infty$, because $fg \geq 0$. Then, suppose that $(fg)'$ is unbounded (towards $-\infty$!) around some fixed $x\in [0,+\infty)$. Since $g' \geq 0$ (and hence $fg' \geq 0$), that means necessarily that $f'g$ is unbounded (towards $-\infty$) around $x$. At the same time, $f'$ and $g$ are bounded around $x$, because $g$ is continuous and $f'$ is Lipschitz.
If I am not mistaken, that should give the result.