Let $f,g:\mathbb R\to\mathbb[0,\infty)$ be nondecreasing and right-continuous and $h:\mathbb R\to\mathbb R$ be right-continuous and of bounded variation with $$\left|h(t)-h(s)\right|\le\sqrt{f(t)-f(s)}\sqrt{g(t)-g(s)}\;\;\;\text{for all }t\ge s\tag1.$$ Then it's well-known$^1$ that $$\int\left|XY\right|\left|{\rm d}h\right|\le\sqrt{\int|X|^2\:{\rm d}f}\sqrt{\int|Y|^2\:{\rm d}g}\tag2$$ for all Borel measurable $X,Y:\mathbb R\to\mathbb R$, where $|{\rm d}h|$ denotes the variation of the Lebesgue-Stieltjes measure induced by $h$.
Now assume, instead of $(1)$, we've got $$\left|h(t)-h(s)\right|\le C\sqrt{f(t)-f(s)}\;\;\;\text{for all }t\ge s\tag3$$ for some $C\ge0$. How can we use $(2)$ in order to conclude $$\int\left|X\right|\left|{\rm d}h\right|\le C\sqrt{\int|X|^2\:{\rm d}f}\tag4$$ for all Borel measurable $X:\mathbb R\to\mathbb R$?
My first intuition was to simply set $g\equiv C$, but (obviously) that doesn't work since then the right-hand side of $(1)$ is $0$ and any integral with respect to ${\rm d}g$ is $0$. So, are we able to use $(2)$ at all or do we need to mimic its proof in order to establish $(4)$?
I don't know how this inequality is called in the nonrandom case, but in stochastic analysis this is the celebrated Kunita-Watanabe inequality.