Converse to Young's classical result on Riemann-Stieltjes integration

Question

A classical result from Young in 1936 says that if $f\in C^\alpha$ and $g\in C^\beta$ with $\alpha+\beta>1$ then $\int f \, dg$ exists as a Riemann-Stieltjes integral.

However, I am interested in the converse. Clearly if the support of $f$ is disjoint with the support of $g$ then they can have as bad of analytic properties as you'd want.

However, can we have $f\in C^\alpha$ and $f$ nowhere locally $C^{\alpha'}$ for any $\alpha'>\alpha$, $g\in C^\beta$ and $g$ nowhere locally $C^{\beta'}$ for any $\beta'>\beta$, with $\alpha+\beta<1$, and have the Riemann Stieltjes integral converge?

Answer 1

$\newcommand\al\alpha\newcommand\be\beta$Yes, of course. Take any $\al>0$ and $\be>0$ such that $\al+\be<1$. For $x\in[0,1]$, let $$f(x):=\sum_{j=1}^\infty 2^{-j}(x-r_j)_+^\al$$ and $$g(x):=\sum_{j=1}^\infty 2^{-j}(x-r_j)_+^\be,$$ where $(r_1,r_2,\dots)$ is an enumeration of the rational numbers in $[0,1)$ and $u_+:=\max(0,u)$.

For any real $u,v$ one has $u_+^\al-v_+^\al\le(u-v)_+^\al$ and hence for any $x,y$ in $[0,1]$ $$f(x)-f(y)\le\sum_{j=1}^\infty 2^{-j}(x-y)_+^\al=(x-y)_+^\al,$$ so that $f\in C^\al[0,1]$. Also, for any $j$
$$\liminf_{x\downarrow r_j}\frac{f(x)-f(r_j)}{(x-r_j)^\al}\ge2^{-j}>0$$ and hence for any real $\al'>\al$ $$\liminf_{x\downarrow r_j}\frac{f(x)-f(r_j)}{(x-r_j)^{\al'}}=\infty,$$ so that $f$ is nowhere $C^{\al'}$ on $[0,1]$. Similarly, $g\in C^\be[0,1]$ but $g$ is nowhere $C^{\be'}$ on $[0,1]$ for any real $\be'>\be$.

Finally, $$\int_0^1 f\,dg=\sum_{j=1}^\infty\sum_{k=1}^\infty 2^{-j}2^{-k}\,I_{j,k},$$ where $$I_{j,k}:=\int_0^1(x-r_j)_+^\al\be(x-r_k)_+^{\be-1}\,dx \le\int_0^1\be(x-r_k)_+^{\be-1}\,dx\le1.$$ So, $\int_0^1 f\,dg\in[0,1]\subset\mathbb R$. $\quad\Box$