I pose this question to seek to prove a statement like the following:
Let $f(x)$ and $g(x)$ be continuous on the interval $[0, 1]$, and let $h(y)=f((y+1)/2)$ where $-1\le y\le 1$. If $f^{(r)}$ is of bounded variation $V$, where $r\ge 1$, and if $|h(y)-g((y+1)/2)|\le\epsilon(V/2^r, ...)$, then $|f(x)-g(x)|\le\epsilon(V, ...)$. In both cases, $\epsilon(...)$ does not depend on $x$ or $y$. (See the following answer by another user; I want to transfer the result it mentions to [0, 1] rather than [-1, 1].)
Proving this statement seems to require proving the following claims. However, although claim 2 is easy, claim 1 is apparently not, although it seems obviously true (by taking the same partition that establishes $f^{(r)}$'s bounded variation and applying it to $h$, appropriately "stretched"). Has a proof of claim 1 been published anywhere?
Claim 1: Let $f(x)$ be continuous on the interval $[a, b]$. If $f^{(r)}$ is of bounded variation $V$, where $r\ge 1$, then $g(x) = f^{(r)}(a+(b-a) (x+1)/2)$ has bounded variation $V\left(\frac{b-a}{2}\right)^r$ on $[-1, 1]$.
Claim 2: Suppose $f(x)$ and $g(x)$ are continuous on $[-1, 1]$. If $|f(x)-g(x)|\le\epsilon$, where $-1\le x\le 1$ and $\epsilon$ does not depend on $x$, then $|f(2y-1)-g(2y-1)|\le\epsilon$, where $0\le y\le 1$.
Proof: Obvious by making a change of variables $x$ to $2y-1$.
This question is not homework or classwork.