Let $f\_n: [0, 1] \to \mathbb R$ be an equicontinuous sequence of functions. Does there exist a continuous function $f$ that dominates $f\_n $ in the following sense?


We say $f$ dominates the sequence $f\_n$ if for all $x \in [0, 1]$, if $\delta > 0, \varepsilon > 0$ are such that $\lvert f(x) - f(y)\rvert < \varepsilon$ for all $y$ such that $\lvert x - y\rvert < \delta$, then also for all $n$, $\lvert f\_n (x) - f\_n (y)\rvert < \varepsilon$ for all $y$ such that $\lvert x - y\rvert < \delta$.



Are equicontinuous function dominated by a continuous function?
It is well known that if $(f\_n)$ are equicontinuous on a compact space, then they are uniformly equicontinuous. Let $\omega(\delta)=\sup\{|f\_n(x)-f\_n(y)|\colon
|x-y|\le\delta; n\in\mathbb N\}$ be the modulus of continuity of the family, so that $\omega(\delta)\to 0$ as $\delta\to 0$. I will assume without loss of generality that the $(f\_n)$ take values in $[0,1]$.


Then we will construct a continuous function $f$ whose modulus of continuity exceeds $\omega(\delta)$, so that it dominates $(f\_n)$ in your sense. The idea is pretty simple: sum together a bunch of periodic functions that are mostly zero, but have steeper and steeper spikes (of decreasing heights).


Let $\delta\_1>\delta\_2>\ldots$ be such that $\omega(\delta)<4^{-k}$ whenever $\delta<\delta\_k$. Then we inductively build functions $g\_k$ for $k=1,\ldots$ such that $g\_k$ is periodic of period $\delta\_{k+1}$ and on each period is zero, followed by a linear branch up to height $4^{1-k}$ followed by a linear branch of the same length back to 0. We let the length of each of the two linear branches be $\ell\_k$, which is chosen to be sufficiently small that the sum of the maximal variations of $g\_1$, $g\_2$, $\ldots$, $g\_{k-1}$ over an interval of length $\ell\_k$ is at most $4^{-k}$.


Finally let $f=\sum g\_k$. I claim this function has the required property. In particular, it is suffices to show that if $\omega(\delta)=\epsilon$, then there exist $x,y\in [0,1]$ such that $|x-y|<\delta$ and $|f(x)-f(y)|>\epsilon$.


Suppose that $\delta\_{k+1}\le \delta<\delta\_k$. Then $\omega(\delta)<4^{-k}$ and we will exhibit $x,y$ such that $|x-y|\le \delta\_{k+1}$, but $|f(x)-f(y)|>4^{-k}$.
In particular, we choose $x$ and $y$ to be the left and right end points of an increasing interval of $g\_k$ with $g\_k(x)=0$ and $g\_k(y)=4^{1-k}=4\cdot 4^{-k}$. By assumption, $$
\sum\_{j=1}^{k-1} |g\_j(x)-g\_j(y)|\le 4^{-k}.
$$
Also
$$
\sum\_{j=k+1}^\infty |g\_j(x)-g\_j(y)|\le \sum\_{j=k+1}^\infty 4^{1-j}=\tfrac 43\cdot 4^{-k}.
$$


It follows that
$$
|f(x)-f(y)|\ge |g\_k(x)-g\_k(y)|-\tfrac 734^{-k}>4^{-k}
$$
as required.


