Are equicontinuous function dominated by a continuous function? 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 $\lvert x - y\rvert < \delta$.
 A: 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.
A: Let $\omega$ be a continuously differentiable modulus of continuity for $(f_n)_n$.  EDIT:  I meant this in the stronger sense that $\omega'$ extends to a continuous function on $[0, 1]$, which might not be possible.  So this is only a partial answer.
Let $M = \sup \{\omega'(t) \mathrel: t \in [0, 1]\}$, and put $f(x) = M x$ for all $x \in [0, 1]$.  Then, for all $x, y \in [0, 1]$ with $x \ne y$, we have that $\frac{\omega(\lvert x - y\rvert)}{\lvert x - y\rvert} = \omega'(t) \le M$ for some $0 < t < \lvert x - y\rvert$, so $\omega(\lvert x - y\rvert) \le M\lvert x - y\rvert = \lvert f(x) - f(y)\rvert$.
Fix $x, y \in [0, 1]$ with $x \ne y$, and $\varepsilon > 0$ such that $\lvert f(x) - f(y)\rvert < \varepsilon$.  Then also $\omega(\lvert x - y\rvert) < \varepsilon$, so $\lvert f_n(x) - f_n(y)\rvert < \varepsilon$.
