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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 $\lvert x - y\rvert < \delta$.

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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.

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    $\begingroup$ Very nice! I was up late last night trying to work out the idea of constructing an $f$ that's ‘worse’ than any given modulus of continuity, but I couldn't make it go. $\endgroup$
    – LSpice
    May 4 at 2:59
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    $\begingroup$ Thank you for this! I’ve been stuck on this question for awhile too. $\endgroup$
    – Nate River
    May 4 at 5:18
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    $\begingroup$ Could you add some details justifying the statement that 'it suffices to show that if $\omega(\delta) = \epsilon$, then there exist $x,y \in [0,1]$ such that [...]'? Given the choice of quantifiers in the definition of domination, would you not instead have to prove something akin to for all $x \in [0,1]$ there is $y \in [0,1]$ with $\lvert x - y \rvert < \delta$ and $\lvert f(x) - f(y) \rvert > \epsilon$? $\endgroup$
    – Leo Moos
    May 4 at 11:44
  • $\begingroup$ @LeoMoos, the quantifier in the original question is over all $x \in [0, 1]$, so the negation should indeed be existential in $x$, right? $\endgroup$
    – LSpice
    May 4 at 13:15
  • $\begingroup$ For what it’s worth you could make an $\epsilon$ modification to the argument to get for all $x$. You let $x$ be given and then choose $y$ such that $g_k(y)$ is as far from $g_k(x)$ as possible within a single period. You’re guaranteed to get at least $2\times4^{-k}$. You would need to play with the constants a bit to make this work (certainly changing 4 to 10 should do the trick). $\endgroup$ May 4 at 16:52
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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 $\lvert f_n(x) - f_n(y)\rvert < \varepsilon$.

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    $\begingroup$ Hm, if you want $\omega$ to be continuously differentiable, is it always possible to have $\omega(0) = 0$? There are functions with Holder moduli of continuity if I’m not mistaken.. $\endgroup$
    – Nate River
    May 3 at 4:37
  • $\begingroup$ I am not an expert, but Wikipedia requires $\omega(0) = 0$, and claims that any modulus of continuity can be smoothed. $\endgroup$
    – LSpice
    May 3 at 5:08
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    $\begingroup$ Ah, yeah it should be smoothable, but then I think $\sup \omega’$ would be $\infty$ even after smoothing, since a function with a Holder modulus of continuity would have derivative of $\omega$ unbounded on $(0, \epsilon)$ for any $\epsilon$. (More precisely $\omega’ \to \infty$ near $0$). $\endgroup$
    – Nate River
    May 3 at 5:11
  • $\begingroup$ Ah, I think you’re right. I suspect that the argument can be adapted by using a more complicated $f$, but don’t immediately see how, so will just leave this partial answer for now. Essentially, I mean to construct a function that “does worse” than any specified modulus of continuity. $\endgroup$
    – LSpice
    May 3 at 5:17
  • $\begingroup$ (In other words, given $\omega$, construct $f$ so the optimal modulus of continuity of $f$ is bounded below by $\omega$.) $\endgroup$
    – LSpice
    May 3 at 5:24

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