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

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.

• 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. May 4 at 2:59
• Thank you for this! I’ve been stuck on this question for awhile too. May 4 at 5:18
• 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 $\lvert f(x) - f(y) \rvert > \epsilon$? May 4 at 11:44
• @LeoMoos, the quantifier in the original question is over all $x \in [0, 1]$, so the negation should indeed be existential in $x$, right? May 4 at 13:15
• 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). May 4 at 16:52

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 $$\lvert f_n(x) - f_n(y)\rvert < \varepsilon$$.

• 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.. May 3 at 4:37
• I am not an expert, but Wikipedia requires $\omega(0) = 0$, and claims that any modulus of continuity can be smoothed. May 3 at 5:08
• 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 $0$). May 3 at 5:11
• 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. May 3 at 5:17
• (In other words, given $\omega$, construct $f$ so the optimal modulus of continuity of $f$ is bounded below by $\omega$.) May 3 at 5:24