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Consider a piecewise constant function $v: [a,b] \rightarrow \mathbb{R}$ defined by a finite partition $a=t_0 < t_1 < t_2 < ... < t_s=b$ of the interval $[a,b]$, and constants $m_1,m_2,...,m_s$ with $v(t)=m_i$ whenever $t \in I_i := [t_{i-1},t_i)$ for $i=1,\dots,s$. Let $\mathcal{P}$ denote the collection of intervals $I_i$ that make up the partition of [a,b].

Now, let $u: [a,b] \rightarrow \mathbb{R}$ be a regulated function (that is, $u$ is the uniform limit of a sequence of piecewise constant functions $u_n: [a,b] \rightarrow \mathbb{R}$) each $u_n$ of which is defined via a corresponding partition $\mathcal{P}_n$ of $[a,b]$ as described above.

Does anyone know of a condition that characterizes when such a function $u$ is $\alpha$-Hölder continuous for some $0<\alpha<1$?

E(up)lio.

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I can't think of a characterization which is not too close to a tautology; a sufficient condition is the following. Denote the modulus of the subdivision $\mathcal{P}$ by $\|\mathcal{P}\|:=\max _ {1\le i\le s} (t _ i-t _ {i-1})$, and by $\mathcal{P}^M$ the set of mid-points of the intervals $I\in \mathcal{P}$.

Assume that

1. $\|\mathcal{P _ n}\|\to0\, ;$

2. ${u _ n} _{|\mathcal{P _ n} }$ are uniformly $\alpha$-Hölder, that is there is $k\ge0$ such that $|u_n(t) - u _ n(s)|\le k|t-s|^\alpha$ holds for any $n\in\mathbb{N}$ and for any $t,s\in\mathcal{P} _ n^M\, .$

Reason: if $\tilde u _ n$ denotes the piece-wise interpolation of the nodes $\mathcal{P} _ n^M$, then by concavity $\tilde u _ n$ has modulus of continuity $k|t|^\alpha$ on $[a,b]$, and $\| u _ n - \tilde u _ n\| _ \infty\le k\|\mathcal{P} _ n\|^\alpha=o(1)$ as $n\to\infty$. Therefore $u$ has the same modulus of continuity $k|t|^\alpha$.

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  • $\begingroup$ First, notice that I edited the question slightly so that the left endpoints are included. Thus, ,instead of $\mathcal{P}_n^0$, let us denote $\mathcal{P}_n^E$ to denote the endpoints (not the midpoints) of the partition intervals. $\endgroup$
    – Euplio M.
    May 25, 2012 at 20:48
  • $\begingroup$ I wonder if changing 2. in the answer above to a similar statement denoted by 2.1 (and stated below) would be enough (under possibly some other conditions on $u$, but only if they are necessary) to allow us to conclude the same result. 2.1 There exists a $k\geq0$ and an $0<\alpha<1$ such that $\abs{u(t)-u(s)} \leq k\abs{t-s}^{\alpha}$ for any $n\in \mathbb{N}$ and for any t,s \in $\mathcal{P}_n^E$. That is, can we replace 2. with a similar statement as 2. but stated in terms of the limit function $u$ instead of for the sequence $u_n$ (that tends to $u$ in the uniform limit). $\endgroup$
    – Euplio M.
    May 25, 2012 at 20:58
  • $\begingroup$ Isn't this the same that saying that the restriction of $u$ to the dense set $D:=\cup_n\mathcal{P}_n^E$ is Hoelder? Then, if $u$ is also continuous, it is certainly Hoelder. So this would somehow make less relevant the role of the sequence $u_n$, that should only satisfy a suitable condition that ensures the continuity of $u$. $\endgroup$
    – Pietro Majer
    May 25, 2012 at 21:13
  • $\begingroup$ Thank you for your comments. Again, I really appreciate having someone with whom to discuss this problem with. In my situation, I do in fact know that $u$ is continuous, so this is a very helpful remark. Do you happen to know of any good books in which I can read about all of the different properties of Holder continuity, like some of the properties that you have stated here? $\endgroup$
    – Euplio M.
    May 25, 2012 at 21:32
  • $\begingroup$ Suppose that we know that $u$ is continuous on $[a,b]$. Moreover, suppose that for every $x \in [a,b)$, we know that there exists a corresponding sequence $x_n \rightarrow x$ approaching $x$ from the right side such that: There exists a $C>0$ and an $0< \alpha <1$, both independent of $x$, such that $\abs{u(x)-u(x_n)} \leq C \abs{x-x_n}^{\alpha}$. I will work to show that $u$ is $C$-uniformly $\alpha$-Holder continuous on $I$. That is, there exists a $C>0$ and $0<\alpha<1$ such that $\abs{u(x)-u(y)} \leq C \abs{x-y}^{\alpha}$ for all $x,y \in I$. $\endgroup$
    – Euplio M.
    May 26, 2012 at 15:03

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