MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $0<\alpha<1$?


share|cite|improve this question
up vote 2 down vote accepted

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 $k|t|^\alpha$.

share|cite|improve this answer
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. – Euplio M. May 25 '12 at 20:48
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). – Euplio M. May 25 '12 at 20:58
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 $u$. – Pietro Majer May 25 '12 at 21:13
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? – Euplio M. May 25 '12 at 21:32
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 $x,y \in I$. – Euplio M. May 26 '12 at 15:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.