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I have changed the title (replaced "well-behaved" by "càdlàg"), since it seems that "a well-behaved function" might be interpreted as "a function of bounded variation" (rather than "a càdlàg function", which I actually meant).

Let $f:[0,1] \to {\bf R}$ have a following property: $f$ is continuous except at the points $x_{k,n} = \frac{{2k - 1}}{{2^n }}$, $k=1,\ldots,2^{n-1}$, $n=1,2,3,\ldots$, where $\lim _{x \downarrow x_{k,n} } f(x) = f(x_{k,n} )$ but $f(x_{k,n} ) - \lim _{x \uparrow x_{k,n} } f(x) = a_n > 0$. Can such a function exist if $a_n > 1/ \log(n)$ for all sufficiently large $n$?

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  • $\begingroup$ Thank you for this comment; I just forgot to define $x_{k,n}$... Will be defined shortly. $\endgroup$
    – Shai Covo
    Commented Nov 14, 2010 at 16:09
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    $\begingroup$ What does "well-behaved" require? Is there more you want to specify about the locations of the jump discontinuities? Presumably points x<sub>k,n</sub> are in [0,1], but do you intend their ordering to be related to the subscripts? We know for each sufficiently large n that the total variation at points x<sub>k,n</sub> will exceed 2<sup>n-1</sup>/log(n), but the jump increases there might be "cancelled" by decreases in the intervening continuous intervals to produce a bounded function. However evidently we cannot have bounded total variation since the sum 2<sup>n-1</sup>/log(n) diverges. $\endgroup$
    – hardmath
    Commented Nov 14, 2010 at 16:10
  • $\begingroup$ @hardmath: the locations of the jump discontinuities are now specified. $\endgroup$
    – Shai Covo
    Commented Nov 14, 2010 at 16:18
  • $\begingroup$ Since they are specified as overlapping, let me revise the argument I gave for why the function cannot be of bounded total variation. For each n = 1,2,3,... there is at least one "binary rational" (2k - 1)/2^n in (0,1) not previously included (by a lower value of n). So the jump discontinuities must contribute a total of at least sum 1/log(n), which diverges. $\endgroup$
    – hardmath
    Commented Nov 14, 2010 at 17:17
  • $\begingroup$ hardmath, doesn't that argument also show that $f$ can't be continuous at $0$? If we take an $\epsilon$-interval around $f(0)$, it will fail to contain those arbitrarily large function values on the binary rationals, so there can't be a $\delta$-interval around $0$ that gets mapped into the $\epsilon$-interval. $\endgroup$
    – kahen
    Commented Nov 14, 2010 at 17:24

1 Answer 1

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If $f\colon[0,1]\to\mathbb{R}$ is cadlag (continu à droite, limites à gauche) then we can ask what the possible jumps are. That is, for what functions $g\colon(0,1]\to\mathbb{R}$ is there a cadlag function $f$ with $g(x)=\Delta f(x)\equiv f(x)-f(x-)$?

The answer is that $g$ occurs as the jumps of a cadlag function if and only if the set $\{x\in(0,1]\colon\vert g(x)\vert > \epsilon\}$ is finite for each $\epsilon > 0$. In particular, there is a cadlag function with jumps as you describe.

First, the necessity: If $S=\{x\in(0,1]\colon\vert g(x)\vert > \epsilon\}$ was not finite, then it would contain a strictly increasing or strictly decreasing sequence $x_n$. Then, $\vert f(x_n)-f(y_n)\vert > \epsilon$ for some $y_n$ chosen arbitrarily close to $x_n$. Replacing $x_n$ by $y_n$ where necessary gives a strictly increasing or strictly decreasing sequence $x_n\in[0,1]$ such that $\vert f(x_{n+1})-f(x_n)\vert > \epsilon/2$. However, as f is cadlag and has left and right limits everywhere, this contradicts the requirement that $f(x_n)$ tends to a limit.

Now, we can show sufficiency: Set $\epsilon_n=2^{-n}$ for $n\ge1$ and $\epsilon_0=\infty$. For each $n\ge1$, consider the finite set $S_n=\{x\in(0,1]\colon \epsilon_{n-1}\ge\vert g(x)\vert > \epsilon_n\}$. We can construct a function $f_n\colon[0,1]\to\mathbb{R}$ such that $\Delta f_n(x)=1_{\{x\in S_n\}}g(x)$ and $\vert f_n(x)\vert\le\epsilon_{n-1}$. The idea is to take $f_n(x)=g(x)$ and $f_n(x-)=0$ for points in $S_n$, and linearly interpolate between these. More precisely, if $S_n$ is empty we set $f_n=0$. Otherwise, $$ f_n(x)= g(a)(b-x)/(b-a) $$ for $a\le x < b$. Here, $a < b$ are consecutive points of $S_n$. Also set $f_n(x)=0$ for $x$ less than the minimum of $S_n$ and $f_n(x)=g(c)$ for $x\ge c=\max S_n$. This has the required jumps $\Delta f_n(x)=1_{\{x\in S_n\}}g(x)$.

Finally set $f(x)=\sum_{n=1}^\infty f_n(x)$. As $\vert f_n\vert \le 2^{1-n}$ for $n\ge 2$ this converges uniformly and, $$ \Delta f(x)=\sum_n\Delta f_n(x)=\sum_n1_{\{x\in S_n\}}g(x)=g(x). $$

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  • $\begingroup$ Thanks George, I will go over your answer tomorrow or so. $\endgroup$
    – Shai Covo
    Commented Nov 15, 2010 at 0:59
  • $\begingroup$ You might wonder why I considered a sequence $a_n > 1/ \log(n)$. The reason is that for the case $a_n > 1/ n^r$, $r > 0$, I found a very simple way to define an explicit $f$ (though, in a binary setting); the $a_n > 1/ \log(n)$ case might be essentially different in this respect. $\endgroup$
    – Shai Covo
    Commented Nov 16, 2010 at 1:00

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