We'll say that a function on $[0,1]$ is uniformly left continuous if for every $\epsilon > 0$ there exists $\delta > 0$ such that $x \in (y - \delta, y)$ implies $|f(x) - f(y)| < \epsilon$ for every $x, y \in [0,1]$. I want to know if the space of all such functions is complete with respect to the uniform norm.

I'm interested in this space because I suspect that it is the completion of the space of piecewise constant functions which are continuous from the left.

Thanks in advance for the help!

EDIT: As was pointed out in the comments, what I wrote above is equivalent to uniform continuity. I think the following revised definition captures what I want. $f$ is uniformly left continuous if for every $\epsilon$ there is a partition $0 = t_0 < t_1 < \ldots < t_n = 1$ such that $|f(x) - f(y)| < \epsilon$ whenever $t_0 \leq x \leq y \leq t_1$ or $t_i < x \leq y \leq t_{i+1}$ for $i > 0$. So with this definition the characteristic function of $(1/2,1]$ is uniformly left continuous while the characteristic function of $[1/2,1]$ is not.

Here are the examples that I'm really trying to kill. Take $f$ to be the function which is 0 on the interval $[0,1/2]$ and let $f(x) = 1/(x-1/2)$ on $(1/2,1]$. This is left continuous, but it shouldn't be uniformly left continuous. Even if you insist that $f$ be bounded, you could set $f(x) = \sin(1/(x-1/2))$ for $x$ in $(1/2,1]$ and get a left continuous function which is not uniformly left continuous. If you can think of a better definition that captures this intuition, let me know.

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    $\begingroup$ Isn't that condition just the same as uniform continuity? $\endgroup$ – Chris Eagle Sep 6 '11 at 20:10
  • $\begingroup$ Isn't the condition "for every $\varepsilon>0$ and $y\in\left[0,1\right]$, there exists $\delta$ such that..."? $\endgroup$ – Davide Giraudo Sep 6 '11 at 20:49
  • $\begingroup$ @Chris Eagle: You're quite right... I edited to give a better definition. $\endgroup$ – Paul Siegel Sep 6 '11 at 21:55
  • $\begingroup$ @Davide Giraudo: I don't think this will be as easy to prove as completeness of $C[0,1]$. In the $C[0,1]$ case, you take a uniformly Cauchy sequence $f_n$, prove that its pointwise limit is continuous, and show that $f_n$ converges to its pointwise limit uniformly. Proving that the pointwise limit of a uniformly Cauchy sequence is continuous is much easier than proving that it's uniformly continuous, but the two statements are equivalent in $C[0,1]$. As my examples above show, this equivalence fails for left continuous functions (whatever uniformly left continuous actually means). $\endgroup$ – Paul Siegel Sep 6 '11 at 22:15

The standard $3\epsilon$ trick works. Suppose $f_n$ is a Cauchy (in uniform norm) sequence of uniformly left continuous functions on $[0,1]$, and $f$ its pointwise limit. Given $\epsilon > 0$, take $n$ large enough that $|f_n(x) - f(x)| < \epsilon/3$ everywhere. Let $0 = t_0 < \ldots < t_k = 1$ be a partition that satisfies your (revised) definition for the function $f_n$ with $\epsilon$ replaced by $\epsilon/3$. Then for $t_0 \le x \le y \le t_1$ or $t_i < x \le y \le t_{i+1}$, $|f(x) - f(y)| \le |f(x) - f_n(x)| + |f_n(x) - f_n(y)| + |f_n(y) - f(y)| < \epsilon$.

  • $\begingroup$ Quite right. I unfortunately didn't go back and try to prove completeness after I revised my definition - maybe I would have discovered this argument myself. Either way, thanks! $\endgroup$ – Paul Siegel Sep 7 '11 at 7:34

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