Let $A$ be the space of almost everywhere differentiable functions $[0,1]\rightarrow [0,1]$, and when differentiable, their derivatives are bounded by $M$.

I'm aware that the space of almost everywhere differentiable function is not complete under uniform norm, as we can construct a sequence that converges into Weierstrass function or the Blancmange curve. However under this construction the derivative explodes, thus I'm wondering whether the additional condition of bounded derivative will make the space complete.

I tried to prove it by definition (using Cauchy sequence) yet I'm fuzzy about "almost everywhere differentible" part.

Thank you!

  • $\begingroup$ Bounty? Detailed canonical answer required to address all concerns? What part of the question is not answered by my (revised) answer? $\endgroup$ – David C. Ullrich Jun 30 '15 at 17:26
  • $\begingroup$ Not to say this twice, trying to figure out why my "at" isn't working. But anyway: Seems to me I answered the question. What's missing? $\endgroup$ – David C. Ullrich Jun 30 '15 at 17:57
  • $\begingroup$ Ah. See latest edit - a newer and more improved new and improved version. $\endgroup$ – David C. Ullrich Jun 30 '15 at 18:54

Three things: An answer to a somewhat different question, that seems more "important" in analysis, a "no" to the original question, and then a result that gives a much stronger version of "no" to the original question.

At first I assumed that the OP intended the functions in $A$ to be continuous, and in fact to be the integral of their derivative. The point being that just saying $f$ is almost everywhere differentiable and $|f'|\le M$ almost everywhere doesn't really say as much as one might think about $f$, for example $\chi_{[0,1/2]} satisfies those conditions.

So first I'm going to address the question with a modified version of $A$, saying that $f$ is in $A$ if $f:[0,1]\to[0,1]$, $f$ is almost everywhere differentiable, $|f'(t)|\le M$, and $f$ is the integral of its derivative: $f(x)-f(0)=\int_0^xf'(t)\,dt.$ (One of various reasons I suspect that's what you meant is that people use similarly loose language all the time, for example when they say a certain Sobolev space is defined by $f,f'\in L^2$.)

In any case, the answer is yes, for my revised version of $A$. This is easy from the following:

Lemma. $f\in A$ if and only if $f:[0,1]\to[0,1]$ and $|f(x)-f(y)|\le M|x-y|$ for all $x,y\in[0,1]$.

Proof: That $f$ in $A$ implies that inequality is clear since $f(x)-f(y)=\int_x^yf'$. The other direction uses a little real analysis: Suppose $f$ satisfies the inequality. Then $f$ is absolutely continuous. Hence $f$ is differentiable almost everywhere and $f$ is the integral of its derivative; this is a big theorem about absolute continuity. And the inequality certainly implies that the derivative is no larger than $M$. QED.

Now completeness should be clear; a uniform or even just pointwise limit of functions satisfying that inequality satisfies the inequality.

Second thing: The OP is still interested in the question with $A$ exactly as defined. The answer is no. In fact if $f:[0,1]\to[0,1]$ is continuous there exist $f_n\in A$ with $f_n\to f$ uniformly:

For $n=1,2,\dots$ and $0\le j<n$ let $I_{n,j}=[j/n,(j+1)/n).$ Define $f_n$ by $$f_n(t)=f(j/n)\quad(t\in I_{n,j})$$and $f_n(1)=f(1)$. Then $f'=0$ almost everywhere so $f_n\in A$. And the fact that $f$ is uniformly continuous shows that $f_n\to f$ uniformly. (Given $\epsilon>0$ choose $\delta>0$ such that etc. Now if $1/n < \delta$ then $|f_n(t)-f(t)|<\epsilon$ for all $t\in[0,1]$.)

Third thing: a new and improved "no". If $f$ is continuous then $f_n\to f$ uniformly, where $f_n$ is continuous and $f_n'=0$ almost everywhere. (We got this in the previous section except $f_n$ was not continuous; continuity makes it more suprising.)

Let $\epsilon>0$. Since $f$ is uniformly continuous we can find $0=x_0<x_1<\dots<x_n=1$ with the following property: If $g$ is any function such that for every $t\in[x_j,x_{j+1}]$ the number $g(t)$ lies in the closed interval with endpoints $f(x_j)$, $f(x_{j+1})$ then $|f(t)-g(t)|<\epsilon$ for all $t$.

So we need only show that there exists a continuous $g$ with $g'=0$ almost everywhere, satisfying the property above.

Let $C$ be the Cantor-Lebesgue function https://en.wikipedia.org/wiki/Cantor_function . In particular $C:[0,1]\to[0,1]$, $C$ is continuous, $C'=0$ almost everywhere, and $C(0)=0$, $C(1)=1$. Now let $g$ be a function such that for each $j$ the restriction of $g$ to $[x_j,x_{j+1}]$ is a rescaled, shifted, possibly vertically flipped version of $C$, so that $g(x_j)=f(x_j)$ and $g(x_{j+1})=f(x_{j+1})$. (I suspect that's easier to follow than a more precise version, which is available on request.)

  • $\begingroup$ Thank you for your response! If I understand it correctly, if I add that $f$ is continuous, does that coincide your revised definition of $A$? $\endgroup$ – iridium Jun 22 '15 at 17:03
  • 1
    $\begingroup$ No, although a counterexample is not as simple. If $f$ is in what I think $A$ "should" be and $f'=0$ almost everywhere then $A$ is constant, right? But there exist non-constant continuous functions $f$ with $f'=0$ almost everywhere! See for example en.wikipedia.org/wiki/Cantor_function . $\endgroup$ – David C. Ullrich Jun 22 '15 at 17:15
  • $\begingroup$ Thanks again! Your answer is quite helpful, yet I'm still interested in the conclusion about my initial specification of $A$. $\endgroup$ – iridium Jun 22 '15 at 17:37
  • $\begingroup$ Hmm... Ah. The answer is no for $A$ exactly as you defined it. Stay tuned. $\endgroup$ – David C. Ullrich Jun 22 '15 at 17:48
  • $\begingroup$ Just take the Weierstrass function as $f$, which is non-differentiable everywhere. $\endgroup$ – user251257 Jun 30 '15 at 15:13

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