I'm looking at the following statement in my textbook:

let $u : (0, 1) → (0, 1)$ the devil's staircase function, aka Cantor-Lebesgue function. Then it's derivative is $u' = 0$ pointwise a.e.

  • We obtain $$\int_0^1 u \, \varphi_k' \to -1 \, \, (k \to \infty)$$ if we choose $(\varphi_k)_k \subset C^\infty_c((0,1))$ suitably with $\varphi_k → \chi_{(0,1)}$ $(k → ∞)$

  • the distributional derivative does not vanish, hence it can't have a weak derivative in $L^1_{\text{loc}}((0,1))$

I get why the the derivative is zero a.e., and why the weak derivative should be zero, if it existed

but I don't understand the reasoning with the $\varphi_k$!

  • Why does this converge to $-1$?
  • And what is "suitably"?

Help is much appreciated!

Edit: I made a mistake in my first version; it should be $\varphi_k'$ in the integral with $\varphi_k \to \chi_{(0,1)}$

Now it's updated correctly


It is hard to write down a precise analytical description of the $\varphi_k$ here, but the basic form they can take isn't too hard to describe. The problem is to find $\varphi_k$ with the property that $$\int_0^1 \varphi_k = 0 \quad \text{and} \quad \int_0^1 u \, \varphi_k' = -1$$for all $k$, and such that $\varphi_k$ converges to $\chi_{(0,1)}$.

Do the following:

  1. let $\varphi_k$ be identically $1$ on $(1/k,1-1/k)$. This takes care of the convergence.
  2. on $(1-1/k,1)$ give $\varphi_k'$ a dip to a large negative number and then a bounce back up to $0$
  3. define $\varphi_k$ on $(0,1/k)$ so that its integral vanishes on $(0,1)$.

Varying the size of the dip will change the value of the integral of $u \, \varphi_k'$ since this functions is weighted more heavily near $1$ than near $0$. The technical challenge is to make the dip just right so that $u \, \varphi_k'$ has integral exactly $-1$. It isn't hard to see this is possible.

Finally this leads to $$\int_0^1 u \, \varphi_k' = -1$$ for all $k$ giving you a nonvanishing distibutional derivative of $u$.

  • 1
    $\begingroup$ So that I understand correctly; We need the convergence $\varphi_k \to \chi_{(0,1)}$ in order to construct a contradiction, ie. We know that $u' = 0$ a.e. hence $\displaystyle \int_0^1 u' \, \mathrm d x =0$, but then $$0 = \int_0^1 u' \, \mathrm d x = \int_0^1 u' \, \chi_{(0,1)} \, \mathrm d x = \lim_{k\to \infty} \int_0^1 u' \, \varphi_k \, \mathrm d x = - \lim_{k \to\ \infty} \int_0^1 u \, \varphi'_k \, \mathrm d x = 1 $$ $\endgroup$ – cesare borgia Jul 6 '17 at 11:38
  • $\begingroup$ i agree, makes sense $\endgroup$ – augustin souchy Jul 6 '17 at 11:48
  • $\begingroup$ im sorry, why does the integral over phi has to vanish (for all k)? $\endgroup$ – cesare borgia Aug 21 '17 at 20:46

This question (and @Umberto P.'s good answer) nicely illustrates how distributions/generalized functions explain that the seemingly paradoxical features of the Cantor-Lebesgue staircase function are not paradoxical after all.

Another approach, is via Sobolev spaces and Sobolev imbedding. That is, for a distribution $u$ such that $u\in L^2[0,1]$ and its distributional derivative $u'$ is also in $L^2[0,1]$, we show not only the Sobolev imbedding $u\in C^{0,{1\over 2}}$ (with Lipschitz index ${1\over 2}$), but that, incidentally, that such functions $u$ do satisfy the fundamental theorem of calculus (which the Cantor-Lebesgue function is designed to fail). Thus, for example, the Cantor-Lebesgue function, while in $L^2$, evidently does not have distributional derivative in $L^2$.


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