10
$\begingroup$

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

$\endgroup$
6
$\begingroup$

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 $u$.

$\endgroup$
  • 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
4
$\begingroup$

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 $L^2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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