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:
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 $u$.