# Prove that $f$ does not have a weak derivative

Consider a function $f:\mathbb{R} \rightarrow [0,1 ]$ defined by:

$\begin{equation*} f(x)=\left\{ \begin{array}{rl}0 & \text{if } x\leq 0,\\ 1 & \text{if } x\geq 1, \\ 1/2 & \text{if } x\in [1/3,2/3] \\ 1/4 & \text{if } x\in [1/9,2/9] \\3/4 & \text{if } x\in [7/9,8/9] \\ \vdots \end{array}\right. \end{equation*}$

I tried to prove this by contradiction. When i tried to integrate i am getting the answer in terms of values of $\phi$ in cantor set where $\phi$ is any measurable $C^{\infty}$ function. I am unable to go further.

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To begin with, suppose $f$ has a weak derivative, say $f'=g$. What properties would $g$ have? For example, what is $g$ on the interval $(1/3,2/3)$? – user53153 Feb 8 '13 at 20:59
Perhaps some clarification about the context of "weak derivative" would be good, since it evidently cannot mean "distributional derivative", which this function will have (an infinite linear combination of Dirac deltas at points $k/3^n$). – paul garrett Feb 19 '13 at 13:24

I think weak derivative of $f$ means $g \in L^1_{loc}(\mathbb{R})$ such that the distribution derivative of f is induced by $g$ i.e. for all test functions $\phi \in C^\infty_c(\mathbb{R})$ we have $$\int_\mathbb{R} g\phi\ dx = -\int_\mathbb{R} f\phi' dx$$ Now if we assume there exists a weak derivative $g$ then clearly $g(x) = f'(x)$ almost everywhere. But if $f$ is the Cantor function then $f'(x) = 0\ a.e$. Hence $g(x) = 0\ a.e$. Now for any $\alpha > 1$ we choose $\phi \in C^\infty_c(\mathbb{R})$ such that $\phi(x) = 1$ for all $x \in [0,1]$ and $\phi(x) = 0$ for all $x \geq \alpha$ Then we find that $\int_\mathbb{R} g(x)\phi(x) dx = 0$ as $g(x) = 0\ a.e$ but $$-\int_\mathbb{R} f(x)\phi'(x) dx = -\int^\alpha_0 f(x)\phi'(x) dx = -\int^\alpha_1\phi'(x) dx = \phi(1)-\phi(\alpha) = 1 \neq 0$$ Hence we have a contradiction.