# Question about to Weak derivative of $|x|$

As I know that the function $$f(x)=|x|$$ is not differentiable.but in the weak sense it has weak derivative

my question is it again weak derivative exists for this function

I.e.,

suppose $$f_1$$ is weak derivative of $$f$$ then is it weak derivative exist for $$f_1$$

I got this relation for $$\int _U f_2(x) g(x)dx=2 g(0) \;\forall g\in C^{\infty}_c(R)$$ and where $$f_2$$ is weak derivative of $$f_1$$

how do we contradict from this

thank you...

• This question is very confused: try rephrasing it in a more precise way. – b00n heT Jan 21 at 15:57
• @b00nheT......is it clear sir.. – learner Jan 21 at 16:04

As you know for strong derivatives, not every function which is once differentiable is also twice differentiable. This also holds for weak derivatives. Your example is a good one, because $$f(x)=|x|$$ has one weak derivative, but not two.

Prove it along this line:

1. Try to find out what the first derivative of $$f$$ is. As a hint, in neighbourhoods where $$f$$ is strongly differentiable, its strong derivative coincides with the weak one. Let's assume you finished that exercise and call the weak derivative $$f_1$$.
2. Now prove that $$f_1$$ does not have a weak derivative. That is a tiny bit more work than the first point. Hint: Assume that $$f_1$$ had a weak derivative. Plug the formula for $$f_1$$ into the definition of a weak derivative (the first large equation appearing in the article https://en.wikipedia.org/wiki/Weak_derivative ). Do it right and you will get a contradiction.

I would suggest you try working these points and keep commenting if you are stuck proving any of them :)

• I got some realtion how we contradicts from this – learner Jan 21 at 16:13

Since the op hasn't shown his result I will lay out a solution:

Consider some open interval $$(-a,a)$$ for $$a \in (0, \infty]$$ and the function $$u(x) := | x |$$ and $$\nu(x) :=$$sgn$$(x)$$. Then $$\nu$$ is the weak derivative of $$u$$: For all test functions $$\phi \in \mathcal{C}_{\text{c}}^{\infty}((-a,a))$$ such that $$\phi(a) = \phi(-a) = 0$$ we have \begin{align*} \int_{-a}^{a} u(x) \phi'(t) \,dx & = \int_{0}^{a} x \phi'(x) \,dx - \int_{-a}^{0} x \phi'(x) \,dx \\ & = \underbrace{\big[-x \phi(x)\big]_{x = -a}^{0}}_{= 0} + \int_{-a}^{0} \phi(x) \,dx + \underbrace{\big[x \phi(x)\big]_{x = 0}^{a}}_{=0} - \int_{0}^{a} \phi(x) \,dx \\ & = - \int_{-a}^{a} \nu(x) \phi(x) \,dx. \end{align*} But, $$\nu$$ is not weakly differentiable, because for all such test functions the following equality for some $$\omega \in L^1_{\text{loc}}((-a,a))$$ must hold: \begin{align} \tag{1} \int_{-a}^{a} \nu(x) \phi'(x) \,dx = 2 \int_{0}^{a} \phi'(x) \,dx \overset{\textrm{FTOC}}{=} - 2 \phi(0) \overset{!}{=} \int_{-a}^{a} \omega(x) \phi(x) \,dx \end{align} Now, you can use the following

Lemma: Let $$u'$$ be the weak derivative of $$u$$ on $$(a,b)$$. Then for all intervals $$(\alpha, \beta) \subset (a,b)$$ it holds that $$u'|_{(\alpha, \beta)}$$ is also the weak derivative of $$u|_{(\alpha, \beta)}$$ on $$(\alpha, \beta)$$.

Proof. Let $$(\alpha, \beta) \subset (a,b)$$ and $$\phi \in \mathcal{C}_{\text{c}}^{\infty}(\alpha, \beta)$$ and define the trivial extension of $$\phi$$ by $$\tilde{\phi} \in \mathcal{C}_{\text{c}}^{\infty}(a,b)$$. Then, we conclude $$\begin{equation*} \int_{\alpha}^{\beta} u(x) \phi'(x) dx = \int_{a}^{b} u(x) \tilde{\phi}'(x) dx = - \int_{a}^{b} u'(x) \tilde{\phi}(x) dx = - \int_{\alpha}^{\beta} u'(x) \phi(x) dx, \end{equation*}$$ which implies the proposition.$$\ \square$$

Back to your question: This implies that the only candidate for the weak derivative of your function has to be $$\begin{equation*} \omega(x) = \begin{cases} 0, & \text{if } x \in (-a,0), \\ 0, & \text{if } x \in (0,a) \end{cases}, \end{equation*}$$ so $$\omega \equiv 0$$ almost everywhere. Because "the integral doesn't see null sets" (the set on which $$\omega \neq 0$$ can only be a null set), we have $$\int_{-a}^{a} \omega(x) \phi(x) \,dx = 0.$$ Now, we can choose a function $$\phi \in \mathcal{C}_{\text{c}}((-a,a))$$ so that $$\phi(-a) = \phi(a) = 0$$ and $$\phi(0) \neq 0$$. Then the equation (1) doesn't hold for all $$\phi$$ and, therefore, $$\nu$$ is not weakly differentiable.