Question about to Weak derivative of $|x|$

Question

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...

Answer 1

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 :)

Answer 2

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.