0
$\begingroup$

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...

$\endgroup$
  • 1
    $\begingroup$ This question is very confused: try rephrasing it in a more precise way. $\endgroup$ – b00n heT Jan 21 at 15:57
  • $\begingroup$ @b00nheT......is it clear sir.. $\endgroup$ – learner Jan 21 at 16:04
0
$\begingroup$

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 :)

$\endgroup$
  • 1
    $\begingroup$ I got some realtion how we contradicts from this $\endgroup$ – learner Jan 21 at 16:13
0
$\begingroup$

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.

$\endgroup$

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.