Why is there no second weak derivative for $f(x) = \lvert x \rvert$? I know that we can find the first weak derivative for $f(x) = \lvert x \rvert$, which is $$w(x) = \begin{cases}-1 &x<0\\0&x=0\\1&x>0\end{cases}.$$But why is there no second weak derivative? If I try to find the first weak derivative of $w(x)$, I get that $$\int_{\mathbb{R}}v(x)\psi(x)\;dx = -\int_{\mathbb{R}}w(x)\psi'(x)\;dx = 2\psi(0).$$What does this mean? Why is there no $v$ such that $Dw = v$? I feel like I'm missing something and I'm not sure how to interpret this result.
Any help is appreciated!
 A: The notion of weak derivative is a subset of the notion of distribution derivative. There is indeed a distribution second derivative to this function and congratulations, you found it! It is $2\delta_0$ where $\delta_0$ is the Dirac mass which verifies
$$\left<\delta_0,\psi\right>=\psi(0).$$
You should have a look on some reference on distribution theory, which contains your weak derivative theory.
A: EDIT: As pointed out in the comments, this answer goes way beyond a simple answer to the question. It touches, without giving details of any sort, a deeper level of insight about what was asked. Some might find it interesting, some confusing.
It depends on your idea of what a derivaive is. Let's see.
-Classical derivatives
Ok, it's clear there cannot be a classical second derivative. Why? Well because in fact the function doesn't have a classical first derivative either! Well, you might say that it "almost" has one, and one can think of the derivative as a step function from  -1 to 1. And you would be right, this is one of the reasons why the weak derivatives where introduced, so that we're not limited by the bas behavior in a single point (i.e. the origin). In fact there's more than this, the weak derivatives are needed for the weak formulations of PDE problems, in which we don't require pointwise equalities but only integral ones , with respect to a specific set of test functions (usually $C^{\infty}_0$ functions)
-Weak derivatives
The whole point here is that you want the derivative of your function to be "something" you can describe and treat reasonably well with respect to integration. The choice here is to consider weak derivatives in a certain $L^p$ space. The reasons are multiple, but basically you have good compactness properties with Sobolev spaces and so you want to restrict to them. Also, in a large number of problems this is enough. In general the weak derivatives are defined to be $L^1 _{loc}$ functions such that the integration by parts hold for every test function $\phi \in C^{\infty} _0$, so for instance your $f$ has a weak derivative $g$ if
$$ \int f \phi' = - \int g \phi \text{ for every } \phi \in C^{\infty}_0$$
and you can verify that in your case $g$ is precisely the step function I was mentioning above.
-Distributional derivatives
This is something more abstract. You can define distributional derivatives of every function in, say, $L^1 _{loc}$. As many as you want. How do you do that? Simply define it like this:
$$ \langle f^{(n)} ,\phi \rangle= (-1)^{n} \int f \phi^{(n)}$$
so you basically discard every derivative on the more regular $\phi$. The distributional derivative lives, in general, in the dual space of $C^{\infty} _0$, so that it's a continuous functional with respect to the convergence in $C^{\infty} _0$. I don't want to enter into details (because I really might tell you something wrong) but you can imagine that you have uniform convergence for every derivative, and for the function itself of course. So there are many functionals which are continuous with respect to this convergence, i.e. the dual is really large.
Let's call $g$ the first weak derivative of $f$. The thing you asked is: is the distributional derivative of $g$ also a weak derivative? Namely: can I represent the linear functional in the dual space above as a function in some $L^p$? The answer in your case is no, because the ditributional derivative of $g$ is a Dirac mass, and there are no functions in $L^1 _{loc}$ that act like a Dirac mass on $C^{\infty}_0$ functions. You can prove this very easily, suppose that such a $v$ existed and prove some absurd. So you need to think of your derivative in a larger space, which in fact wil not be the whole dual of $C^{\infty}_0$, that's too much, but it will be $H^{-1}$. But this is another story.
So, to sum up: your functions does have a second distributional derivative. As all $L^1 _{loc}$ functions. But it does not have a weak derivative because you cannot represent that distributional derivative as a $L^{1}_{loc}$ function, but only as a distribution which in this case is a Dirac mass.
