A function $u: \Bbb R \longrightarrow \Bbb R$ is weakly differentiable with weak derivative $v$ if there exists a function $v: \Bbb R \longrightarrow \Bbb R$ such that
$$\int_{-\infty}^\infty u \phi' ~dx = – \int_{-\infty}^\infty v \phi ~dx$$
for all smooth functions $\phi: \Bbb R \longrightarrow \Bbb R$ that vanish outside some bounded set.
Functions which are not differentiable can still be weakly differentiable. For example,
$$u(x) = |x|$$
is not differentiable because of the corner at $x = 0$, but it does have weak derivative
$$v(x) = \begin{cases} -1, & x < 0, \\ 0, & x = 0, \\ 1, & x > 0, \end{cases}$$
My question is does the weak derivative of the absolute value have a weak derivative itself? In other words does absolute value function have a second week derivative?
Best Answer
No, the function $$v(x) = \begin{cases} -1, & x < 0, \\ 0, & x = 0, \\ 1, & x > 0, \end{cases}$$ is not weakly differentiable. Indeed, we have
$$- \int v(x) \phi'(x) dx = 2 \phi(0)$$ for all smooth $\phi$ with compact support. Note that there is no functions $w$ so that $$\int w(x) \phi(x) dx = 2 \phi(0)$$ for all $\phi\in C^\infty _0(\mathbb R)$.
However, for any integrable function $v$, one can define a "distributional derivative" $Dv$, which is a linear functional $Dv : C^\infty_0(\mathbb R) \to \mathbb R$ given by $$Dv( \phi ) = - \int v(x) \phi'(x) dx.$$ So in our case, $Dv(\phi) = 2\phi(0)$ is a bounded linear functional on $C_c(\mathbb R)$, which corresponds to (twice) the Dirac delta measure at $0$. Note that a function is weakly differentiable if and only if the distributional derivative can be represented by a integrable function.
Lastly, if $v$ is weakly differentiable, then it is absolutely continuous and in particular it is a continuous function. This is another way to see that your $v$ is not weakly differentiable.