Weak derivative

Remarks

The weak derivative, when it exists, is unique up to a set of measure zero. If $u$ is of class $C^{\abs{\alpha}}$ on $U$, then its classical partial derivative $\partial^\alpha u$ is also its weak derivative, by integration by parts.

Example: on $U = (-1, 1)$, the function $u(x) = \abs{x}$ has weak derivative $v = \sgn x$. For $\varphi \in C_c^\infty(-1, 1)$, integration by parts on $(-1, 0)$ and on $(0, 1)$ gives $\int_{-1}^{1} \abs{x} \varphi'(x) \, dx = \int_{-1}^{0} \varphi \, dx - \int_{0}^{1} \varphi \, dx = -\int_{-1}^{1} \sgn(x) \varphi(x) \, dx$, since the boundary terms at $0$ vanish.

Non-example: on $(-1, 1)$, the function $\sgn x$ has no weak derivative. For any $\varphi \in C_c^\infty(-1, 1)$ one finds $\int \sgn(x) \varphi'(x) \, dx = -2 \varphi(0)$. A locally integrable $v$ with $\int v \varphi \, dx = 2 \varphi(0)$ for all $\varphi$ would vanish almost everywhere on $(-1, 0)$ and on $(0, 1)$, and then the integral would be $0$, not $2 \varphi(0)$. In the language of distributions, the derivative of $\sgn x$ is $2\delta$, which is not a function.