# Weak derivative of $\operatorname{sgn}(x_1)$

Let $x\in \mathbb{R}^{n}, x = (x_1,\ldots,x_n)$, and $f(x) = \operatorname{sgn}(x_{1})$. Is $f$ weakly differentiable on $U = B(0,1)$, i.e. unit ball in $\mathbb{R}^{n}$, and what is the weak derivative?

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What have you tried? –  Davide Giraudo Oct 29 '12 at 10:46
What exactly do you mean by "weakly differentiable" here? There are several possible definitions in different contexts. Do you want the weak derivative to be a function, a distribution, ...? –  Nate Eldredge Oct 29 '12 at 12:40

We see that $sgn(x) = 2H(x) -1$ where $H$ is the Heaviside function with $H(0) = \frac{1}{2}$. Then distribution derivative of $sgn$ would be $2\delta_x$ which is not induced by any function. So $f$ is not weakly differentiable.