I am new to distributions and I want to make sure I understand how to do computations with them. So I was trying out this problem:
Problem: Let $u(x)=\vert x^2-1\vert$ be a function on $\mathbb R$. Find its distributional derivative $u''.$
My Attempt: Let $\psi\in\mathcal C_c^\infty(\mathbb R)$. By the definition of the derivative of a distributio and using integration by parts we have, \begin{align*} u'(\psi)&=-u(\psi')=-\int_{-\infty}^\infty\vert x^2-1\vert\psi'(x)\,dx\\ &=-\int_{-\infty}^{-1}(x^2-1)\psi'(x)\,dx+\int_{-1}^1(x^2-1)\psi'(x)\,dx-\int_1^\infty(x^2-1)\psi'(x)\,dx\\ &=\int_{-\infty}^{-1}2x\psi(x)\,dx-\int_{-1}^12x\psi(x)\,dx+\int_1^\infty 2x\psi(x)\,dx. \end{align*} Then using integration by parts again we get, \begin{align*} u''(\psi)&=-\int_{-\infty}^{-1}2x\psi'(x)\,dx+\int_{-1}^12x\psi'(x)\,dx-\int_1^\infty 2x\psi'(x)\,dx\\ &=2\left[2\psi(1)+2\psi(-1)+\int_{\vert x\vert>1}\psi(x)\,dx-\int_{-1}^1\psi(x)\,dx\right]\\ &=2\left[2\delta_1(\psi)+2\delta_{-1}(\psi)+\int_{\vert x\vert>1}\psi(x)\,dx-\int_{-1}^1\psi(x)\,dx\right]. \end{align*} We conclude that $$u''=4\delta_1+4\delta_{-1}+2\cdot\mathbf1_{\{\vert x\vert>1\}}-2\cdot\mathbf1_{[-1,1]}.$$
Do you agree with my computation above?
Any feedback is much appreciated. Thank you for your time.