Distributional derivatives

Question

I need to compute derivatives as distributions of following functions:

$f(x) =$

1. $|x|$
2. $|x^2 - 1|$
3. $\mathrm{sgn}(x)$
4. $4$

Where $f : \mathbb{R} \to \mathbb{R}$.

ad 1) $|x|$ is continuous, so it belongs to $L^1_{loc}(\Omega) \Rightarrow$ its derivative exists.

I know, that

$$ \left< F', \phi \right> = - \left< F, \phi' \right> = -\int_\Omega f(x) \varphi'(x) dx $$

So now, I'm trying to compute the derivative of $|x|$:

\begin{align} - \left< F, \varphi' \right> &= - \int_\Omega |x| \varphi'(x) dx \\ &= - \int^0_{- \infty} -x \cdot \varphi'(x) dx - \int^{\infty}_0 x \cdot \varphi'(x) dx \\ &= \left[ x \cdot \varphi(x) \right]^0_{- \infty} - \int^0_{- \infty} 1 \cdot \varphi(x) dx - \left[ x \cdot \varphi(x) \right]^{+ \infty}_0 + \int^{+ \infty}_0 1 \cdot \varphi(x) dx \end{align}

And here I'm hopelessly lost. The result is supposed to be

$$ \int^{+ \infty}_{- \infty} g(x)\varphi(x) dx, $$

where

$$ g(x) = \begin{cases} 1, & x>0\\ -1, & x< 0 \end{cases}, $$

but I have literally no idea, how should I proceed. E.g.

$$ \left[ x \cdot \varphi(x) \right]^0_{- \infty} = 0 - \left( -\infty \cdot \varphi(-\infty) \right) $$

is supposed to be $0$, but still, I have no idea why...

So, could you, please, describe me, how to compute this derivative of a distribution and explain it a little?

---

Edit:

Ok, Thanks to the Zachary Selk's answer I understand why the result of square brackets is 0. Then we have just the sum of two remaining integrals:

\begin{align} - \int^0_{- \infty} 1 \cdot \varphi(x) dx + \int^{\infty}_{0} 1 \cdot \varphi(x) dx &= \int^0_{- \infty} -1 \cdot \varphi(x) dx + \int^{\infty}_{0} 1 \cdot \varphi(x) dx\\ &= \int^{\infty}_{- \infty} g(x) sgn(x) dx\\ g(x) &= \begin{cases} 1, & x > 0\\ -1, & x < 0 \end{cases} \end{align}

Ok, I'm able to understand this.

---

ad 2)

$$ |x^2 - 1| = \begin{cases} x^2 - 1, & x < -1\\ 1 - x^2, & x \in \left< -1;1 \right>\\ x^2 -1, & x > 1 \end{cases} $$

I can see, that when $x < -1$ or $x > 1$ the integral does not converge. So, the only one remaining to compute is

\begin{align} \int^{1}_{-1} (1-x^2) \varphi'(x) dx &= 2 \int^1_0 (1-x^2)\varphi'(x) dx\\ &= 2 \left( [(1-x^2)\varphi(x)]^1_0 - \int_0^1 -2x \cdot \varphi(x) dx \right)\\ &= -2 \varphi(0) +4 \int^1_0 x \varphi(x) dx \end{align}

And here I'm lost again, I think that only per partes will not help me with this integral.

Answer 1

Hint: Remember that $\varphi \in S(\Bbb{R})$ or $\varphi \in C_c^{\infty}(\Bbb{R})$. Either way $\lim\limits_{x\to\pm\infty} x\varphi(x)=0$