I need to compute derivatives as distributions of following functions:
$f(x) =$
- $|x|$
- $|x^2 – 1|$
- $\mathrm{sgn}(x)$
- $4$
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.
Best Answer
Hint: Remember that $\varphi \in S(\Bbb{R})$ or $\varphi \in C_c^{\infty}(\Bbb{R})$. Either way $\lim\limits_{x\to\pm\infty} x\varphi(x)=0$