How to calculate the distributionderivatives of abs(x)?

Question

Lets say we have $f(x) = |x|$.

I want to calculate $f'$ and $f''$, how would I go about this?

I understand that this is not defined at $x = 0$, so it will have to be done in two steps.

Answer 1

I think that for a thorough understanding of distributions, it is necessary at the beginning (and sometimes a long time after, when faced to tricky situations) to stick to the definitions, in the following way.

Let $T=[f]$ denote the distribution associated with locally integrable function $f$. Then for any $\varphi$ in the set of infinitely derivable functions with compact support:

$\langle T',\varphi\rangle=-\langle T,\varphi'\rangle =\int_{-\infty}^{+\infty} |x| \varphi'(x)dx=\int_{-\infty}^0(-x)\varphi'(x)dx+\int_0^{\infty}x\varphi'(x)dx$

Then integrate by parts,

$-\int_{-\infty}^0\varphi(x)dx+\int_0^{\infty}\varphi'(x)dx=<-1 \ on \ \mathbb{R_{-}} ,\varphi>+<+1 \ on \ \mathbb{R_{+}} ,\varphi>$

giving:

$$< T', \varphi> \ = \ <-1 \ on \ \mathbb{R_{-}} +1 \ on \ \mathbb{R_{+}} ,\varphi>$$

i.e.,

$$ T'=-1 \ on \ \mathbb{R_{-}} +1 \ on \ \mathbb{R_{+}}$$

Otherwise said: $[f]'(x)=2H(x)-1$ where $H$ is the Heaviside step function.

Thus, more directly, $f''(x)=2\delta \ \ (1)$ because of the jump of intensity +2 in $0$.

(see math.mit.edu/~stevenj/18.303/delta-notes.pdf).

A good exercise would be to establish (1) using definitions like done for the first derivative.

Answer 2

To compute the second derivative, you have to express $$ \int_{\Bbb R}|x|·φ''(x)dx $$ in terms of the values of any test function $φ\in C_c^\infty(\Bbb R)$. Indeed this is done via partial integration on the half-axes. $$ \int_0^∞ x·φ''(x)dx = [x·φ']_0^∞-\int_0^∞ φ'(x) dx=φ(0), $$ and $$ \int_{-∞}^0 (-x)·φ''(x)dx = -[x·φ']_{-∞}^0+\int_{-∞}^0 φ'(x) dx=φ(0). $$

Answer 3

The derivative of $f$ is a distribution defined, for any test function $\phi$, by $\langle f',\phi\rangle=-\langle f,\phi'\rangle$.

$-<f,\phi'>=-\int_{\mathbb{R}}|x|\phi'(x)dx=-\left(\int_{-\infty}^{0}-x\phi'(x)dx+\int_0^{+\infty}x\phi'(x)dx\right)$

Integrating by parts, we get:

$\left[x\phi(x)\right]_{-\infty}^0-\int_{-\infty}^0\phi(x)dx-\left[x\phi(x)\right]_{0}^{+\infty}+\int_0^{+\infty}\phi(x)dx$

As $\phi$'s support is compact, we obtain $<H(x),\phi>-<H(-x),\phi>$ where $H$ is the heaviside distribution:

$$<H,\phi>=\int_0^{+\infty}\phi(x)dx$$

Then, knowing that the derivative of $H$ is $\delta$, you can compute $f''$.