# What is the derivatibe of 0

1. Jul 8, 2010

### Amok

I would have said it is 0, but then why is it that a twice derivable function is a function like, for example, f(x) = x. I've been studying maths for a while, but I had never asked myself this question until I came across the Heaviside function:

$$H(x) = 1 \ if\ x\geq 0\ and\ H(x)=0\ if\ x<0$$

The derivative of this function (in the distributional sense) is the Dirac delta function:

Let $$\varphi \in S$$, the Schwartz space then:

$$\langle T_{H}^{(1)},\varphi \rangle = - \int_{- \infty}^{+ \infty} H(x) \varphi '(x) dx = - \int_{0}^{+ \infty} \varphi '(x) dx = -[\varphi (x)]_{0}^{\infty} = \varphi (0) = \langle \delta, \varphi \rangle = \delta$$

And not,

$$\langle T_{H}^{(1)},\varphi \rangle = \int_{- \infty}^{+ \infty} H'(x) \varphi (x) dx = \int_{0}^{+ \infty} 0 \cdot \varphi (x) dx + \int_{- \infty}^{0} 0 \cdot \varphi (x) dx = 0$$

Which implies that H(x) is not derivable in the 'normal' sense (which implies 0 doesn't have a derivative).

EDIT: I just realized the Heaviside function is not differentiable at 0 (it is not even continuous)!

Last edited: Jul 8, 2010
2. Jul 8, 2010

### arildno

No, it doesn't imply that 0(x) doesn't have a derivative at all.
It implies that the Heaviside function doesn't have a derivative in the normal sense.

The dirac delta is the WEAK derivative of the Heaviside function, H'(x) in your second line is a meaningless expression for a function that was to be defined at ALL points.