# Derivative of step function is Dirac delta function

The derivative of a locally integrable function $f$ is by definition the linear functional

$$\phi \mapsto -\int_\mathbb{R} \phi'(x)f(x) dx$$

Using this definition, why is it that the derivative of the step function is the dirac delta function?

Since the derivative is a linear functional, I don't understand why we should get a function like the dirac delta. Linear functionals and functions are not the same...

• Hint: $\int \phi^{'}(x) f(x) dx = \int_0^\infty \phi^{'}(x)dx = \dots$. Remember $\phi$ is drawn from $C_c^\infty (\mathbb{R})$. Commented Dec 4, 2013 at 4:47
• @ChrisJanjigian I'm confused even with the definition. Please see my edit. Commented Dec 4, 2013 at 4:51
• @ChrisJanjigian But following your hint, I get $\phi(0)$. How does that imply the derivative is the Dirac delta? Commented Dec 4, 2013 at 4:53
• Just a quick comment. The Dirac delta is emphatically not a function. If any function is zero everywhere except a point, then its integral is zero, so there is no hope of finding a function which behaves like a delta. It only makes sense as a distribution or a measure, both of which are defined as linear functionals on an appropriate space. Commented Dec 4, 2013 at 15:41

For simplicity let's assume the step function $f$ is $f = \chi_{[0,\infty)}$. Let $\phi$ be a test function, then
$$(Df)(\phi) = -\int \phi'(x) f(x) dx = -\int_0^\infty \phi'(x) dx = - \phi(C) + \phi(0) = \phi(0)$$
where $C$ is any large number such that $\phi(x) = 0$ for all $x\geq C$. Thus $Df$ is the delta function at $0$.