# I Heavyside step function chain rule

Tags:
1. Mar 21, 2017

### Black_Hole_93

Hi,

I have a probably very stupid question:

Suppose that there is an expression of the form $$\frac{d}{da}ln(f(ax))$$ with domain in the positive reals and real parameter a. Now subtract a fraction $\alpha$ of f(ax) in an interval within the interval $[ x_1, x_2 ]$, i.e.

$$f(ax) \rightarrow f(ax) \left[\theta(a(x-x_1)) \theta(a(x_2-x))+ (1-\alpha) \theta(a(x_1-x)) \theta(a(x-x_2)) \right]$$

Now in the case that $a>0$ one can use $\theta(ax) = \theta(a)\theta(x)$ and after doing a case analysis for $x \in [x_1,x_2]$ and $x\not\in [x_1,x_2]$ above expression becomes

$$\frac{d}{da} ln(f(ax)) \rightarrow \frac{d}{da}ln(f(ax)) + \frac{d}{da} ln(\theta(a)) + \frac{d}{da} ln(\theta(a)) \theta(x-x_1) \theta(x_2-x)$$

I have two questions:
i) Are the steps above legitimate, as one is dealing with functionals?
It does not seem right to me that the result does not depend on $\alpha$...
ii) How can one evaluate $\frac{d}{da} ln(\theta(a))$ ? Is the typical chain rule applicable there (which would yield 0 for a > 0)?

Thank you and best greetings :)

Last edited: Mar 21, 2017
2. Mar 21, 2017

### andrewkirk

The derivative does not depend on $\alpha$ because the expression does not depend on it.

The expression written above gives $\log f(ax)$ for $x\in[x_1,x_2]$ and 0 elsewhere.

That's because the product of Heaviside functions in the second term is always zero, since it is the indicator function of the set
$$\{x\ :\ x<x_1\wedge x>x_2\}$$
which is the empty set, since $x_1\leq x_2$.

Have another go at writing the coefficient with Heaviside functions so that it gives $(1-\alpha)$ for $x\in[x_1,x_2]$ and 1 elsewhere.

3. Mar 22, 2017

### Black_Hole_93

$$f(ax) \rightarrow f(ax) \left{(1-\alpha) [\theta(a(x-x_1)) \theta(a(x_2-x))]+ \theta(a(x_1-x)) + \theta(a(x-x_2)) \right}$$