How can you show that $f\delta′=f(0)\delta′−f′(0)\delta$ for a function f that is infinitely differentiable?

Question

Assume that $f$ is infinitely differentiable. Let $\delta$ be the (Dirac) delta functional.

I know that $f\delta = f(0)\delta$, but I'm not sure how to derive the equation $f\delta′=f(0)\delta′−f′(0)\delta$.

Answer 1

I'm assuming you mean the delta function(al) $\delta(x)$.

If so, you should use formal integration by parts. $$\int g\times (f\delta') =\int -(fg)'\delta = \int -(f'(0)g(x)+g'(x)f(0))\delta = -f'(0) \int g\delta-f(0)\int g'\delta$$

Integrating the last term by parts and stripping off the $\int g\times$ gives the result.

Answer 2

You're missing an $f$ in the left hand side.

Compute $(f\delta)'$ in two different ways.

On one hand $(f\delta)' = f'\delta + f\delta' = f'(0)\delta + f\delta'$.

On the other hand $(f\delta)' = (f(0)\delta)' = f(0)\delta'$.

Hence $f\delta' = f(0)\delta' - f'(0)\delta$.

Answer 3

(I suppose that with $\delta$ you mean the Dirac function.) Just evaluate $(f(t)\delta(t))'$ in two ways.

First, use Leibniz' rule: $$ (f(t)\delta(t))'=f(t)\delta'(t)+f'(t)\delta(t)=f(t)\delta'(t)+f'(0)\delta(t). $$ We also have $$ (f(t)\delta(t))'=(f(0)\delta(t))'=f(0)\delta'(t). $$

Since the two expressions must be equal, we have $$ f(t)\delta'(t)=f(0)\delta'(t)-f'(0)\delta(t). $$

Answer 4

you can also expand $f$ into Taylor series

$f(t)=f(0)+f'(0)t + \frac{f''(0)}{2!}t^2+\cdots$

and consider

$(t \delta(t))' = 0 = \delta(t) + t\delta'(t)$

So we have

$t\delta'(t) =- \delta(t)$

all high order terms involving $t^n \delta'(t)=0, n\ge2$.

Finally

$f\delta′=f(0)\delta′−f′(0)\delta$.