Dirac Delta Function Identity

1. Sep 23, 2013

Poligon

1. The problem statement, all variables and given/known data

Good day.

May I know, for Dirac Delta Function,
Is δ(x+y)=δ(x-y)?

3. The attempt at a solution
Since δ(x)=δ(-x), I would say δ(x+y)=δ(x-y). Am I correct?

2. Sep 23, 2013

vanhees71

No, that's not correct! The $\delta$ distribution (not function!) is defined by how it is acting on a function in the sense of a linear functional, written as a formal integral,
$$\int_{\mathbb{R}} \mathrm{d} x \delta(x) f(x)=f(0),$$
where $f$ is a function of an appropriate space of functions (e.g., the Schwartz space of quickly vanishing functions or the space of functions with compact support, etc.).

You can prove properties of the distribution by formally handling this kind of integral as if the $\delta$ distribution were a usual function. E.g., you have
$$\int_{\mathbb{R}} \mathrm{d} x \delta(x-y) f(x) = \int_{\mathbb{R}} \mathrm{d} x' \delta(x') f(x'+y)=f(0+y)=f(y).$$
Now you can evaluate yourself, what $\delta(x+y)$ must be and also prove the (correct) statement $\delta(-x)=\delta(x)$.

3. Sep 23, 2013

arildno

In order for two delta functions to be equal, their singularities must occur at the same place IDENTICALLY.

Is x+y=x-y an identity or an equation?

4. Sep 23, 2013

Poligon

Hi vanhees71 and arildno,

I see. So ∫dx δ(x+y)f(x) will equal to f(-y),

if f(x) is an even function, then
∫dx δ(x+y)f(x) = f(y)

if f(x) is an odd functoin, then
∫dx δ(x+y)f(x) = -f(y)

Am I correct? I am sorry if I still not able to get the concept, do allow me to make some mistakes so I can learn from it. While for arildno's question, I am sorry that I am not able to answer as it's little deep for me (I can't understand what equation and identity are in this context). I will dig into that.

5. Sep 24, 2013

vanhees71

Yes, that's correct. So you see that in general $\delta(x+y) \neq \delta(x-y)$.