No, that's not correct! The [itex]\delta[/itex] 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,
[tex]\int_{\mathbb{R}} \mathrm{d} x \delta(x) f(x)=f(0),[/tex]
where [itex]f[/itex] 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 [itex]\delta[/itex] distribution were a usual function. E.g., you have
[tex]\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).[/tex]
Now you can evaluate yourself, what [itex]\delta(x+y)[/itex] must be and also prove the (correct) statement [itex]\delta(-x)=\delta(x)[/itex].
In order for two delta functions to be equal, their singularities must occur at the same place IDENTICALLY.
Thus, you must ask yourself:
Is x+y=x-y an identity or an equation?
#4
Poligon
23
0
Hi vanhees71 and arildno,
Thanks a lot for your prompt reply!
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.