Does the equality below hold?
$f(x)=\int_{-\infty}^{\infty}f(x')\delta(x-x')dx=\int_{-\infty}^{\infty}f(x')\delta(x'-x)dx$
Does the equality below hold?
$f(x)=\int_{-\infty}^{\infty}f(x')\delta(x-x')dx=\int_{-\infty}^{\infty}f(x')\delta(x'-x)dx$
Yes since the argument of the delta distribution equals zero when x=x' in either case, and the result is f(x) for either integral, assuming dx is changed to dx'.
The correct form should be:
$f(y)=\int_{-\infty}^{\infty}f(x) \delta(x-y)dx = \int_{-\infty}^{\infty}f(x) \delta(y-x)dx$
When $x=y$, we have $\delta(y-y)=\delta(0)$ and this performs a sampling on the $f(x)$ function at the point $x=y$.
Well, $\int_{\mathbb R} f(x)\delta (x - a)dx = {f(a)\int_{- \infty }^\infty {\delta (x - a)dx}} = f(a) = {f(a)\int_{- \infty }^\infty {\delta (a - x)dx}}$ so YES, if you change your $dx$ to $dx'$.