I have dirac delta defined as $\delta(f)=f(0)$, where $f(x)$ is an arbitrary function. I have defined convolution of distribution and function as $T\ast f=T(\tilde{f}\ast\varphi)$, where $\tilde{f}(x)=\mathrm{d}_{-1}f(x)=f(-x)$ and $\varphi\in\mathscr{S}$ is a test schwartz function and $T$ is a distribution.
I need to prove, that
$$\delta\ast f=f$$
using my definition of convolution and $\delta$.
I know how to prove it informally, but I can't formulate a formal proof.
$(f\ast\phi=\int_{\mathbb{R}^n}f(y-x)\phi(x)\mathrm{d}x)$ if $f,\phi$ are functions.
[Math] Convolution with dirac delta – proof
convolutiondirac deltadistribution-theory
Best Answer
Going to take a stab at it, based on http://en.wikipedia.org/wiki/Distribution_%28mathematics%29#Functions_and_measures_as_distributions
By the definition, $$\langle \delta\ast f,\varphi\rangle = \delta(\tilde{f}\ast\varphi) = \delta\left(t\mapsto\int_{\mathbb{R}}\tilde{f}(\tau)\varphi(t-\tau)\,d\tau\right) = \int_{\mathbb{R}}f(-\tau)\varphi(-\tau)\,d\tau = \int_{\mathbb{R}}f(\tau)\varphi(\tau)\,d\tau = \langle f, \varphi\rangle,$$ since integrating over the whole real line backwards is the same as going forwards.
Now, isn't there some kind of identification of functions $f$ with distributions $T_f$?Since $\langle \delta\ast f,\varphi\rangle = \langle f,\varphi\rangle$ for all $\varphi$, we see that the distributions $\delta\ast f$ and $f$ are equal.