Convolution with dirac delta - proof

Question

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.

Answer 1

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.

Answer 2

Also, depending what you want/need, it is true that compactly-supported distributions act reasonably by "convolution" on smooth functions on $\mathbb R^n$ (and on real Lie groups...) It may be worthwhile to look at the mechanism here, to be confident that it's all working alright.

First, we need to know that the space of compactly supported distributions is the (continuous) dual to the space of smooth functions. Also, the translation action of the group $\mathbb R^n$) (or real Lie group...) by $T_g\varphi(h)=\varphi(hg)$ is a continuous linear map of the space of smooth functions to itself. Then, for compactly supported distribution $u$, the map $g\to u(T_g\varphi)$ is smooth... so is back in the space on which compactly supported distributions act.

Writing the action as "convolution" is very traditional, I know, but is a little misleading about the asymmetry between the actor and the acted-upon. That is, the compactly-supported distributions act... on smooth functions. Forgetting this asymmetry can lead to confusion about why, for example, there's the classic non-associativity $$ 1 \;=\; \delta * 1 \;=\; (H * \delta') * 1 \not= H * (delta'*1) \;=\; H*0 \;=\; 0 $$

And, in case we forget, equality of two continuous functions as distributions assures their pointwise equality. :)