Convolution with dirac delta - proof

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.

• "informally ... formulate ... formal" – Unit Dec 20 '14 at 20:02
• what do you mean? – user74200 Dec 20 '14 at 20:09
• You should clean up the context of the question. Do you know about regularization? And when $f$ is a smooth, compactly supported function, and $u$ a distribution, then $u*f$ is actually a smooth function and given by the formula $u*f(x)=(u(y),f(x-y))$. Your result would follow immediately. However the previously stated formula requires a bit of effort. – Matt Dec 20 '14 at 20:48
• No, I don't know about regularization. What context do you mean? – user74200 Dec 20 '14 at 21:23
• @user74200 $T\ast f = T(\bar{f}\ast\varphi$ is NOT a valid definition, as the right hand side depends on $\varphi$, the left does not. In fact, on the left side you have your $T_{\delta\ast f}$... – Vobo Dec 21 '14 at 13:52

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.
• Well, this proves that $\delta \ast f=T_f$(which is not the question), but is there a way to prove $\delta \ast f=f$ or it is not true? – user74200 Dec 20 '14 at 20:19
• Given en.wikipedia.org/wiki/…, I think the convolution of a distribution and a function is a distribution. By $\delta\ast f$ you really mean the distribution $\delta\ast f$ defined by $\langle\delta\ast f,\varphi\rangle = \langle\delta,\tilde{f}\ast\varphi\rangle$. – Unit Dec 20 '14 at 20:31
• "Conventionally, one abuses notation by identifying $T_f$ with $f$, and thus the pairing between $T_f$ and $\varphi$ is often written \langle f,\varphi \rangle = \langle T_f,\varphi \rangle." – Unit Dec 20 '14 at 20:39
• Is it a distribution? It would make sense, because I am investigating it because of the partial differential operators and we often look for a distributive solution over all functions. It is posible the writer of my book I got this from meant $T_f$ instead of $f$ since it is a physics book(that even defines $\delta$ as $\delta(x)$). – user74200 Dec 20 '14 at 21:00