# 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, 2014 at 20:02
• what do you mean? Dec 20, 2014 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, 2014 at 20:48
• No, I don't know about regularization. What context do you mean? Dec 20, 2014 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, 2014 at 13:52

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.

• 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? Dec 20, 2014 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, 2014 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, 2014 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 $\delta(x)$). Dec 20, 2014 at 21:00
• I don't know. Have a look at math.stackexchange.com/questions/639786/…
– Unit
Dec 20, 2014 at 21:05

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. :)