# Proving $(T * S )(x-a)=T * S (x-a)= T (x-a)* S$ where $S,T \in \mathcal{D'}(\mathbb{R})$

Considering the definition of convolution:

If $$T,S \in \mathcal{D'(\mathbb{R})}$$ and at least one of them has compact support, $$\forall \varphi \in \mathcal{D}(\mathbb{R})$$ $$\langle T*S, \varphi\rangle= \langle T(y), \langle S(x), \varphi(x+y)\rangle \rangle$$

and the translation property: $$\langle T(y-a), \varphi\rangle = \langle T(y),\varphi(y+a) \rangle$$

I am trying to prove the following property:

$$(T * S )(x-a)=T * S (x-a)= T (x-a)* S$$ where $$S, T$$ are distributions: $$( S,T \in \mathcal{D'}(\mathbb{R}))$$

My try: I take a test function $$\varphi \in \mathcal{D}$$

Using the property of translation:

$$\langle (T*S)(y-a), \varphi\rangle =\langle (T*S)(y), \varphi(y+a) \rangle .....(1)$$ calling $$\varphi(y+a)=\psi(y)$$:

$$=\langle (T*S)(y), \psi(y) \rangle$$

and by definition of convolution of distributions:

$$= \langle T(y), \langle S(x), \psi(x+y)\rangle \rangle$$ $$= \langle T(y), \langle S(x), \varphi(x+y+a)\rangle \rangle$$ using translation again: $$= \langle T(y), \langle S(x-a), \varphi(x+y)\rangle \rangle .....(2)$$

and now I wish I could call this $$T * S (x-a)$$, but for that I guess I would instead need to have $$= \langle T(y), \langle S(x-a), \varphi(x+y-a)\rangle \rangle$$, because according to the definition of convolution, I need $$\varphi$$ to have as argument the sum of the arguments of the $$T$$ and $$S$$.

Is this correct? I have the following doubts :

(1) I am not very sure if $$(T*S)(y-a)$$ means $$T(y-a)*S(y-a)$$ or if the $$(y-a)$$ applies just to S , so that $$(T*S)(y-a)= T(x)*S(y-a)$$. I used the second one, because according to the definition of convolution, T and S take different variables: $$\langle T*S, \varphi\rangle= \langle T(y), \langle S(x), \psi(x+y)\rangle \rangle$$

I am also starting with a $$y$$ variable, because otherwise I don't get the $$x-a$$ in the final step The fact that using variables with T and S is an abuse of notation also is making it confusing, because I don't know if I am abusing correctly

(2) As I mentioned at the end of the proof can I really conclude :$$T * S (x-a)$$ ? For that I guess I would instead need to have $$= \langle T(y), \langle S(x-a), \varphi(x+y-a)\rangle \rangle$$

## 1 Answer

Define the operator $$\tau_a$$ on $$\mathcal{D}(\mathbb R)$$ by $$(\tau_a \varphi)(x) = \varphi(x-a)$$ and on $$\mathcal{D}'(\mathbb R)$$ by $$\langle \tau_a T, \varphi \rangle := \langle T, \tau_{-a}\varphi \rangle$$.

We can then write $$\langle \tau_a(T*S), \varphi \rangle = \langle T*S, \tau_{-a}\varphi \rangle = \langle T(t), \langle S(s), (\tau_{-a}\varphi)(t+s) \rangle \rangle = \langle T(t), \langle S, \tau_{-a-t}\varphi \rangle \rangle = \langle T(t), \langle \tau_a S, \tau_{-t}\varphi \rangle \rangle = \langle T(t), \langle (\tau_a S)(s), \varphi(s+t) \rangle \rangle = \langle T*\tau_a S, \varphi \rangle,$$ i.e. $$\tau_a(T*S) = T*\tau_a S.$$

• Thanks, very clear, I am trying to translate back with the notation they used in my lectures, Before you applied the definition of convolution, how should $T*S$ be understood :$T(t)*S(t),$ $T(t)*S(s)$ or $T(s)*S(s)$ ? Sep 1, 2020 at 19:23
• $T*S$ is a function of only one variable, so possibly $T(t)*S(t)$ or $T(s)*S(s),$ not $T(t)*S(s).$ The best certainly is to write $(T*S)(x)$ without setting variables on the individual factors. Sep 1, 2020 at 19:53