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

Question

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 $

Answer 1

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. $\tau_a(T*S) = T*\tau_a S.$