What is the result of $x(at) * δ(t-k)$ It seems rational  that $x(at) * δ(t-k) = x(at - ak)$. 
I tried to prove this: 
Let  $x_1(t) = x(at)$ then $x_1(t) * δ(t-k) = x_1(t-k) = x(at - ak)$. 
But I thought this as well:
$$x(at) * δ(t-k) = x(at) * δ\left(\frac{at-ak} a \right)  = a(x(at) * δ(at-ak)) = ax(at-ak)$$
Well I know that this may sound silly but I can't find where I'm wrong. 
 A: Dirac Delta is not zero at $t = k$ hence
$$x(at)\delta(t-k) = x(ak)$$
By the general distribution theory
$$f(x)\delta(x-a) = f(a)$$
Convolution
When it's about convolution, a simple rule reads:
\begin{align}
f(t)*\delta(t-a)=\int_0^t f(t-s)\delta(s-a)\,ds&=\begin{cases} 0, &t<a,\\ f(t-a) &t\ge a,\end{cases}=f(t-a)\theta(t-a),
\end{align}
where $\theta(t)$ denotes the unit step function.
Can you solve it for your function?
The fastest solution
Call $t-k = p$ in order to have
$$x(a(p+k))*\delta(p) = x(a(\tau + k))$$
In which we made use of the convolution identity
$$F[x\pm a]*\delta(x) = F[\tau \pm a]$$
A: The way you "prove" $x(at)*\delta(t-k)=x(at-kt)$ is not correct. Note that, starting from the signal $x(t)$, two operations are performed in series:


*

*A time scaling operation to the signal $x(t)$, resulting in the signal $y(t)=x(at)$

*A shift of the signal $y(t)$ by $k$, resulting in the signal $z(t)=y(t)*\delta(t-k)=y(t-k)=x(a(t-k))=x(at-ak)$.


The second equation in the "alternative" proof does not hold.
