How to show a piecewise binary operation over functions is associative?

Question

Given some set of functions $S(X) = \{f:[0,a]\to X : f(0)=f(a)=x_0 , a \geq 0\}$, define the binary operator $*$ over two functions $f:[0,a]\to X$ and $g:[0,b]\to X$ as: $$(f * g)(t) = \begin{cases} f(t), & \text{if $0\leq t \leq a$} \\ g(t-a), & \text{if $a\leq t\leq a+b$} \end{cases}$$ in particular such that $f * g : [0, a+b]\to X$. I want to show that this operation is associative, i.e.: $$((f * g) * h)(t) = (f * (g * h))(t)$$ I've written out the domains of each function composed with each other function, but I don't know if that's useful. My approach was initially to take it in two cases: when $t$ is small, and when $t$ is large. In the case where $t$ is small, and the operation just evaluates to the first function's value at $t$, I have shown associativity quite easily: $$((f * g) * h)(t) = (f * (g * h))(t)$$ $$(f * h)(t) = (f * g)(t)$$ $$f(t) = f(t)$$ However, trying to do that case where $t$ is large and the operation evaluates to the second function of $t - a$, I don't see how I can proceed. Is my approach on the right track?

Answer 1

\begin{align} (f*(g*h))(t) &=\begin{cases}f(t) & 0\le t\le a\\(g*h)(t-a) & a\le t\le a+b+c\end{cases} \\&=\begin{cases}f(t) & 0\le t\le a\\g(t-a) & 0\le t-a\le b \\ h(t-a-b) &b\le t-a\le b+c\end{cases} \end{align} Now, do the same thing for $((f*g)*h)$, and show you get the same result.