Given two function $f(t)$ and $g(t)$ and denote the convultion of the functions by $h(t)$, that is $$f(t)*g(t)=h(t).$$ What is $f(t-1)*g(t+1)$ in terms of $h(t)$?
Have no idea how to treat this question, everywhere I read, it is said, that this isn't a Ņonventional way of writing convolution.
The only thing that Im thinking about is, if we denote $T_kf(t)$ as$f(t-k)$ (shifting operator), we can write: $$T_1f * T_{-1}g$$
And then what?
Given a function $s\colon\mathbb R\to\mathbb R$ such that $\int_{-\infty}^\infty s(\tau)\,\mathrm d\tau$ exists, note that for any $z\in\mathbb R$ we have $$ \int_{-\infty}^\infty s(\tau)\,\mathrm d\tau = \int_{-\infty}^\infty s(\tau-z)\,\mathrm d\tau $$ by substitution.
Hence, for the convolution: \begin{align*} f(t) * g(t) &= \int_{-\infty}^\infty f(\tau)\,g(t-\tau)\,\mathrm d\tau \\ &= \int_{-\infty}^\infty f(\tau-z)\,g(t-\tau+z)\,\mathrm d\tau \\ &= f(t-z)*g(t+z). \end{align*}
For $z=1$ this yields $f(t)*g(t) = f(t-1)*g(t+1)$.
I agree that "$f(t) * g(t)$" can be considered a bad notation, since $f(t)$ is just a real number, not a function. Here we used $f(t)$ as a shorthand for the function $t\mapsto f(t)$ and $f(t-z)$ as a shorthand for the function $t\mapsto f(t-z)$.
In the notation you suggested, this result might be written as $$ f * g = T_z f * T_{-z} g, $$ where $(T_z s)(t) := s(t-z)$.
