$\partial_t[x(t)]$ is a function, $\partial_t,$ whose input is real-valued functions, just as $x(t)$ is a function of real-valued numbers.
Well, I'd like to understand the implications of calculus with respect to function-space valued inputs of the standard differential operator.
What is the derivative of the differential operator mapping with respect to the input of elements of a Banach space, $x \in X$? Let's denote the derivative with respect to the standard differential operator's input as $\partial_x$ and its (whatever) derivative as $\partial'_t$
I would expect this to follow a few properties,
$\partial_x[x] = 1,$
$\partial_x[\partial_t\partial_t[x]] = (\partial' \partial_t + \partial_t\partial'_t)[x]$ (product rule)
$\partial_x[\partial^2_t] = 2\partial'_t\partial_t[x]$ (power rule).
Does this framework make sense? And if so, what is $\partial'_t$ with respect to $t$?
Best Answer
The functional derivative of the derivative operator $D$ at the function $f$ is $g \mapsto \lim_{\varepsilon \to 0} \frac{D[f+\varepsilon g]-D[f]}{\varepsilon}=g \mapsto \lim_{\varepsilon \to 0} D[g]=g \mapsto D[g]$. So $D$ is its own functional derivative. This always happens with an operator that is already linear.
I don't really understand what you're asking in the rest of your question, maybe you can clarify.