What would be the correct notation to define a functional that also uses a scalar variable?

Question

I'm wondering how I should define a functional operator (a map from a space of functions to another space of functions) that also takes in a scalar valued variable as an argument.

Question: How should I define an operator $G$ that maps a function $u$ and a scalar valued variable $d$ to a function in $V$?

I know that if $G$ is an operator and $u$ is a function, you can write $G(u)$ to denote the function in the output space, and $G(u)(y)$ to denote the value of the output function evaluated at the point $y$. How would you denote an operator that takes in a scalar valued variable as well? Could you just write $G(u, d)$, where $d$ denotes the scalar valued parameter? And then could you write $G(u, d)(y)$ to denote the value of the output function evaluated at the point $y$?

Thanks!

Answer 1

Yes, $G(u, d)$ would work just fine. If $F$ and $F'$ are spaces of functions and $\mathbb{k}$ is your set of scalars, then $G$ would be a function $F \times \mathbb{k} \to F'$.

But of course any notation you like is possible. If it is natural to think of $G$ as a family of operators $F \to F'$ parameterized by the scalar $d \in \mathbb{k}$, then you might choose to use the notation $G_d(u)$.

In any case, since say $G_d(u)$ is then again a function, people are happy to accept that $G_d(u)(y)$ is just evaluation at $y$. If you're doing lots of calculations which involve evaluation this might be a little notationally ugly (though not wrong), and you could choose e.g. to use different brackets for the function argument to make it slightly clearer which argument is which: i.e. $G_d[u](y)$.

Ultimately, it's your call to choose notation which is the best mixture of clear and conventional/well known. (Probably $\overset{u}{\underset{d}{G}}(y)$ is not a good idea :).)