How combine pure functions of several slots?

Question

How can one define in a functional way a 1st-order linear differential operator involving several independent variables that can then be applied to a function of that many variables?

Consider an example with two variables, where one wants to form D[f[x, y], x] + 3 D[f[x, y], y] from a function f. Of course one could define

    diff[f_][x_, y_] := D[f[x, y], x] + 3 D[f[x, y], y]

so that for a function such as

    g[x_, y_] := x^2 y + Cos[x + 2 y]

we just evaluate:

    diff[g][x, y]
2 x y + 3 (x^2 - 2 Sin[x + 2 y]) - Sin[x + 2 y]   (* desired final output *)

But how can such an operator be defined functionally, that is, without explicitly using variables initially?

We could try

    diffOp[y_] := Derivative[1, 0][y] + 3 Derivative[0, 1][y]

and then

    diffOp[g]
3 (-2 Sin[#1 + 2 #2] + #1^2 &) + (-Sin[#1 + 2 #2] + 2 #1 #2 &)

But now how does one use such a combination of pure functions of several variables so as to produce the same result as from diff[g][x,y]?

The crux of the difficulty appears in the following simpler problem. Consider two functions of two variables:

    g1 = (#1^2 + #2) &
    g2 = Cos[#1 #2] &

How can one produce the same result as, say,

    g1[x, y] + 3 g2[x, y]
(* x^2 + y + 3 Cos[x y] *)

directly from the functional linear combination g1 + 3 g2 -- by forming an expression of the form oper[g1 + 3 g2][x, y]?

By contrast with the single-variable situation, where a simple Through would serve as the oper',Through` will not work in the multi-variable situation here:

   Through[(g1 + 3 g2)[x, y]]
(* x^2 + y + (3 (Cos[#1 #2] &))[x, y] *)

A pure function embedded in that output.

Note that a simple sum g1 + g2 instead of the linear combination g1 + 3 g2, Through will work (just as it does for a single variable):

    Through[(g1 + g2)[x, y]]
(* x^2 + y + Cos[x y] *)

Answer 1

In Mathematics, you can define new functions using operation on their images. Your example, $g_1+3 g_2$ is effectively defined as the function $x\mapsto g_1(x)+3g_2(x)$. Mathematicians use the following notation for functions:

The $f$ before the semicolon is just a label.

In Mathematica, an assignment such as g1 = (#1^2 + #2) & introduces the label $g1$ but more importantly it creates the rule $( \#1, \#2 )\mapsto \#1^2+\#2$.

Paraphrasing your question: how can we create, out of $\#\#\mapsto g1(\#\#)$ and $\#\#\mapsto g2(\#\#)$, a new rule $\#\#\mapsto whateverLabel(\#\#)$ without explicitly giving $whateverLabel(\#\#)$ yet keeping it rather arbitrary.

Some operations on functions, (such as multiplication by a number $α f$), have a natural meaning in terms of the images of the functions; but natural is not the same as nonexistent. And in the end, you need to have the equivalent of the line $x\mapsto α f(x)$.

Therefore, your quest is impossible. You could, for example, limit your quest and keep the operations limited to a small subset (sums of functions, multiplication by a number, multiplication of functions). In that case, we might be able to help.

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PS. Think twice before replying that one could define the operations on the images by the operations on the labels.

Answer 2

(Sorry, this is too long to fit into a comment) A possible workaround is to use fresh unevaluated symbols to represent your expression of free functions

expr = 3 f1 + 2Exp[-f2]

g1 = (#1^2 + #2) &;
g2 = Cos[#1 #2] &;

This will produce a list with the functions of said variables.

funlist = {g1, g2}; arglist = {x, y};
dummylist = {f1, f2};
funrule = #@(Sequence @@ arglist) & /@ funlist

{x^2+y, Cos[x y]}

Then you can use a replacement rule, as suggested in one of the comments:

expr /. Thread[dummylist -> funrule]

3(x^2+y) + 2 Exp[-Cos[x y]

You might automate this into a procedure that could generate the unique identifiers by parsing an unevaluated (held) expr so that when you pass expr[g1,g2], you'll end up with expr[f1,f2] in the body of the procedure.