Split real and imag parts of function with derivative

Question

Consider a complex function y with a real argument x and the function y itself is an argument of another function f, which takes the 2nd-derivative of y with respect to x

f[y_] := y''[x]

It seems that ComplexExpand gives the right answer

ComplexExpand[f[yr + I*yi]]

(*((I yi + yr)^\[Prime]\[Prime])[x]*)

Well, when I try to extract its real and imaginary parts separately, the following approaches failed.

ComplexExpand[Re@f[yr + I*yi]]
(*((I yi + yr)^\[Prime]\[Prime])[x]*)

Re@ComplexExpand[f[yr + I*yi]]

(*Re[((I yi + yr)^\[Prime]\[Prime])[x]]*)

ComplexExpand[f[yr + I*yi]] // Re

(*Re[((I yi + yr)^\[Prime]\[Prime])[x]]*)

I expect to obtain something like yr''[x] and yi''[x] for the real and imaginary parts.

Can anyone help with this?

Answer 1

You have to tell Mathematica what symbols are functions, by explicitly stating them with an argument. Say you have a "function" $f: \mathbb{R} \rightarrow \mathbb{R}$ and you try

D[f, x]
(* 0 *)

Since Mathematica does not know that f is a "function". However, the following gives the desired output

D[f[x], x]
(* f'[x] *)

For your problem my recommended way to go would be via pure functions, e.g.

ClearAll[f]
f[y_, x_] := y''[x]
f[Function[{x}, yr[x] + I yi[x]], x]
(* I (yi^\[Prime]\[Prime])[x] + (yr^\[Prime]\[Prime])[x] *)

instead of Function you could also use the lambda-style syntax where the arguments are denoted by Slots (#) and the function is finalized by a &, e.g.

#^2&[x] == Function[{x}, x^2][x]
(* True *)