I'm using value-defined function as set of parameter, i.e.

f[1] ===> first parameter
f[2] ===> second parameter
f[3] ===> third parameter

etc. I would like to tell Mathematica that all these parameters are positive. I tried something like

FullSimplify[Abs[f[1]],Assumptions-> {f[x_]>0}]

and I get


instead of the desired


Of course here I posted just an example, the function I have to simplify in my case is much more complicated. How can I do?

  • $\begingroup$ Have you tried creating a list of assumptions like Thread[Table[f[i], {i, fmin, fmax}] > 0]? $\endgroup$
    – Feyre
    Commented Aug 14, 2016 at 11:41
  • $\begingroup$ Interestingly using f[1] > 0 in Assumptions does work. Doesn't pattern matching work here, and if so, why? $\endgroup$
    – kirma
    Commented Aug 14, 2016 at 11:45
  • $\begingroup$ Of course this is a solution but in my program I don't know the range of the parameters. I want to tell Mathematica that every value of the function is positive. $\endgroup$
    – MaPo
    Commented Aug 14, 2016 at 11:47
  • 1
    $\begingroup$ @MaPo A way I see (not sure whether it fits your need in your more general case) would be to define an UpValue for f, either with f /: Abs[fun : f[_]] := fun or with Abs[fun : f[_]] ^:= fun. This will evaluate Abs[f[1]] to f[1], for instance, without the need of using Assumptions and FullSimplify. $\endgroup$
    – user31159
    Commented Aug 14, 2016 at 14:44
  • 2
    $\begingroup$ Related: (6182), (42607), (58271), (67343), (79301), (79756), (94983) $\endgroup$
    – Mr.Wizard
    Commented Aug 15, 2016 at 1:05

1 Answer 1


EDIT: Simplified per suggestion from @BobHanlon.

This constructs assumptions by finding all occurrences of pattern f[_] in the expression being simplified:

FullSimplify[#, Cases[#, v : f[_] :> v > 0, Infinity]] &[
 Abs[f[f[1]]] + Abs[f[x]]]

(* f[x] + f[f[1]] *)
  • 2
    $\begingroup$ +1 Or more succinctly: Simplify[#, Cases[#, v : f[_] :> v > 0, Infinity]] &[ Abs[f[f[1]]] + Abs[f[x]]] $\endgroup$
    – Bob Hanlon
    Commented Aug 14, 2016 at 14:57
  • 2
    $\begingroup$ I find it somewhat surprising that Mathematica doesn't have support for a more concise way of expressing this. $\endgroup$
    – mikado
    Commented Aug 14, 2016 at 20:32
  • $\begingroup$ @mikado I wouldn't be surprised if I there would actually be a way, but I wouldn't be aware of it. Thankfully the above form is not that long, just slightly inelegant. $\endgroup$
    – kirma
    Commented Aug 15, 2016 at 3:36

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