How does one specify the assumption that a function is real for real arguments?


Simplify[Re[I f[x]], Assumptions -> f[x_] ∈ Reals]


Simplify[Re[I f[x]], Assumptions -> x ∈ Reals && f[x_/;x ∈ Reals] ∈ Reals]

How do I get zero in the second case?

  • 1
    $\begingroup$ Well, I know it does work in some cases, but pattern-matching-based assumption is never documented… $\endgroup$ – xzczd Sep 12 '19 at 10:24
  • $\begingroup$ @xzczd How would you otherwise formulate the mathematical fact of this kind about the given function? $\endgroup$ – yarchik Sep 12 '19 at 11:31
  • $\begingroup$ The first case will be Simplify[Re[I f[x]], Assumptions -> f[x] ∈ Reals], but sadly I've no idea how to express 2nd case. $\endgroup$ – xzczd Sep 12 '19 at 11:47
  • $\begingroup$ Re[I f[x]] // ComplexExpand $\endgroup$ – Bob Hanlon Sep 13 '19 at 4:56
  • $\begingroup$ @BobHanlon No, ComplexExpand unconditionally assumes that all symbols are real. I need to impose a condition. For instance ComplexExpand[Re[I (f[x] + f[I x])]] gives me zero. But if f[z] is Sin[z]---it is a false statement. $\endgroup$ – yarchik Sep 13 '19 at 5:36

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