Is there a continuous function on R such that $f(f(x))=e^{-x}$?

Question

Is there a continuous function on R such that $f(f(x))=e^{-x}$? I have tried to take derivative of the two sides,but I can't get anything I want.what can I do?

Note $f(f(x))=e^x$ has been discussed at http://math.stackexchange.com/questions/65876/thoughts-about-ffx-ex and at http://math.stackexchange.com/questions/296745/there-exist-a-function-such-that-f-circ-fx-ex — Gerry Myerson, Oct 10 '13 at 11:44

score 7 · Accepted Answer · answered Oct 10 '13 at 11:39

7

No. Hint: An injective continuous function is monotonic and for any monotonic $f(x)$ the function $f(f(x))$ should be increasing.

answered Oct 10 '13 at 11:39

njguliyev

14,473
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This is only if f required to be real-valued. If it is allowed being complex-valued (but continuous and defined on R), it can exist. – Anixx Nov 19 '14 at 16:49

score 0 · Answer 2 · answered Nov 19 '14 at 16:55

0

If the function f is allowed to take complex values on R, such function exists.

answered Nov 19 '14 at 16:55

Anixx

9,119

Is there a continuous function on R such that $f(f(x))=e^{-x}$?

2 Answers2