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I have three properties for a function:

Assumptions -> 0 <= x <= 1

f[1 - x] == 1 - f[x]
f[x/3] == 1/2 f[x]
Assuming[0 <= a <= b <= 1, f[a] <= f[b]]

And I would like to solve for the following: f[6/7]

I tried using RSolve, but I don't know how to put those equations in with a domain constraint as well.

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  • $\begingroup$ Even though I don't think you can find an exact solution, it turns out f[x_] := ArcSin[x]/Sqrt[2 x] almost works satisfying the equations within about 0.10 error over most of the domain and has the increasing property too. Of course that error will quickly blow up after a few iterations. $\endgroup$ – flinty Nov 23 '20 at 16:51
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Consider the so-called Cantor function.

Plot[CantorStaircase[x], {x, 0, 1}] 

As the following plot confirms it does have the functional properties.

Plot[{CantorStaircase[1-x]-(1-CantorStaircase[x]),CantorStaircase[x/3]-CantorStaircase[x]/2}, {x, 0, 1}] 

Moreover, it is CantorStaircase[6/7] equal to $3/4$.

It can be shown (see Chalice, D. R. "A Characterization of the Cantor Function." Amer. Math. Monthly 98, 255-258, 1991.) that this is the only real-valued increasing function satisfying your equations.

I'm not sure if RSolve is able to recover that kind of function.

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There can be no solution, because the both equations contradict each other.

First get solution for second equation.

eqs = {f[1 - x] == 1 - f[x], f[x/3] == 1/2 f[x]};

fs = f /. First@RSolve[eqs[[2]], f, x]

(*   Function[{x}, 2^(-1 + Log[x]/Log[3]) C[1]]   *)

Solve for the constant C[1] with equation 1 and you see,that C[1] depends on x, but it should be constant.

sol = First@Solve[eqs[[1]] /. f -> fs, C[1]]

(*   {C[1] -> 2/(2^(Log[1 - x]/Log[3]) + 2^(Log[x]/Log[3]))}   *)

Plot[C[1] /. sol, {x, 0, 1}]
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