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Which functional equations and constraints specify

f(x)=x where F:[0,1] to [0,1], without directly specifying that the function is continuous or sur-for f:[0,1] to [0,1]jective.

Apparently if the Cauchy equation holds for all reals

  1. F(x+y)=F(x)+F(y),

and the field automorphism equation also does for all reals

  1. F(xy)=F(x)F(y) holds for real values (at least in [0,1]), F(x)=x; or that F(x)=o , and automatic continuity is bestowed.

I presume then that if in addition, something like 3. holds.

ie, 3.F(1)=1 or something a little stronger. ie that there exists some value xneq0, x>0,x<1 such that F(x) is defined F(x)>0, F(x)neq=0, (and maybe F(x)<1 ) on top. Ie to rule out the F(x)=0 constant function.

Then automatic continuity would be bestowed and F(x)=x, directly without any further continuity requirement; or at least again that F(x)=o or F(x)=x

I find it a little difficult to believe that the above two constraints (or three i suppose F(1)=1 or F(1)=1 and some minor additions like those just mentioned literally induce F(x)=x!

Or entail, at very least that F(x)=0 or F(x)=x over all reals or in this context in [0,1] and thus automatic continuity is already entailed .

Ie that they essentially induce that F(x)=Ax, already where A\in{0,1}, and thus entail continuity directly.

My issues are (1) how could one capture just one (transcedental) irrational number directly then, without reference to F(x)=x, or any direct reference to any theorem linking them to continuity, ie F(1/pi)=1/pi etc.

That is, Using (1) and (2) and (3), in full generality. If indeed this is what they entail. One should be able to at least do this, to some degree, at least for a singular transcendental. or is just an artefact of our real valued system of analysis (archimedean constraints on real numbers) and measurability constraints on algebras, rings

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  • $\begingroup$ (2) if this is correct, how does one establish that (2) holds; ie the multiplicative equation but just the above three constraints, Ie it would appear that the field equations (1+2) would not be able to capture transcendental numbers directly F(1/pi)=1/pi for example, of which there are presumably many (uncountably many more in [0,1] then the rationals and the algebraic irrationals combined). Maybe there is something I am missing. $\endgroup$ – William Balthes Apr 18 '17 at 5:53
  • $\begingroup$ ie F(x+y)=F(x)+F(y) with F(1)=1) can be used to establish both the values of the rational numbers and I suppose the rational homogeneity/ additivity of the irrational numbers; where cauchy's equations holds for all reals (at least in [0,1], and is generalizable to all rational combinations and multiplicative combinations ie F(delta x)=delta F(x) where x is real valued and delta rational (in [0,1] at least). ). $\endgroup$ – William Balthes Apr 18 '17 at 5:53

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