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
- F(x+y)=F(x)+F(y),
and the field automorphism equation also does for all reals
- 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