Is the following function F:[0,1] to [0,1]
F strictly monotonic increasing F(1)=1,(i presume this unnecessary as its specified by the first two
(1)ie x+y=1 if and only iof F(x)+F(y)=1 F(x)+F(1-x)=1; F-1(x)+F-1(1-x)=1
(2)F homomorphic(2) x+y+z=1 if and only F(x)+F(y)+F(m)= 1
(3) x+y+z+m=1; if and only ifF(x)+F(y)+F(z)+F(M)=1
Give F(x)=x and F continuous (it appears to entail cauchys equation with F(1)=1 over ethe unit triangle due tot he common term). F(x+y)+F(z)=1, F(x)+F(y)+F(z)=1, etc, F(x+y)=1-F(z)=F(x)+F(y)
.
I
t appears that the third equation is redundant and I presuming any further equation where they sum to one (any given four, any given five, which sum to one is derivable from the first two) at least given that the first two entail some form of cauchy additivity. Is that correct
e
the above equation for any (4)given four points in the domain which sum to one
Does this generalize to Cauchy equation over the unit triangle. Ithough that one had to show that it held for any arbitary number of components not just 2 and three (although it appeas that the equations just generalize
F:[0,1] to [0,1]
F(1)=1 F(0)=1
F strictly monotonic increasing
(1)ie F(x)+F(1-x)=1; F-1(x)+F-1(1-x)=1
(2)F homomorphic(2) x+y+z if and only F(x)+F(y)+F(m)= 1
(3)x+y+z=2 if and only if F(x)+F(y)+F(z)=2, and th
(4) x+y+m+z=1 if and only F(x)+F(y)+F(z)+F(M)=1
Are the latter two redundant, ie do the first two generalize to any arbitrary factor; as the first two appears to entail cauchy equations of the unit trial
F(x+y+M)+F(Z)=1 F(Z)+F(x+y)+F((M)=1
so F(X+y+M)=F(x+y)+F(M)
but also x+y+z+m=1 entails
((x+y)+(1-x+y)) entails
F(x)+F(Y)+F(1-x+y) by (1) and (2)
F(x+y)+F(1-x-y)=1, so by cancelling out common terms F(X)+F(Y)=F(x+Y)
and so
F(x+y+M)=F(x+y)+F(M)=F(X)+F(y)+F(M)
z+ (x+y+M)=1 entials F(z)+F(x+y+M)=1 so F(x+y+m)=1-F(z)
so 1-F(Z)=F(X+Y)+M)=F(X)+F(Y)+F(M), and so F(X)+F(Y)+F(Z)+F(M)=1 x+y+z+m=1, appears in addition to a form of cauchy equation s
These appear to entail those first two fixed pints
These appear to entail F(x+y)=F(x)+F(y), over x,y,(x +y)\in in [0,1],;
I thought one required that any for any n=2.........infinity which sum to one. But I presume this is entailed by these first both collectively working, as any given four example which sum to one can always be express as (x+y) +z+m = 1F(x+y)+F(m)+F(z)=1;
F(x+y+M)+F(Z)=1 so F(x+y+M)=F(x+y)+F(M) so -F(x+y)=-F(X+y+
F(M)+F(Z)=1-F(x+y)
F(x+y+m,)+F(Z)=1,
F(x+y+M)=F(x+y)+F(M)
yet 1-F(1-(X+y+M)=F(Z) as so 1-F(1-x+
x+y+M+z=1 z=1-(x+y+M), then F((x+y+M))+F(z)=1, F(z)=1-F(x+y+z) so as z=1-x-m-y F(z)=F(1-(x+y-+M)=1-F(x+y+M)=F(z)
so 1-F(x+y+M)=1-F(x+y)-F(M), F(1-x+y+=F(M_M= ;1-F(x+y)=F(M)+F(Z)
(2) appears to just say this ie F(1-x-y)=1-F(x)-F(y)
which given 2 entails F(0)=0, F(1)=1; ie (1) specifies F(0.5)=0.5, and F(0)+F(1)=1 and F(0.5)+F(0.5)=1
(2) entails that 0 +0.5+0.5=1 so that F(0.5)+F(0.5)+F(0)=1,
so that 2 times 0.5+F(0)=1; 1+F(0)=1,which gives, F(0)=0
which by subsitution into into one gives F(1)=
and that (2) F(1/3)=1/3 however when one has both
Moreover if(A) (x+y) +z =1 then (B)x+y+z =1 so clearly by (1) on A, (2) on (B)
F(x+y)+F(z)=1 and F(x)+F(y)+F(z)=1,
F(x+y)+F(z)=F(x)+F(y)+F(z) but F(z) is common
so \forall (x)(y)\in [0,1] so long as x+y is in [0,1]F(x+y)=F(x)+F(y)
where F(1)=1 and F non negative and strictly monotone increasing. Is this a pair wise result. It appears to generalize to F(x)=x for all rationals as far i can see
ie F(10)=2F(20), as 10+10+80=1 so F(10)+F(10)+F(80)=1, so 2F(10)+F(80)=1 and by the first equation 20 +80=1 so F(20)+F(80)=1 so that
F(20)=2F(10)
and F(30)=F(10)+F(20) use 10+20+70=1=F(10)+F(20)+F(70)=1 and F(30)+F(70)=1 and one can 10 F(10)=1=0.1 and one can seeiming building this standard sequence, It appears to entail F(x)=x for all rations even x>0..5 and F(x+y) generalizes to arbitarily many components if x+y in [0,1[
Or is just a cauchy function on the restricted interval with with strictly monononitc and positive, on [0,1] to [0,1] then F(x)=x
forall(x,y)
They appear t
and otherwise x+y+z>1 then F(x)+F(y)+F(z)>1 and x+y+z1
x+y =1 iff F(x)+F(y)=1
where otherwise x+y>1 iff F(x+y)>1, and x+y1; x+y+z+m1
x+y+z=2 if and only F(x)+F(y)+F(z)=2
where x+y+z>2 if and only if F(x)+F(y)+F(z)>2
x+y+z
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