integrals

integrals

[Peter Freyd]

Let  I  be a closed interval and let  C(X)  denote the bipointed 
midpoint algebra of continuous functions from a space  X  to  I.
I would not have guessed the following:

1.
   Any map from  C(X)  to  C(Y)  that preserves midpoints, top, and
   bottom is monotonic and preserves constant functions.

The order preservation is an immediate consequence of:

    f =< g   iff  there exists  h  such that  f|top = g|h

If we specialize to the case that  X  and  Y  are single points, the
order preservation implies the only map from  I  to  I  that preserves
midpoints, top and bottom, is the identity map (after first noticing 
that it must have a dense subset of fixed points corresponding to the
dyadic rationals). Let  I -> C(X)  be the "K-map" that assigns 
constant maps. For any  y:Y  let  C(Y) -> I  be its evaluation map.
Then for any map  C(X) -> C(Y)  that preserves midpoints, top, and 
bottom we have that  I -> C(X) -> C(Y) -> I  is the identity map. 
And from that we can easily conclude that  C(X) -> C(Y)  preserves 
constant maps.

In a 22 Dec posting I said that the "mean-value" function  M:C(I) -> I
can be characterized from its order preservation together with what it
does to constant functions and
                                                MxM
         C(I v I) -->  C(I + I) --> C(I) x C(I) ---> I x I
  
        C(F) |                                         | m
             v                                         v         
                                 M
           C(I)  ----------------------------------->  I

where  F:I -> I v I  is the coalgebra structure and  m  is the 
midpoint operation. I went on to say that if  C(F)  is inverted then 
this diagram can be read as a fixed-point definition of  M: "It's the
unique fixed-point of an operator acting on the set of all those 
order-preserving maps from  C(I)  to  R  that do the right thing on 
constant functions." It's much better to change the  R  to  I  and say
just that it's acting on the set of bipointed midpoint-algebra
homomorphisms.

To recapitulate (using  \  for lambda, T  for top, B  for bottom):

2.
    The mean-value operator,  M:C(I) --> I, is the unique map 
    such that:

                   M(\x.B) = B
                   M(\x.T) = T
                 (Mf)|(Mg) = M(\x.(fx)|(gx))
                        Mf = (M(\x.f(B|x)))|(M(\x.f(x|T)))

No inequalities, no limits, only equations.


Let me take the occasion to do a little clean-up. In a 31 Dec posting
I gave a curse-of-analysis proof for:

3.
    Midpoint-preserving functions between intervals are monotonic.

Here's a much better proof. Monotonicity is equivalent to the
preservation of betweenness. Suppose  f  is a midpoint-preserving map 
between intervals. It's conceptually useful to take the target to be, 
instead, the real line and show that a failure to preserve betweenness
forces the values of  f  to be unbounded. So let  a,b,c  be points in 
the source interval with  b  between  a  and  c  which fact will be 
denoted as  {a.b.c}  and suppose that  fb  is not between  fa  and  fc.
Without loss of generality we can assume that  {fa.fc.fb}.  There 
exists  {a.b'.c}  such that either  b = a|b'  or  b = c|b'.  Each of
the equations  fb = fa|fb'  and  fb = fc|fb'  implies   {fa.fc.fb'}.
So  {a.b'.c}  is still an example but the first equation implies that
the distance from  fc  to  fb'  is twice -- and the second equation 
implies it is _at least_ twice -- the distance from  fc  to  fb.  By
iteration we may thus increase that distance beyond any bound.

A point that the previous proof didn't really cover is dispatched
by this corollary:

4.
    Midpoint-preserving functions between intervals are either
    one-to-one or constant.

Suppose  a  and  c  are distinct but  fa = fc.  If there's  c'  such
that  c = a|c'  then we may infer that  fa = fc'.  By replacing  c
with  c'  in this manner as long as possible we may assume that  a  
and  c  are such that there is no solution to the equation  c = a|x 
and from that we may infer that if  c'  is chosen to be the endpoint
such that  {a.c.c'}  then there is  {a,b,c}  such that  c = b|c'. By
the last result we have  fa = fb = fc  hence  fc = fb = fc'.  Thus we
may assume that  a  and  c  are distinct with  fa = fc  and  c  an 
endpoint. Similarly we may assume that  a  is  an endpoint and finish
by applying, once again, the last result.

Besides injectivity we have this corollary on surjectivity obtained
by combining 1 and 3:

5.
    The image of any midpoint preserving function between intervals
    is a closed interval. 

PS

>From the proof of  1  it is clear that order preservation doesn't 
require preservation of bottom. A similar proof uses preservation of
bottom instead of top, but if both conditions are dropped the map 
needn't even be  monotonic. Let  I = [-1,1]  and let  C(I) -> I  be
the map that sends  f  to  f(-1)|(-f1).  It sends all constant 
functions to  0, it sends the identity map to  -1  and it sends the
negating map to  1.)

The proof of  3  suggests that if the target interval is replaced by 
the reals then there do exist midpoint-preserving maps that are not 
monotonic. And that is, indeed, the case in the presence of the axiom
of choice: view the reals as a rational vector-space and count the
number of endomorphisms (each of which preserves midpoints) to find 
that there are  2^{2^N} of them. But there are only  2^N  monotonic
self maps (midpoint preserving or not).

Do we need the axiom of choice? If, instead, we add the axiom of 
measurability, then the counterexamples disappear: let  f  be a 
measurable midpoint-preserving map from  R  to  R;  we can easily
specialize to the case that  f0 = 0, hence  f  is a measurable group
endomorphism; for any real  a  consider the induced group homomorphism
from  R/aZ  to  R/(fa)Z;  any measurable group character is continuous
and that's enough to force  f  to be continuous.