Conference rusure::math

The statement is true.  Here is an "elegant" :-) proof I came up with.
However, I strongly believe a more elementry proof must exist.

Proof:

Let A be bounded in value and bounded in degree.  Let N = max   deg(p)
                                                          p in A
                               oo
Now consider the Banach space L  ([a, b]).  Clearly, 

                                   oo
A is a subset of the Banach space L  ([a, b]).  Then A is bounded

                                                       oo 
in value is equivalent of saying that A is bounded in L  ([a, b]).

                            oo                                   2     N
Now let V be a subspace of L  ([a, b]) such that V = span{1, x, x ... x  }

Then dim(V) is finite.  By the result 1038.0 and 1038.11, All norms on 

                                                                 oo
V are equivalent.  Since A is a subset of V and A is bounded in L  ([a, b]),

                 2
A is bounded in L ([a, b]).  Let E = {e , e ,..., e } be an orthonormal basis
                                       1   2       N

                       2                                                
of V as a subspace in L ([a, b]).  Let a be an element of A.  

          N
Then a = sum a  e   ==>
         i=1  i  i

           2         N          2       N           2
      ||a||   =  || sum a  e  ||    =  sum ||a  e ||  
           2        i=1  i  i   2      i=1    i  i  2

The last equality is true because the basis is orthonormal.  Hence,

     2    N      2        2    N      2
||a||  = sum |a | * ||e ||  = sum |a |
     2   i=1   i       i  2   i=1   i

                       2               2    N      2
Since A is bounded in L ([a, b]), ||a||  = sum |a |  < M for
                                       2   i=1   i

all a in A.  Now e  is obviously a polynomial for all 1 <= i <= N
                  i          

Since E is finite, E is bounded in coefficient.  Let that bound be

M1.  Then the coefficients of the polynomials in A are bounded by 

M1 * M * N.  Q.E.D

Eugene

     Oops.  Let me also add the condition that a < b, otherwise
     .0 already included a counterexample.  With that
     restriction, the answer is that the statement is indeed
     true.  Eugene (that was quick!) you could have at least let
     me do the problem myself first. :-)  I have to go study .1
     now.
     
     Dan

     re .1
     
     The key step, which I didn't follow, seems to be
     
          "Then A is bounded in value is equivalent of
                                        oo
           saying that A is bounded in L  ([a,b])."
     
     Why?
     
     Dan

    re -1
                                                                   oo
    The following theorem is almost direct from the definition of L
                          oo
    Theorem:  If a is in L  ([a, b]), then ||a||   = max         |a(x)|
                                                oo   x in [a, b]       
    
              provided a is continuous.
    
    Eugene
    

    Here is another way of looking at the .1.   Forget about where V
    sits.  Just think of V as a normed space itself with two norms.
    
    The ||*||    and ||*||   norms defined as:
             oo           2                   
    
    Let v in V.  Let  ||v||   = max        |a(x)| 
                           oo   x in [a,b]        
    
    
    
                   /b
    ||v||  =      (         2      1/2
         2     (   )  |v(x)|  dx  )
                  / a           
    
    Since V has finite dimension, V is closed and complete.  Hence,
    V is a Banach space.  Moreoever, these two norms are equivalent.
    
    Moreoever, V with the ||*||
                               2
    
    is also a Hilbert space....                           
    
                               
    Eugene

    I deleted .3 because I just realized that what is said in .3 (about
    a minor fixable problem in .1) is not correct.  The proof in .1
    does not have the problem mentioned previously in .3.
    
    Eugene

     re .1

>>   By the result of 1038.0 and 1038.11, All norms on V are equivalent.
     
                                           oo      2
     For those of us who don't understand L   and L  as well as
     they do finite dimensional spaces, this can be used more
     directly.  If all of the polynomials in A have degree less
     than N, then consider the polynomials with real coefficients
     and degree less than N as a vector space of dimension N over
     the reals.  Show that the following two define norms: the
     maximum of the absolute values of the coefficients of a
     polynomial; and the maximum of the absolute values of the
     values of a polynomial for x in [a,b].  We are given that
     the second norm over polynomials in A is bounded; then use
     the result quoted above.
     
     Dan
     

    answer after CR
    
       0

    ba, ma, bs, bse, ms, phd