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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

1051.0. "A Problem about Polynomials" by AITG::DERAMO (Daniel V. {AITG,ZFC}:: D'Eramo) Thu Apr 06 1989 20:08

	This problem deals with polynomials with real coefficients,
	in a single real variable, restricted to some closed interval
	[a,b] = {x | a <= x <= b}.

	Suppose A is a set of such polynomials.  We say that the
	coefficients of A are bounded if there is a bound M such
	that any coefficient a of a polynomial p(x) in A satisfies
	|a| <= M.  A is bounded in degree if there is an n such that
	every polynomial p(x) in A has degree <= n.  The values of
	A are bounded if there is a bound R such that for every
	polynomial p(x) in A and every x in [a,b], | p(x) | <= R.

	For example, let A = {1, x, x^2, x^3, x^4, ...}.  A is not
	bounded in degree.  The coefficients of A are bounded.  The
	values of A are bounded if -1 <= a <= b <= 1, but are unbounded
	otherwise.  Or let B = {0, x, 2x, 3x, 4x, 5x, ...}.  B is
	bounded in degree, B's coefficients are not bounded, and B's
	values are not bounded unless a = b = 0.  Finally, if C is any
	finite set of polynomials, then its coefficients, values, and
	degree are all bounded.  (The values of any one polynomial p(x)
	will always be bounded for x limited to a closed bounded
	interval [a,b].)

	The problem is to prove the following or provide a counterexample:

		If A is a set of polynomials bounded in degree and values,
		then A must also be bounded in coefficients.

	Dan
T.RTitleUserPersonal
Name		DateLines
1051.1HPSTEK::XIAThu Apr 06 1989 23:1859
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
1051.2AITG::DERAMODaniel V. {AITG,ZFC}:: D&#039;EramoThu Apr 06 1989 23:198
     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
1051.4where did the N+1st basis element go? :-)AITG::DERAMODaniel V. {AITG,ZFC}:: D&#039;EramoThu Apr 06 1989 23:4611
     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
1051.5HPSTEK::XIAThu Apr 06 1989 23:5511
    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
1051.6HPSTEK::XIAFri Apr 07 1989 01:1626
    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
1051.7HPSTEK::XIAFri Apr 07 1989 01:275
    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
1051.8in Hilbert space, no one can hear you screamAITG::DERAMODaniel V. {AITG,ZFC}:: D&#039;EramoSun Apr 09 1989 23:3919
     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
1051.9which polynomial has no degree ?STAR::ABBASIi^(-i) = SQRT(exp(PI))Sun Jun 14 1992 16:573
    answer after CR
    
       0
1051.10Define degreeAUSSIE::GARSONMon Jun 15 1992 01:280
1051.11can't resistMOCA::BELDIN_RAll&#039;s well that endsMon Jun 15 1992 10:151
    ba, ma, bs, bse, ms, phd
1051.12bse=bovine spongiform encephalopathy?AUSSIE::GARSONTue Jun 16 1992 23:460