Conference rusure::math

>>    (In other words, all norms on a finite dimensional vector space
>>    are equivalent).
     
     Did you really intend to make such a sweeping statement?  Or
     is the context supposed to include the restriction "... over
     the field of complex numbers C"?  Is there extra credit for
     finding either a counterexample with the restriction removed
     or a proof that the restriction isn't needed?
     
     Dan

    re .1
    
    I think it gotta be over some nice fields like C or R; however,
    I am not sure how nice the field got to be, but if you can find
    a proof for all fields (nice or not), or (Hint:  this is an exclusive
    or :-)) find a counter-example for some not so nice field, I will buy 
    you a cheap drink :-) :-).
    
    Eugene

     I'd rather have a cookie. :-)
     
     A norm on a vector space over R or C is a function from the
     vector space to the real numbers satisfying:
     
          a)  || x || >= 0 for all vectors x
          b)  || x || = 0 if and only if x is the 0 vector
          c)  || ax || = |a| || x || for all scalars a and all vectors x
          d)  || x+y || <= || x || + || y || for all vectors x and y
     
     Dan

     Note that part (c) of the definition of a norm is
     assuming that the elements of the underlying field have an
     absolute value |*| operation defined, so if not R or C the
     field must still be very nice.  The absolute value would
     map the field into the reals, obeying
     
          1) |a| >= 0
          2) |a| = 0 if and only if a=0
          3) |ab| = |a| |b|
          4) |a+b| <= |a| + |b|
     
     which makes the absolute value a norm itself, treating the
     field as a one dimensional vector space over itself.
     
     There are absolute values over the rational numbers that are
     not equivalent to the "standard" |a| = if a >= 0 then a else -a.
     But fixing a particular absolute value to define the norms
     with, I believe any two norms will still be equivalent.
     
     Dan

    re .4
    
    I looked over my proof again, and I didn't any place in that proof
    where I use the fact that the scalar field is R or C, so you are
    probably right.  As a matter of fact, I did not see any place in
    my proof that uses the fact that the elements in the scalar field
    is commutitive, so probably R-module will do in place of the field.
    Oh well, that is really going beyond the call of duty.      
    In any case, I am not absolutely sure about it, so just do it for
    C.  Chances are your solution won't require the scalar to be R or
    C.
    
    Eugene

   Eugene, there's something wrong there.

   Suppose the vector space is just C, and take the point x to be the origin,
0.  Then, by one of the properties of the norm, ||x|| , and ||x||  are both
zero.                                                1           2

   Thus, you obtain that for all A and B in R, A||x|| = ||x|| = B||x|| = 0.
                                                     1       2        1

   Perhaps you wanted a less-than-or-equal in there?


					-mjg

    re -1.
    
    Yea Mike, you are right.  It should be less than or equal to (by the
    way how do you do that in notes?  Maybe the C convention i.e. <=).
            
    Eugene

   As long as you don't get "a => b" (a is equal to or greater than b) and
"p => q" (p implies q) confused, then no problem!

   Are you sure this is right?  You seem to be implying that every norm in
a vector space is equivalent (well, you came out and said it.)  Moving this
topolically over, this would be equivalent to having every metric spaces of
equal dimensionality always being equivalent.  Is this true?  I was trying
to construct a counter example with one of the norms being the euclidean
distance measure (the "normal" norm) and the other being ||(x,y)||=|x+y|, but
realized to my chagrin that those do form equivalent metric spaces, which
while I'm still missing the formality for it implies to me that the norms are
equivalent.  But I can't say I remember seeing something about metric spaces
of equal dimensionality being equivalent in all cases.

   Still puzzling a bit on this one...


					-mjg

    re -1
    
    It only works with normed finite dimensional vector spaces.  I can
    easily give you two non equivalent metric spaces on the real--
    One of them is the usual metric (i.e. the Euclidean metric), and
    the other the discrete metric.  These two metric spaces are obviously
    not equivalent.  However, the discrete metric does not form a norm.
    
    Eugene

   I haven't had any courses yet which merged lin. alg and topology, and it
was beginning to worry me that it all collapsed!

					-mjg (sleeping easier now)

     Here is an outline of the steps in my proof.
     
     1)  Let e1,...,en be a basis for the finite dimensional
     vector space (which I'll call X) over the complex numbers C.
     If x = a1 e1 + ... + an en (where a1,...,an are in C) define
     
          <<x>> = max(|a1|,...,|an|)
     
     2) <<*>> as defined in 1 is a norm
     
     3) Let ||*|| be an arbitrary norm.  Using the triangle
     inequality and induction gives for x = a1 e1 + ... + an en
     ||x|| <= ||a1 e1|| + ... + ||an en||.  Using the property of
     norms of scalar multiples, we have ||a1 e1|| = |a1| ||e1||,...
     and so ||x|| <= |a1| ||e1|| + ... + |an| ||en||
                  <= max(|a1|,...,|an|)(||e1|| + ... + ||en||)
     So ||x|| <= <<x>> (||e1|| + ... + ||en||) for any norm ||*||.
     
     4)  Let D(x,y) = <<x-y>>.  Then D is a metric for X.
     
     5)  Treating X as a metric space with metric D and given the
     reals the usual topology, the function ||*|| from X into R
     is continuous.
     
     Sketch of proof of 5:  Use the epsilon-delta version of
     continuity for functions between metric spaces.  Let x be a
     vector and epsilon a positive real.  Find a delta so that
     for all vectors y with D(x,y) < delta, then | ||y|| - ||x|| |
     < epsilon.
     
     The choice for delta is epsilon/(||e1|| + ... + ||en||).
     
     Now, ||y|| = ||x+(y-x)|| <= ||x|| + ||y-x|| = ||x|| + ||x-y||
     so ||y|| - ||x|| <= ||x-y||.
     
     Likewise ||x|| = ||y+(x-y)|| <= ||y|| + ||x-y|| so
     ||x|| - ||y|| <= ||x-y||.  Now if A < B and -A < B then |A| < B,
     so | ||y|| - ||x|| | <= ||x-y||.  But by part 3,
     ||x-y|| <= <<x,y>>(||e1|| + ... + ||en||) and by part 4
     D(x,y) = <<x-y>>.  So if D(x,y) < delta, then
     | ||y|| - ||x|| | < epsilon and so ||*|| is continuous.
     
     6) Let B be the subspace of (X,D) consisting of those
     vectors x such that <<x>> = 1.
     
     7) (B,D) is compact.
     
     8) ||*|| is continuous and (B,D) is compact so the image of
     B under ||*|| is a compact subset of the reals.  The image
     has a lower bound (e.g., 0) and thus a greatest lower bound
     which I'll call b.  The image is closed so it contains b, so
     there is some vector y in B such that ||y|| = b.  Since y is
     in B we know <<y>> = 1, so y is not zero, so ||y|| = b is
     not zero.
     
     9)  Let x be any non-zero vector.  Let a = 1/<<x>>.  Then
     <<ax>> = <<x>>/<<x>> = 1 and so ax is in B.  Therefore
     b <= ||ax|| = |a| ||x|| = ||x|| / <<x>> and so
     b<<x>> <= ||x||.  This inequality also holds when x is zero.
     
     10) By parts 3 and 9 there are positive reals a and b,
     a = ||e1|| + ... + ||en|| and b as in part 8, such that
     b<<x>> <= ||x|| <= a<<x>> for all vectors x.
     
     11)  If ||*||  and ||*||  are any norms on X, then by part 10
                  1          2
     
     ((b^2)/a)<<x>> <= (b/a)||x||  <= b<<x>> <= ||x||
                                 1                   2
     
     and ||x||  <= a<<x>> <= (a/b)||x||  <= ((a^2)/b)<<x>>
              2                        1
     
     so (b/a)||x||   <= ||x||   <= (a/b)||x||
                  1          2               1
     
     and both b/a and a/b are positive quantities.  So any two
     norms are equivalent.
     
     Dan
     
     P.S.  The definition of norms or absolute values being
     equivalent that I knew was that two of them were equivalent
     if they gave rise to the same Cauchy sequences.  But this
     definition is equivalent to the one in .0.

    re -1
    
    Wow! Dan, you are very good.  By the way, how long did it take you
    to figure it out?  By the way, you don't really need induction in
    step 3, but that is no big deal.
    
    Eugene

    Show that there exist positive real numbers A and B such that 
    
    
    A||x||   <= ||x||   <= B||x||     for all x in X
          1         2          1                  
    
---------

Assume the contradiction.  Therefore either of the following
cases can be true.

case 1.  There is a vector x1 such that  ||x1|| < A||x1||.
				               2           1
This leads to the contradiction that ||0|| = ||0.x1|| = 0||x1||
	     			          2          2         2
< 0.A.||x1|| = 0.  So ||.|| is not norm.
           1               2

case2.  There is a vector x2 such that ||x2||  >  B||x2||.  This
				             2            1
says ||0|| > 0, and so ||.|| is not a norm.
          2                 2

That's all.  It will be nice to hear comments if there is any hole
in the above arguments.

/Snehamay K.

     re .12:
     
     Thanks, Eugene.  My first pass was to try to prove all norms
     gave the same Cauchy sequences and then to prove that
     condition equivalent to the inequality in .0.  I soon got to
     the inequality ||x|| <= <<x>>(||e1|| + ... + ||en||) and
     that ||*|| was continuous relative to the metric obtained
     from <<*>>, then I didn't see how to go on so I stopped. 
     When I came back to it I thought of the argument from the
     compactness of {x | <<x>> = 1} yielding a lower limit for
     the positive constant.  The elapsed time would be from the
     posting of .1 to a little earlier in the day Sunday than
     .11.  But I did do other stuff during that time. :-)
     
     re .13:
     
>>   case 1.  There is a vector x1 such that  ||x1|| < A||x1||.
>>     				                    2         1
>>   This leads to the contradiction that ||0|| = ||0.x1|| = 0||x1||
>>	     		         	       2          2         2
>>   < 0.A.||x1|| = 0.  So ||.|| is not norm.
>>              1               2
     
     There is a flaw in the reasoning here.  When dealing with
     inequalities amon the reals,  you cannot go from
     
                         a < b
     
     to
     
                        ac < bc
     
     If c is positive, you can.  If c is negative, however, you
     get ac > bc.  If c is zero, then you end up with ac = bc = 0.
     
     In .13 you multipled an inequality a < b by c = 0 and then
     incorrectly kept the <.
     
     Dan