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Proof: Norms converge imply Coordinates converge

  1. Feb 3, 2012 #1
    Theorem: Let (X, norm) be a m-dimensional normed linear space. Let {xn} and x be expressed by Xn=λn1*e1+...+λnm*em and X=λ1*e1+...+λm*em. Then xn→x IFF λnk→λk for k=1,...,m.

    The proof for λnk→λk for k=1,...,m implies xn→x is rather straight forward. I am having trouble with the proof for λnk→λk implies xn→x.

    Here is the proof from class lectures.
    View attachment 43442

    This is a proof by contradiction.

    First, since the coordinates do not converge to 0, then at least one set of λnj's do not converge to 0.

    Then a subsequence is built with each element greater than some positive number, r, which must exist since at least one of the λnj's do not converge to 0.

    Observe that each element of this subsequence is bounded below by r and above by the greatest λnk, called Mi.

    Construct yi as xni/Mi. Then yi→0 because xni→0.

    Then I get lost.

    1. I am not sure why |μik|<=1.
    2. I am not sure why sum(|μik|)>=1.

    Can anyone help me fill in these missing parts? I've worked through this proof for over an hour and haven't gotten any further.

    Thanks in advance,

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    Last edited: Feb 3, 2012
  2. jcsd
  3. Feb 3, 2012 #2
    |μik|<=1 because Mi>=xni. So by definition, |μik| would be <=1. Got it.

    Still stuck on the second part though.
    Last edited: Feb 3, 2012
  4. Feb 3, 2012 #3


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    Fact (2) is equivalent to

    [tex]\sum_{k=1}^n |\lambda_k^{n_i}|\geq M_i[/tex]

    This is true since [itex]M_i=\{|\lambda_k^{n_i}|~\vert~1\leq k\leq n\}[/itex]. So there is an element j such that [itex]|\lambda_j^{n_i}|=M_i[/itex]. So if we add more terms to the left side, then we have [itex]\geq[/itex].
  5. Feb 3, 2012 #4


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    you are asking why, two vectors are close if and only if all their coordinates are close, or equivalently, why a vector is small iff all its coordinates are small.

    you hav to look at the definition of the size of a vector in terms of the sizes of its coordinates.

    Usually the size of a vector is defined in a way that its size is always larger than the size of any coordinate, so if the vector is small, then all coordinates are even smaller.

    in the other direction, the size of a vector is usually less that the sum of the sizes of the coords, so if they are all small, and there is only a finite number of them, say 10, then the vector is at most 10 times as large. This is enough.

    So if you just have an abstract norm there are two ways to go. One way is cheating: i.e. use the fact that all norms on a finite diml space are equivalent, and then it follows from proving it for a known standard norm like the euclidean norm, or easier the sum norm.

    or you could actually prove all norms are equivalent, which is pretty much what you need to do for this abstract argument i guess. I.e. we want to show that if | | is any norm say on R^2, then there is a constant C such that |x| ≤ C|x1| + C|x2|, where

    x = (x1,x2). (It turns out this is the easy direcction.)

    now how did this go when i saw it back in 1964? It probably uses compactness of the unit ball in R^2.

    ok i better recall what a norm is, a subadditive, homogeneous, non negative functional.

    so |a1e1+a2e2| ≤ |a1||e1| + |a2||e2|, where e1, e2 is a basis.

    that seems to do it. i.e. if M = max{ |e1|, |e2|}, and x = (a1,a2), then

    (*) |x| ≤ M(|a1| + |a2|). Thus if both a1 and a2 -->0, so will |x|. (Yes this was the easy direction.)

    Now for the other direction. assume |a1e1 + a2e2| is small and try to prove that both a1 and a2 are small.

    i guess i'll use compactness of the unit ball. Ok, think of the usual, or sum norm, on R^2. In that norm the unit ball is closed and bounded and compact.

    now the inequality (*) proved above implies that the unknown norm | | is continuous in the sum norm, hence | | has a minimum value K on the unit ball in the sum norm. (Now I need the assumption that a norm is non zero on non zero vectors, hence K > 0.)

    If |x| can be arbitrarily small without its coordinates x = (a1,a2) being small, then we can find x = (a1,a2) with |x| arbitrarily small and yet one of a1 or a2, say a1, is at least 1.

    This places x on or outside the (sum norm) unit ball which says that |x| ≥ K.

    but that is a contradiction to |x| being arbitrarily small.

    (In your argument, saying that some coordinate was ≥ r, means all your vectors were on or outside the max norm ball of radius r, and the abstract norm is also continuous and bounded below on that ball.)

    this sort of pickiness is why i am not an analyst. of course an analyst would make this look easy.

    aha! micromass to the rescue. Forgive me for not actually reading your proof, it is always easier to make ones own. which you might try yourself instead of reading the one in the lecture.
    Last edited: Feb 3, 2012
  6. Feb 3, 2012 #5
    You both have been amazingly helpful. I'm not sure how I missed this step in class, or again when trying to work through the proof myself.

    Thanks again.
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