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Bilinear functions

  1. Jun 29, 2007 #1
    1. The problem statement, all variables and given/known data
    Let f: R^n X R^m --> R^p be a bilinear function.
    Prove that |f(h, k)|/|(h, k)| --> 0 as (h, k) --> 0 (zero vector in R^(n+m)).

    2. Relevant equations
    If f: R^n X R^m --> R^p is bilinear, then for x, x1, x2 in R^n, y, y1, y2 in R^m, a in R:
    a) f(ax, y) = f(x,ay) = af(x,y)
    b) f(x1+x2,y) = f(x1,y) + f(x2,y)
    c) f(x,y1+y2) = f(x,y1) + f(x,y2)

    3. The attempt at a solution
    Straight from delta-epsilon definition of limit, we have:
    for every epsilon>0, there exist delta>0 s.t. |(h,k)|<delta -> |f(h,k)|/|(h,k)|<epsilon
    That means delta>|f(h,k)|/epsilon

    I'm stuck on this. How do I find a bound on |f(h,k)| using the bilinear properties given?
    Last edited: Jun 30, 2007
  2. jcsd
  3. Jul 1, 2007 #2


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    If you can show there is some constant A such that |f(h,k)|<A for all h,k on the unit sphere, then you have for any h,k:

    [tex] |f(h,k)| = |h||k| |f(h/|h|,k/|k|)| < A|h||k| [/tex]

    from which you should be able to get what you need.
  4. Jul 1, 2007 #3
    Thanks for the reply.

    From what was given in the question, I don't see how to get a bound on |f(u,v)| where u, v are the unit vector for h and k respectively.

    First of all, I'm a bit confused by the notion of |(h,k)|. Let's say h is in R^2 and k is in R^3, then what does |((3,4), (1,2,3)| evaluate to?
  5. Jul 1, 2007 #4


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    Use linearity. For example, if you had an ordinary linear functional f, its values are determined by its values on a basis e_i, since any vector v can be written a_1 e_1 + ... +a_n e_n, so that f(v) = a_1 f(e_1) + ... +a_n f(e_n). Then by the triangle inequality:

    [tex] |f(v)| \leq |a_1| |f(e_1)| + ... + |a_n| |f(e_n)| [/tex]

    The advantage of this is that there are only finitely many values |f(e_i)|, so this set definitely does have a max. I'll let you try to finish the proof and extend it to the case of bilinear functions.

    And from what you said, it sounds like they want you to consider the pairs of vectors as vectors in Rn+m, so the norm would be:

    [tex] |(h,k)|^2 = {h_1}^2 + ... + {h_n}^2 + {k_1}^2 + ... + {k_m}^2 = |h|^2 + |k|^2 [/tex]
  6. Jul 3, 2007 #5
    Well, I understand the linear case but I can't quite figure out the bilinear case. Would you enlighten me on that?

    However, looking at another angle, wouldn't f(x,y) be in a general form of f(x,y)=Ax+By+C, where A is p*n matrix, B is p*m matrix and C is in R^p? Therefore, express P in terms of elements in A, B, C, and h, k, then definitely, |f(x,y)| has a bound.

    However, this is quite trivial, and doesn't involve the bilinear property. Therefore, I wonder if it's the right path?
  7. Jul 5, 2007 #6


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    No, a bilinear function satisfies f(ax,y)=f(x,ay)=af(x,y), which your form doesn't (incidentally, ignoring the C, which shouldn't be there since for linear functions (in the vector space sense) zero goes to zero, that's the form of a (singly) linear function from Rn+m to Rp).

    Put aijk equal to the kth component of f(e_i,e_j), where e_i=(0,...,0,1,0,...,0), with the one in the ith place. Then:

    [tex] f(x,y) = f(\sum_i x_i e_i,\sum_j y_j e_j) [/tex]

    [tex] = \sum_i x_i f(e_i,\sum_j y_j e_j) = \sum_i x_i \sum_j y_j f(e_i, e_j) [/tex]

    Then since [itex]f(e_i, e_j) = \sum_k a_{ijk} e_k[/itex], we get:

    [tex] f(x,y) = \sum_{ijk} a_{ijk} x_i y_j e_k [/tex]

    So if you want to use matrices, you'll need to use 3 dimensional ones. But my suggestion was to write:

    [tex] |f(x,y)| = |\sum_{ijk} a_{ijk} x_i y_j e_k |[/tex]

    [tex] \leq \sum_{ijk} |a_{ijk}| |x_i| |y_j| |e_k|[/tex]

    and use this to get a bound.
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