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||L||, transformations

  1. Sep 18, 2008 #1
    1. The problem statement, all variables and given/known data

    for L, M: V -> W, L, M, linear let||L|| = sup{|L(v)|: v in V, |v| <= 1}

    show ||L + M|| < ||L|| + ||M||

    2. Relevant equations

    3. The attempt at a solution

    so is it true that if |L(x) + M(x)| defines a sup for L + M (x for which |L(x) + M(x)| is the sup), then it also defines a sup for L and sup for M, as L and M are both defined on V? im getting caught up on this and im thinking maybe theres a simpler way, i.e., either defining sets and using the standard results for sup on sets, or to exploit the linearity in a clever way

    think im overlooking something simple
  2. jcsd
  3. Sep 18, 2008 #2


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    Homework Helper

    Pick a vector [tex] \mathbf v [/tex] such that [tex] \|\mathbf v \| \le 1 [/tex].
    Then what can you do with this?

    |L(\mathbf v) + M(\mathbf v)| \le |L(\mathbf v)| + |M(\mathbf v)|

    (think about the given norm and the r.h.s. first, then think again about the definition of this norm and the l.h.s. remember that [tex] \mathbf v [/tex] is arbitrary )
  4. Sep 18, 2008 #3
    suppose we have v satisfying the conditions. then |L(v) + M(v)| <= |L(v)| + |M(v)| <= ||L|| + ||M|| so sup(|L(v) + M(v)|) = ||L + M|| <= ||L|| + ||M||.
  5. Sep 18, 2008 #4


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    Homework Helper


    One minor quibble - you would usually see your final line written in this order:

    ||L + M || = \sup_{||\mathbf v || =1}|(L+M)\mathbf v| \le ||L|| + ||M||
  6. Sep 18, 2008 #5
    i got mthis solution in few second but am spending several hours thinking about it

    but i do not have confidence i am thinking "this is silly something i did wrong, this norm can be complex structure etc" thinking about something extra etc.etc

    okay thanks for this very much :)

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