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Homework Help: Min(|v|) and max(|v|) in relation to norms of a vector

  1. Sep 18, 2014 #1
    1. The problem statement, all variables and given/known data
    I have a homework problem in honors calculus III that I'm having a little trouble with. Given these three qualities of norms in Rn:

    1) f(v)[itex]\geq[/itex]0, with equality iff v=0
    2) f(av)=|a|f(v) for any scalar a
    3) f(v+w) [itex]\leq[/itex] f(v)+f(w)

    we were given a set of 3 functions and told that 2 were norms and 1 was not a norm. I very easily classified that f(v1,...,vn)=|v1|+...+|vn| was a norm using these three properties. the two left were g(v1,...vn) = min(|v1|,...,|vn|) and h(v1,...vn) = max(|v1|,...,|vn|). I determined through some research that the maximum function was a norm and the minimum function was not. But I don't know why that is. I tried using the triangle inequality 3) to prove this, but I came up with the inequality being true for both the max and min. I'm really not sure what to do from here as I believe that both of the first 2 properties of norms work for the max and min. I was reliant on the triangle inequality being the counterexample I needed for the minimum function. If someone could help me out I'd greatly appreciate it!
    Last edited: Sep 18, 2014
  2. jcsd
  3. Sep 18, 2014 #2


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    There is a counterexample for the triangle inequality, you just need some clever choice of vectors to test. Two vectors with two components each are sufficient, if you just test some cases you should find a counterexample.
  4. Sep 18, 2014 #3

    D H

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    Try that again. Surely you can come up with a non-zero vector whose norm is zero per [itex]||\mathbf v|| = \min (|v_1|, |v_2|, \cdots, |v_n|)[/itex].
  5. Sep 18, 2014 #4


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    Consider [itex]v = (1,0, \dots, 0)[/itex]. Then [itex]v \neq 0[/itex] so by (1) its norm must be strictly positive.
  6. Sep 18, 2014 #5
    Thanks guys! I got it now I believe. I used the triangle inequality with v=(1,0) and w=(0,1). That way (after some simplification) it turns into:
    Therefore, 2<0. Which is completely untrue.
  7. Sep 18, 2014 #6


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    That's more complicated than necessary, but it certainly works.
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