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

[infinitylord]

Homework Statement

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)\geq0, with equality iff v=0
2) f(av)=|a|f(v) for any scalar a
3) f(v+w) \leq 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!

 

[mfb]

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.

 

[D H]

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.

Try that again. Surely you can come up with a non-zero vector whose norm is zero per ||\mathbf v|| = \min (|v_1|, |v_2|, \cdots, |v_n|).

 

[pasmith]

Homework Statement

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)\geq0, with equality iff v=0
2) f(av)=|a|f(v) for any scalar a
3) f(v+w) \leq 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) = man(|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.

Consider v = (1,0, \dots, 0). Then v \neq 0 so by (1) its norm must be strictly positive.

 

[infinitylord]

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:
v₁+w₂\leqv₁-v₁+w₂-w₂.
Therefore, 2<0. Which is completely untrue.

 

[mfb]

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