Inequality proof involving Infs and Sups

  • Thread starter Thread starter homegrown898
  • Start date Start date
  • Tags Tags
    Inequality Proof
homegrown898
Messages
16
Reaction score
0

Homework Statement


Let F and G be bounded functions on S. If f(x) <= g(x) for all x in S prove that inf{f(x):x belongs to S} <= inf{g(x):x belongs to S}


Homework Equations


None

The Attempt at a Solution


Basically the idea is to let L0 = inf{f(x):x belongs to S} and L1 = inf{g(x):x belongs to S} and show that L0 <=L1
 
Physics news on Phys.org
I suggest contradiction. Assume there is some value of x where g(x) is less than the inf of all possible f(x)'s, then see if you can show why that just don't make no sense.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top