Why Does Boundedness of a Function Affect Its Oscillation?

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Suppose that f is bounded by M. Prove that ω(f^2,[a,b])≤2Mω(f,[a,b]).

I can show that ω(f,[a,b])≤2M and that ω(f^2,[a,b])≤M^2 but this procedure is getting me nowhere. I also have a similar problem that likely calls for the same approach:

Suppose that f is bounded below by m and that m is a positive number. Prove that ω(1/f,[a,b])≤ω(f,[a,b])/m^2.

This one I think I have right but my instructor is telling me that it's wrong. Since all values are positive, by the nature of \frac{1}{x}, \displaystyle\sup f = \frac{1}{\displaystyle\inf f} and \displaystyle\inf f = \frac{1}{\displaystyle\sup f}. We can now analyze the oscillation as follows:

ω(1/f,[a,b])=\frac{1}{\displaystyle\inf f}-\frac{1}{\displaystyle\sup f}=\frac{ω(f,[a,b])}{(\displaystyle\inf f)(\displaystyle\sup f)}≤\frac{ω(f,[a,b])}{m^2}
 
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Arkuski said:
Suppose that f is bounded by M. Prove that ω(f^2,[a,b])≤2Mω(f,[a,b]).
It's probably not the most elegant, but you could try breaking it into separate cases according to the signs of sup f and inf f.
\displaystyle\sup f = \frac{1}{\displaystyle\inf f}
I think you mean \displaystyle\sup \frac{1}{f} = \frac{1}{\displaystyle\inf f} etc. Other than that, your proof looks fine.
 
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...
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