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Showing sup{f(x)} - inf{f(x)} >= sup{|f(x)|} - inf{|f(x)|}

  • Thread starter jdinatale
  • Start date
  • #1
155
0
Hello, below I have typed the problem and drew a diagram to help me think about the problem.

MMMMM.png


It seems intuitive to me that [itex]M \leq M'[/itex] since M' might be the absolute value of the most extreme negative f value that is greater in magnitude than the most extreme positive f value. Also, [itex]m \leq m' [/itex] since [itex]m' \geq 0[/itex] and m could be negative.

Is this type of problem best handled with cases such as Case 1: [itex]f(x) \leq 0[/itex]?
 

Answers and Replies

  • #2
Bacle2
Science Advisor
1,089
10
I think working out the individual cases may do it. It seems to come down to showing
that:

A-B ≥ |A|-|B|

So that you get equality. Only real case is when A,B have different signs. Maybe
you can show it geometrically, using |X| asthe distance from X to a fixed value
(thinking of 0 in the real line).
 

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