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Continuous bounded function - analysis |
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| Jan31-10, 09:08 AM | #1 |
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Continuous bounded function - analysis
1. The problem statement, all variables and given/known data
Assume the theorem that a continuous bounded function on a closed interval is bounded and attains its bounds. Prove that if f: R -> R is continuous and tends to +[tex]\infty[/tex] as x tends to +/- [tex]\infty[/tex] then there exists an x0 in R such that f(x) [tex]\geq[/tex] f(x0) for all x in R. 2. Relevant equations 3. The attempt at a solution I think I understand the basic idea behind this but I'm not sure that my proof is rigorous enough. R is not closed or bounded, however as x tends to +/- infinity f(x) tends to infinity, so won't affect its minimum value (do I need to prove this? If so, how?) So to consider the minimum value of f we can consider a closed bounded interval [a,b] a,b in R By the assumed theorem, f is bounded on this interval and attains its bounds, so there exists an x0 in [a,b] such that f(x0) = inf f(x) in the interval [a,b] By definition of an infimum we then know that f(x) [tex]\geq[/tex] f(x0) for all x in R. |
| Jan31-10, 09:59 AM | #2 |
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But that's easily fixed. Take "a" to be any real number. Since f goes to infinity as x goes to either infinity or negative infinity, there exist a positive number, N, such that if x> N or x< -N, f(x)> f(a). NOW use your proof on the interval [-N, N]. |
| Jan31-10, 10:01 AM | #3 |
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But that's easily fixed. Take "a" to be any real number. Since f goes to infinity as x goes to either infinity or negative infinity, there exist a positive number, N, such that if x> N or x< -N, f(x)> f(a). NOW use your proof on the interval [-N, N]. |
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