
#1
Jan3110, 09:08 AM

P: 146

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 x_{0} in R such that f(x) [tex]\geq[/tex] f(x_{0}) 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 x_{0} in [a,b] such that f(x_{0}) = inf f(x) in the interval [a,b] By definition of an infimum we then know that f(x) [tex]\geq[/tex] f(x_{0}) for all x in R. 



#2
Jan3110, 09:59 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

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]. 



#3
Jan3110, 10:01 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

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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