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Continuous bounded function - analysis

by Kate2010
Tags: analysis, bounded, continuous, function
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Kate2010
#1
Jan31-10, 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 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.
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HallsofIvy
#2
Jan31-10, 09:59 AM
Math
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P: 39,348
Quote Quote by Kate2010 View Post
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.
There is a fundamental problem with your proof. You have taken x0 such that [itex]f(x)\ge f(x_0)[/itex] in [a, b]. But [a, b] is just "some" interval. There is no reason to think that f does not have smaller values off [a,b]. Also notice that you have NOT used the fact that f goes to infinity.

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].
HallsofIvy
#3
Jan31-10, 10:01 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,348
Quote Quote by Kate2010 View Post
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?)
I have no idea what you even mean by that!

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.
There is a fundamental problem with your proof. You have taken x0 such that [itex]f(x)\ge f(x_0)[/itex] in [a, b]. But [a, b] is just "some" interval. There is no reason to think that f does not have smaller values off [a,b]. Also notice that you have NOT used the fact that f goes to infinity.

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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