# Continuous bounded function - analysis

by Kate2010
Tags: analysis, bounded, continuous, function
 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 +$$\infty$$ as x tends to +/- $$\infty$$ then there exists an x0 in R such that f(x) $$\geq$$ 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) $$\geq$$ f(x0) for all x in R.
Math
Emeritus
Thanks
PF Gold
P: 39,682
 Quote by Kate2010 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 +$$\infty$$ as x tends to +/- $$\infty$$ then there exists an x0 in R such that f(x) $$\geq$$ 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) $$\geq$$ f(x0) for all x in R.
There is a fundamental problem with your proof. You have taken x0 such that $f(x)\ge f(x_0)$ 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].
Math
Emeritus
Thanks
PF Gold
P: 39,682
 Quote by Kate2010 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 +$$\infty$$ as x tends to +/- $$\infty$$ then there exists an x0 in R such that f(x) $$\geq$$ 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) $$\geq$$ f(x0) for all x in R.
There is a fundamental problem with your proof. You have taken x0 such that $f(x)\ge f(x_0)$ 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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