Continuous bounded function - analysis

[Kate2010]

Homework Statement

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.

Homework Equations

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.

 

[HallsofIvy]

Homework Statement

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.

Homework Equations

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].

 

[HallsofIvy]

Homework Statement

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.

Homework Equations

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].