Continuous and open function and local extrema

[mahler1]

Homework Statement .
Let $f:ℝ→ℝ$ be an open and continuous function. Prove that f doesn't have local extrema The attempt at a solution.

I suppose there is some $x_0 \in ℝ$ and some $ε>0$ such that $f(x_0)≤f(x)$ for all $x \in (x_0-ε,x_0+ε)$ (the proof for relative maximum is analogue to this one). Now, I consider the open interval $(x_0-ε,x_0+ε)$. Then, $f( {(x_0-ε,x_0+ε)} )$ is open by our hypothesis. Then, there exists $δ>0$ such that $B(f(x_0),δ) \subset f({(x_0-ε,x_0+ε)})$. I want to come to an absurd, this means, find some $z \in (x_0-ε,x_0+ε)$ such that $f(z)<f(x_0)$, but I don't know how to get here, I must use continuity at some point. How could I get the wanted z?

By the way, sorry for the notation $f({(x_0-ε,x_0+ε)})$, it should be f({(x_0-ε,x_0+ε)}), with brackets, but I couldn't put it this way with latex.

 

[pasmith]

Homework Statement .
Let $f:ℝ→ℝ$ be an open and continuous function. Prove that f doesn't have local extrema


The attempt at a solution.

I suppose there is some $x_0 \in ℝ$ and some $ε>0$ such that $f(x_0)≤f(x)$ for all $x \in (x_0-ε,x_0+ε)$ (the proof for relative maximum is analogue to this one). Now, I consider the open interval $(x_0-ε,x_0+ε)$. Then, $f( {(x_0-ε,x_0+ε)} )$ is open by our hypothesis. Then, there exists $δ>0$ such that $B(f(x_0),δ) \subset f({(x_0-ε,x_0+ε)})$. I want to come to an absurd, this means, find some $z \in (x_0-ε,x_0+ε)$ such that $f(z)<f(x_0)$, but I don't know how to get here, I must use continuity at some point. How could I get the wanted z?

You have $f(x_0) - \frac12\delta \in B(f(x_0),\delta) \subset f((x_0 - \epsilon, x_0 + \epsilon))$. Hence $f(x_0) - \frac12\delta \in f((x_0 - \epsilon, x_0 + \epsilon))$.

By the way, sorry for the notation $f({(x_0-ε,x_0+ε)})$, it should be f({(x_0-ε,x_0+ε)}), with brackets, but I couldn't put it this way with latex.

You don't need the brackets here. $(x_0 - \epsilon, x_0 + \epsilon)$ is a subset of $\mathbb{R}$ so its image under f is $f((x_0 - \epsilon, x_0 + \epsilon))$. $\{(x_0 - \epsilon, x_0 + \epsilon)\}$ is a singleton set whose only member is a subset of $\mathbb{R}$, and is not a itself a subset of $\mathbb{R}$.

For future reference, set brackets in LaTeX are produced by escaping: \{ \} gives $\{ \}$.

 

[mahler1]

You have $f(x_0) - \frac12\delta \in B(f(x_0),\delta) \subset f((x_0 - \epsilon, x_0 + \epsilon))$. Hence $f(x_0) - \frac12\delta \in f((x_0 - \epsilon, x_0 + \epsilon))$.




You don't need the brackets here. $(x_0 - \epsilon, x_0 + \epsilon)$ is a subset of $\mathbb{R}$ so its image under f is $f((x_0 - \epsilon, x_0 + \epsilon))$. $\{(x_0 - \epsilon, x_0 + \epsilon)\}$ is a singleton set whose only member is a subset of $\mathbb{R}$, and is not a itself a subset of $\mathbb{R}$.

For future reference, set brackets in LaTeX are produced by escaping: \{ \} gives $\{ \}$.

Oh, I've mixed up the notation.

 

[brmath]

Suppose f does have a minimum at $x_0$. Then there are $\epsilon_1 \text{ and } \epsilon_2$ such that $f(x_0-\epsilon_1) = f(x_0+\epsilon_2)$. That is because f is continuous.

Now what does this do to your openness?

 

[mahler1]

Suppose f does have a minimum at $x_0$. Then there are $\epsilon_1 \text{ and } \epsilon_2$ such that $f(x_0-\epsilon_1) = f(x_0+\epsilon_2)$. That is because f is continuous.

Now what does this do to your openness?

Sorry, maybe it's obvious, but I don't see why continuity would imply $f(x_0-\epsilon_1) = f(x_0+\epsilon_2)$

 

[brmath]

Hi,

It's a local minimum, which means the function is rising on both sides of it, as least in some interval. And the function is continuous which means it satisfies the intermediate value condition -- that is it can't get from here to there without passing thru everything in between. So on both the left and right it is going to go continuously thru a bunch of values, and there will be some equalities.

Does that help? I did think it was obvious, and am finding it hard to explain. Perhaps you could draw a picture? Try y = $x^2$. Then try something not so symmetric.

 

[mahler1]

Hi,

It's a local minimum, which means the function is rising on both sides of it, as least in some interval. And the function is continuous which means it satisfies the intermediate value condition -- that is it can't get from here to there without passing thru everything in between. So on both the left and right it is going to go continuously thru a bunch of values, and there will be some equalities.

Does that help? I did think it was obvious, and am finding it hard to explain. Perhaps you could draw a picture? Try y = $x^2$. Then try something not so symmetric.

It certainly does help, now I get it but I realize I have to study some more single variable calculus.

 

[brmath]

It certainly does help, now I get it but I realize I have to study some more single variable calculus.

You don't need a lot of calculus in this case -- you just need to understand what a continuous function on an interval is.

Would you agree that if f(q) = f(r) for q,r in the interval than f cannot be open?