Proof about continuous function related to balls and sets

[Hernaner28]

Homework Statement

Let \displaystyle f:{{\mathbb{R}}^{n}}\to \mathbb{R} a continuous function. Proove that:

If \displaystyle f\left( p \right)>0 then there's a ball \displaystyle {{B}_{p}} centered at p such that \displaystyle \forall x\in {{B}_{p}} we have \displaystyle f\left( x \right)>0.

Homework Equations

The Attempt at a Solution

f is continuous, that is:

\displaystyle \forall \varepsilon >0 \displaystyle \exists \delta >0 such that \displaystyle f\left( B\left( p,\delta \right) \right)\subset B\left( f\left( p \right),\varepsilon \right).

Let f(p)>0, then I cannot conclude anthing with the above hypothesis. What's missing?

Thank you!

 

[LCKurtz]

Let's talk for a moment about a continuous f(x) on the real line. If f(p) > 0 do you see how to find an interval about p where f(x) > 0? What happens if $\epsilon = \frac {f(p)}{2}$?

 

[Hernaner28]

I don't understand why you ask that and why you say that at the end. Is there seomthing wrong with my attempt?

Thank you!

 

[HallsofIvy]

He is making a specific choice for \epsilon. While there are many possible, f(p)/2 is one choice that will help here. Do you see why?

 

[LCKurtz]

Let's talk for a moment about a continuous f(x) on the real line. If f(p) > 0 do you see how to find an interval about p where f(x) > 0? What happens if $\epsilon = \frac {f(p)}{2}$?

I don't understand why you ask that and why you say that at the end. Is there seomthing wrong with my attempt?

Thank you!

Because I am trying to help you see how to work your problem by working a simpler one. How about addressing my questions? You might learn something in the process.

 

[Hernaner28]

Hmm.. alright. Answering your question I honestly don't remember exactly how to find an interval about p where f(x) > 0. Could you please remind me?

Thank you both!

 

[LCKurtz]

Hmm.. alright. Answering your question I honestly don't remember exactly how to find an interval about p where f(x) > 0. Could you please remind me?

Thank you both!

If $\epsilon = \frac {f(p)}{2}$, write out carefully what the $\delta,\epsilon$ definition of continuity of $f$ at $p$ tells you.

 

[Hernaner28]

Yes. It would be:

\displaystyle \forall \varepsilon >0 \displaystyle \exists \delta >0 such that if: \displaystyle |x-p|<\delta then \displaystyle |f\left( x \right)-f\left( p \right)|<\frac{f\left( p \right)}{2}.

But what is this for? Just setting \displaystyle \varepsilon =\frac{f\left( p \right)}{2} it's not enough to find an interval about p such that f(x)>0. It also depends on delta... doesn't it?

Thank you

 

[Hernaner28]

Ahhh I see it now!. If you choose that epsilon you get forced to choose a delta such that the interval about p of radious delta has positive image. Is it right?

 

[LCKurtz]

Yes. It would be:

\displaystyle \forall \varepsilon >0

In particular for $\epsilon = \frac{f(p)}{2}$

\displaystyle \exists \delta >0 such that if: \displaystyle |x-p|<\delta

So you have a $\delta$ that works for this value of $\epsilon$. And doesn't $|x-p|<\delta$ describe an interval about $p$ that goes from $p-\delta$ to $p+\delta$?

then \displaystyle |f\left( x \right)-f\left( p \right)|&lt;\frac{f\left( p \right)}{2}.

What does that last inequality tell you about $f(x)$? Unscramble it without the absolute value signs.

 

[Hernaner28]

Yeah, now I get it. It implies that:

\displaystyle \frac{f\left( p \right)}{2}&lt;f\left( x \right)&lt;\frac{3}{2}f\left( p \right)

That is: f(x)>0 . So the interval about p exists and it's the one of radious delta for that epsilon. Now it's almost the same but for balls, isn't it?

Thank you very much!

Edit. I think the definition of continuity I used in my attempt is not useful in this demonstration. I think I should use the equivalent definition with distances, shouln't I?

 

[LCKurtz]

Yeah, now I get it. It implies that:

\displaystyle \frac{f\left( p \right)}{2}&lt;f\left( x \right)&lt;\frac{3}{2}f\left( p \right)

That is: f(x)>0 . So the interval about p exists and it's the one of radious delta for that epsilon. Now it's almost the same but for balls, isn't it?

Thank you very much!

Yes, it's almost the same proof. It is frequently a good idea, when trying to prove something for $R^n$ to make sure you understand it first for $R^1$ or $R^2$.

You're welcome.

 

[Hernaner28]

Yeah, I think the problem was that we were given three equivalent definitions and in my attempt I used the definition of continuityn (for several variables) involving balls but I think in this case I should have used the definition involving distances or norm, those are closer to the one for one variable.