# Proof about continuous function related to balls and sets

1. Aug 25, 2012

### Hernaner28

1. The problem statement, all variables and given/known data
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 $$\displaystyle f\left( x \right)>0$$.

2. Relevant equations

3. 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!

2. Aug 25, 2012

### 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 $\epsilon = \frac {f(p)}{2}$?

3. Aug 25, 2012

### 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!

4. Aug 25, 2012

### 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?

5. Aug 25, 2012

### LCKurtz

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.

6. Aug 25, 2012

### 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!

7. Aug 25, 2012

### LCKurtz

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

8. Aug 25, 2012

### 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

9. Aug 25, 2012

### 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?

10. Aug 25, 2012

### LCKurtz

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

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

11. Aug 25, 2012

### Hernaner28

Yeah, now I get it. It implies that:

$$\displaystyle \frac{f\left( p \right)}{2}<f\left( x \right)<\frac{3}{2}f\left( p \right)$$

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?

12. Aug 25, 2012

### LCKurtz

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.

13. Aug 25, 2012

### 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.