Proof about continuous function related to balls and sets

In summary: Do you think that would make a difference in the conclusion?In summary, the homework statement is that if a continuous function f(x)>0 there exists a ball centered at p such that f(x)>0.
  • #1
Hernaner28
263
0

Homework Statement


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

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

Homework Equations


The Attempt at a Solution



f is continuous, that is:

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

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

Thank you!
 
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  • #2
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
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
He is making a specific choice for [itex]\epsilon[/itex]. While there are many possible, f(p)/2 is one choice that will help here. Do you see why?
 
  • #5
LCKurtz said:
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 said:
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.
 
  • #6
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
Hernaner28 said:
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.
 
  • #8
Yes. It would be:

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

But what is this for? Just setting [tex] \displaystyle \varepsilon =\frac{f\left( p \right)}{2}[/tex] 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
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
Hernaner28 said:
Yes. It would be:

[tex] \displaystyle \forall \varepsilon >0[/tex]

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

[tex] \displaystyle \exists \delta >0[/tex] such that if: [tex] \displaystyle |x-p|<\delta [/tex]
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 [tex] \displaystyle |f\left( x \right)-f\left( p \right)|<\frac{f\left( p \right)}{2}[/tex].

What does that last inequality tell you about ##f(x)##? Unscramble it without the absolute value signs.
 
  • #11
Yeah, now I get it. It implies that:

[tex] \displaystyle \frac{f\left( p \right)}{2}<f\left( x \right)<\frac{3}{2}f\left( p \right)[/tex]

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
Hernaner28 said:
Yeah, now I get it. It implies that:

[tex] \displaystyle \frac{f\left( p \right)}{2}<f\left( x \right)<\frac{3}{2}f\left( p \right)[/tex]

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.
 
  • #13
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.
 

What is a continuous function?

A continuous function is a type of mathematical function that maintains a consistent and unbroken graph. This means that there are no sudden jumps or breaks in the graph, and the output values change smoothly as the input values change. In other words, small changes in the input result in small changes in the output.

How is continuity related to balls and sets?

In mathematics, a ball is defined as a set of points that are all within a certain distance from a given point, called the center. A set is a collection of elements. Continuity is related to balls and sets because it describes how the elements in a set or the points in a ball are connected and how they behave as the input values change.

What is the definition of a continuous function?

A function f is continuous at a point c if the limit of f(x) as x approaches c is equal to f(c). In other words, the function is continuous if the value of the function at a specific point is equal to the limit of the function as the input approaches that point.

How do you prove that a function is continuous?

To prove that a function is continuous, you must show that it satisfies the definition of continuity. This can be done by using the epsilon-delta definition, which states that for any small number ε, there exists a corresponding small number δ such that if the distance between the input values is less than δ, the difference between the output values is less than ε. This proves that the function is continuous at a specific point.

What is the importance of continuity in mathematics and science?

Continuity is important in mathematics and science because many real-world phenomena can be described and modeled using continuous functions. It allows us to make predictions and understand the behavior of these phenomena. Additionally, the concept of continuity is fundamental in the development of calculus and other branches of mathematics.

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