Help with an epsilon-delta proof

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Homework Statement:

With the definition of continuity and the formal definition of the limit value prove that: Let f:R→R be a function f that is continuous in x=0 and has the property that if x≠0 then f(x)≥0. Then f(0)≥0.

Relevant Equations:

The formal definiton of the limit value
Definition of continuity
I have been struggling with this problem and also my friends. We are not the best at epsilon-delta proof and we have not found an understandable solution to this problem.
 
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  • #2
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I would approach this indirectly. Assume ##f(0)=:c <0##. Then you get a positive gap between the points approaching zero which does not vanish, regardless how close to zero you are. That's the idea.

Continuity in the limit definition means: ##\lim_{x \to 0} f(x) = f(0)##. Continuity in the ##\varepsilon-\delta-##version means: For any ##\varepsilon > 0## there is a ##\delta > 0## such that ##|x|<\delta## implies ##|f(x)-f(0)|<\varepsilon##. Now if ##c<0##, does either of the definitions hold?
 
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  • #3
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I would approach this indirectly. Assume ##f(0)=:c <0##. Then you get a positive gap between the points approaching zero which does not vanish, regardless how close to zero you are. That's the idea.

Continuity in the limit definition means: ##\lim_{x \to 0} f(x) = f(0)##. Continuity in the ##\varepsilon-\delta-##version means: For any ##\varepsilon > 0## there is a ##\delta > 0## such that ##|x|<\delta## implies ##|f(x)-f(0)|<\varepsilon##. Now if ##c<0##, does either of the definitions hold?
I'm not sure if I understand what you mean by the last question.
 
  • #4
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If we have ##f(0)=c < 0## and choose ##\varepsilon := \dfrac{|c|}{3}##, will we find that ##f(x) \stackrel{x\to 0}{\longrightarrow}c## gets arbitrary close (in the limit definition), or will we find a ##\delta > 0## such that ##|f(x)-c|< \varepsilon =c/3## for ##|x|<\delta ## in the other definition?

The ##\varepsilon-\delta## definition of continuity is easier here, since regardless how small ##|x|## is, ##|f(x)-c|## will always exceed ##\varepsilon##.
 
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1568591426669.png

I have done this, so is this wrong? And if not, how do I continue?
 
  • #6
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The idea looks ok, but what do you want to show with it? There is no structure, i.e. no quantifiers.
##|f(x)-f(0)|<\varepsilon## means ##-\varepsilon < f(x)-f(0) < \varepsilon##. Now why can't this be for any small ##|x|## and for which value of ##\varepsilon##? Or is it always true, regardless what ##f(0)## is?
 
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  • #7
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@fresh_42 gave an indirect proof. Here are two direct proofs (maybe only the first proof is relevant to you).

Proof 1:

Fix ##\epsilon >0##. Choose ##\delta>0## such that ##|f(y)-f(0)|< \epsilon## whenever ##|y|<\delta##. Taking ##y=\delta/2## (or another strictly positive value smaller than delta), we see that ##|f(0)|=|f(y)-f(0)|<\epsilon##.

Conclusion: we have proven that

$$\forall \epsilon >0: |f(0)|<\epsilon$$

Hence, it follows that ##|f(0)|=0##, thus ##f(0)=0##.

Proof 2:

Since ##f## is continuous at ##0##, we have

$$f(0)=\lim_{x\to 0, x \neq 0} f(x) = 0$$
 
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  • #8
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@fresh_42 gave an indirect proof. Here are two direct proofs (maybe only the first proof is relevant to you).

Proof 1:

Fix ##\epsilon >0##. Choose ##\delta>0## such that ##|f(y)-f(0)|< \epsilon## whenever ##|y|<\delta##. Taking ##y=\delta/2## (or another strictly positive value smaller than delta), we see that ##|f(0)|=|f(y)-f(0)|<\epsilon##.

Conclusion: we have proven that

$$\forall \epsilon >0: |f(0)|<\epsilon$$

Hence, it follows that ##|f(0)|=0##, thus ##f(0)=0##.

Proof 2:

Since ##f## is continuous at ##0##, we have


$$f(0)=\lim_{x\to 0, x \neq 0} f(x) = 0$$
If I use the first proof, why is y=delta/2?
 
  • #9
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If I use the first proof, why is y=delta/2?
It is just a choice I made.

For any ##y\in \mathbb{R}## with ##|y|<\delta##, it holds that ##|f(y)-f(0)|<\epsilon##. In particular, it is true that ##|f(\delta/2)-f(0)|< \epsilon##.

(But also ##\delta/3, \delta/100## or any number ##y## with ##0<|y|<\delta## will make the proof work.)
 
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  • #10
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One way to prove it is with the help of the following lemma:

Lemma: if ##\lim_{x \to x_0}f(x)=L\neq 0## then there exists a neighborhood ##\Pi## of ##x_0## such that ##f(x)## and ##L## have the same sign for ##x\in \Pi##.

You can prove this lemma with an epsilon-delta proof. This lemma holds even if the function is not continuous at ##x_0##

Now since we know the function is continuous at ##x_0=0## it is

##\lim_{x \to 0} f(x)=f(0)##

We take cases:
1) ##f(0)=0##, got nothing to prove here
2) ##f(0)\neq 0##. You can use the lemma and conclude that ##f(0)## will be positive.
 
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  • #11
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@fresh_42 gave an indirect proof. Here are two direct proofs (maybe only the first proof is relevant to you).

Proof 1:

Fix ##\epsilon >0##. Choose ##\delta>0## such that ##|f(y)-f(0)|< \epsilon## whenever ##|y|<\delta##. Taking ##y=\delta/2## (or another strictly positive value smaller than delta), we see that ##|f(0)|=|f(y)-f(0)|<\epsilon##.

Conclusion: we have proven that

$$\forall \epsilon >0: |f(0)|<\epsilon$$

Hence, it follows that ##|f(0)|=0##, thus ##f(0)=0##.
I don't follow this at all. Why must ##f(0)=0##? We were only to show ##f(0) \ge 0##.
 
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  • #12
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I don't follow this at all. Why must ##f(0)=0##? We were only to show ##f(0) \ge 0##.
I misread the question. I thought it said that ##f(x)=0## for all ##x\neq 0##. Thanks for noticing.
 
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  • #13
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Homework Statement: With the definition of continuity and the formal definition of the limit value prove that: Let f:R→R be a function f that is continuous in x=0 and has the property that if x≠0 then f(x)≥0. Then f(0)≥0.
Homework Equations: The formal definiton of the limit value
Definition of continuity

I have been struggling with this problem and also my friends. We are not the best at epsilon-delta proof and we have not found an understandable solution to this problem.
In general, for a proof like this you have to decide how to approach it. There are two obvious strategies:

A) A direct proof. You have:

##f## is continuous at ##x = 0## and ##f(x) \ge 0## for ##x \ne 0##

You can use these two premises to show directly that ##f(0) \ge 0##

B) Proof by contradiction. In this case, you assume:

##f## is continuous at ##x = 0##, ##f(x) \ge 0## for ##x \ne 0## and ##f(0) < 0##

From these three premises you achieve a contradiction, which means that if the first two hold, the third must be false. (As a side note, in this case you will also have proved that if any two hold, then the third must be false. E.g. if ##f(x) \ge 0## for ##x \ne 0## and ##f(0) < 0##, then ##f## is not continuous at ##x =0##. But, let's not get side-tracked.)

That's your first decision.

Next, if you are stuck, then you could try to find a counter-example. It's a very good idea to try to do this. You might learn a lot about why something is true if you try to find a function that shows it is false. In this case, what happens if you try to find a function with ##f(0) < 0##? Why can't you find such a function?

Finally, if something is ##< 0##, then it must have some definite negative value. It's often a good idea to introduce a term to denote this. So, for example, you could say:

If ##f(0) < 0##, then let ##f(0) = -c##, where ##c > 0##.

That gives you something to work with. Of course, it's possible just to work with ##f(0)## as a negative number, and ##-f(0)## as a positive number. But, introducing a simple number like ##c## can sometimes clarify what you are doing.
 
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  • #14
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Thank you so much for your help everyone! I have finally solved it! :)
 
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