Help with an epsilon-delta proof

[Kolika28]

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.

 

[fresh_42]

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?

 

[Kolika28]

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.

 

[fresh_42]

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 $\varepsilon$.

 

[Kolika28]

I have done this, so is this wrong? And if not, how do I continue?

 

[fresh_42]

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?

 

[member 587159]

@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

$$f(0)=\lim_{x\to 0, x \neq 0} f(x) = 0$$

 

[Kolika28]

@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?

 

[member 587159]

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.)

 

[Delta2]

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.

 

[PeroK]

@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 $f(0) \ge 0$.

 

[member 587159]

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.

 

[PeroK]

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.

 

[Kolika28]

Thank you so much for your help everyone! I have finally solved it! :)