2
$\begingroup$

The exercise is as follows: Given a function $f : D \mapsto \mathbb R$ Are the following conditions stronger, weaker or not comparable with continuity?

$$\forall a \in D, \exists \epsilon \gt 0, \forall \delta \gt 0, \forall x \in D: |x-a| \lt \delta \Longrightarrow |f(x) - f(a)|\lt \epsilon $$

The only thing that is different from the $\epsilon$-$\delta$-definition is that the quantifiers are contrary for the $\epsilon$ and $\delta$.

Does it suffice to say that the statement is weaker than continuity because the $\exists$-quantifier is weaker than the $\forall$?

I can't come up with better arguments for it.

That was what we were taught in class but it's so self-explanatory that it shouldn't even be in our exercise lessons.

Another statement is the following:

$$ \forall \epsilon \gt 0, \exists \delta \gt 0,\forall a \in D, \forall x \in D: |x-a| \lt \delta \Longrightarrow |f(x) - f(a)|\lt \epsilon $$

It is a stronger statement and the argument that was used here is :

$Q(a,\delta) \Rightarrow \forall a \qquad \exists \delta \Rightarrow Q(\mu, \delta)$

I don't remember what the $\mu$ stands for? I think the Q is just a symbol to say it's a statement.

Please clarify this.

$\endgroup$
2
$\begingroup$

It is not comparable. A function that is continuous on D need not satisfy that statement, and a function that satisfies that statement need not be continuous.

In general, a statement $A$ is considered to be stronger than $B$ if the statement $A \to B$ is true, but $B \to A$ is false.

As an example, differentiability is stronger than continuity. If a function is differentiable at a point, then it is continuous at that point, but the reverse is not necessarily true.

EDIT- In response to your comment, I don't know what to make of the argument that was used. I don't know what $\mu$ is intended to mean either. It is stronger because a function that satisfies the 2nd statement will necessarily be continuous, but a continuous function need not satisfy the 2nd statement. The 2nd statement is actually the definition of "uniform continuity". As an interesting side note I will mention that when D is a closed bounded interval, the 2nd statement is equivalent to continuity, but in general for arbitrary D uniform continuity will be stronger (for the reasons mentioned above).

$\endgroup$
  • 1
    $\begingroup$ You need to compare the two statements presented here with the definition of continuity (So, $\forall a \in D \forall \epsilon \gt 0 \exists \delta \gt 0 \forall x \in D |x-a|\lt \delta \Rightarrow |f(x) -f(a)|\lt \epsilon$ . Not compare these two statements with eachother. $\endgroup$ – Anonymous196 Jan 17 '18 at 21:05
  • $\begingroup$ O weird because my Professor assistant told us it was weaker or it was my mistake. $\endgroup$ – Anonymous196 Jan 17 '18 at 21:08
  • $\begingroup$ Thx for the example it makes more sense now. $\endgroup$ – Anonymous196 Jan 17 '18 at 21:09
  • $\begingroup$ And do you know the purpose of using the argument for the 2nd statement? $\endgroup$ – Anonymous196 Jan 17 '18 at 21:15
  • $\begingroup$ @AnonymousI Yes and No. See edit $\endgroup$ – David Reed Jan 17 '18 at 21:25
2
$\begingroup$

The first statement:

$$\forall a \in D, \exists \varepsilon > 0, \forall \delta > 0, \forall x \in D: |x-a| < \delta \implies |f(x) - f(a)| < \varepsilon$$

If you think about it hard enough, all it is saying is that the function is bounded.

So, consider the function $f(x) = \begin{cases} 0 & x < 0 \\ 1 & x \ge 0 \end{cases}$. One can verify that $f$ satisfies the first statement (by letting $\varepsilon = 2$ for every $a \in D$). However, it is not continuous.

Consider the function $g(x) = x$. It is continuous, but does not satisfy the first statement.

Therefore, the first statement is not comparable with continuity.


The second statement:

$$\forall \varepsilon > 0, \exists \delta > 0,\forall a \in D, \forall x \in D: |x-a| < \delta \implies |f(x) - f(a)| < \varepsilon$$

This statement is the definition of uniform continuity. It is a standard exercise to prove that every uniformly continuous function is continuous, but that there are continuous functions which are not uniformly continuous, so the second statement is stronger than continuity.

$\endgroup$
  • $\begingroup$ That's very good explained with easy to understand examples. Thank you. $\endgroup$ – Anonymous196 Jan 17 '18 at 22:20
  • $\begingroup$ +1. I think this question was on the Math GRE a few years ago $\endgroup$ – qbert Jan 17 '18 at 22:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.