A game with $\delta$, $\epsilon$ and uniform continuity.

Question

UPDATE: Bounty awarded, but it is still shady about what f) is.

In Makarov's Selected Problems in Real Analysis there's this challenging problem:

Describe the set of functions $f: \mathbb R \rightarrow \mathbb R$ having the following properties ($\epsilon, \delta,x_1,x_2 \in \mathbb R$) :

a) $\forall \epsilon \qquad\qquad, \exists \delta>0 \qquad, |x_1-x_2| < \delta \Rightarrow |f(x_1)-f(x_2)|<\epsilon$

b) $\forall \epsilon >0 \qquad, \exists \delta \qquad \qquad, |x_1-x_2| < \delta \Rightarrow |f(x_1)-f(x_2)|<\epsilon$

c) $\forall \epsilon >0 \qquad, \exists \delta>0 \qquad, (x_1-x_2) < \delta \Rightarrow |f(x_1)-f(x_2)|<\epsilon$

d) $\forall \epsilon >0 \qquad, \forall \delta>0 \qquad, |x_1-x_2| < \delta \Rightarrow |f(x_1)-f(x_2)|<\epsilon$

e) $\forall \epsilon >0 \qquad, \exists \delta>0 \qquad, |x_1-x_2| < \delta \Rightarrow |f(x_1)-f(x_2)|>\epsilon$

f) $\forall \epsilon >0 \qquad, \exists \delta>0 \qquad, |x_1-x_2| < \epsilon \Rightarrow |f(x_1)-f(x_2)|<\delta$

g) $\forall \epsilon >0 \qquad, \exists \delta>0 \qquad, |f(x_1)-f(x_2)| > \epsilon \Rightarrow |x_1-x_2|> \delta$

h) $\exists \epsilon >0 \qquad, \forall \delta>0 \qquad, |x_1-x_2| < \delta \Rightarrow |f(x_1)-f(x_2)|<\epsilon$

i) $\forall \epsilon >0 \qquad, \exists \delta>0 \qquad, x_1-x_2 < \delta \Rightarrow f(x_1)-f(x_2)<\epsilon$

Here's what everybody got so far:

a) $\{ \}$

b) every functions

c) constant functions

d) constant functions

e) $\{ \}$

f) functions that are bounded on any closed interval (not sure)

g) uniform continous functions

h) bounded functions

i) Non-decreasing and uniformly continuous.

Thanks for your suggestions.

Answer 1

For (h), you are right.

If $f$ is in the set of all functions with property (h), then there exists $\varepsilon_f > 0$ s.t. $|x_1 - x_2| < \delta $ implies $|f(x_1)-f(x_2)| < \varepsilon_f $ for all $\delta >0$. Let $x \in \mathbb{R}$ be given. So, you can set $\delta =2|x|$. Then, we have $|x-0| < \delta$, and thus $|f(x)-f(0)| < \varepsilon_f $ and we have $$ |f(x)| < \varepsilon_f + |f(0)|.$$

On the other hand, if $f$ is bounded, then we can take $\varepsilon = 2M$ where $|f(x)| \leq M$ for all $x \in \mathbb{R}$.Then $f$ is in the set of all functions with property (h).

Answer 2

I don't agree with you on e)

Let's take for example the function $f = \mathbb 1_\mathbb Q$. That means $\forall x \in \mathbb Q, f(x) = 1$ and $\forall x \in \mathbb R -\mathbb Q, f(x) = 0$. Let's take $\epsilon = 2$, we can see that $\forall x_1,x_2 \in \mathbb R,\forall \delta \gt 0, \lvert f(x_1) - f(x_2) \rvert \lt \epsilon$, so this everywhere discontinuous function does not fit.

Answer 3

I think (i) might be non-decreasing, uniformly continuous functions.

Suppose there exists $b \gt a$ such that $f(b) \lt f(a)$ (i.e., $f$ is, somewhere, decreasing).

Take $\epsilon = \frac{ f(a) - f(b)}{2}$. No matter what $\delta$ is chosen, $a - b \lt \delta$ because $a \lt b$ and so $a - b \lt 0 \lt \delta$. But $f(a) - f(b) = 2\epsilon > \epsilon$, so no suitable $\delta$ can exist. Therefore no function that decreases can satisfy the criteria.

If we have a non-decreasing function, any choice of $x_1 , x_2$ with $x_1 \lt x_2$ gives us a negative value for $f(x_1) - f(x_2)$ which will be less than any positive epsilon. It remains , then, only to be able to choose $\delta$ such that the $\epsilon$ condition holds when $x_1 \gt x_2$. But in that situation, both $x_1 - x_2 = |x_1 - x_2|$ and $f(x_1) - f(x_2) = |f(x_1) - f(x_2)|$, and we're just looking at the definition of uniform continuity.

I think.

It's been about 20 years since I've had my hands in this stuff :).

Answer 4

I think I finally got the proof that (e) is the empty set (see the edit history of this comment for how I stumbled around on the way to the answer).

I believe all you have to do is choose $x_2 = x_1$, right? No positive $\epsilon$ will be able to satisfy $|f(x_1) - f(x_2)| \gt \epsilon$