# A game with δ\delta, ϵ\epsilon and uniform continuity.

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 $$f:R→Rf: \mathbb R \rightarrow \mathbb R$$ having the following properties ($$ϵ,δ,x1,x2∈R\epsilon, \delta,x_1,x_2 \in \mathbb R$$) :

a) $$∀ϵ,∃δ>0,|x1−x2|<δ⇒|f(x1)−f(x2)|<ϵ\forall \epsilon \qquad\qquad, \exists \delta>0 \qquad, |x_1-x_2| < \delta \Rightarrow |f(x_1)-f(x_2)|<\epsilon$$

b) $$∀ϵ>0,∃δ,|x1−x2|<δ⇒|f(x1)−f(x2)|<ϵ\forall \epsilon >0 \qquad, \exists \delta \qquad \qquad, |x_1-x_2| < \delta \Rightarrow |f(x_1)-f(x_2)|<\epsilon$$

c) $$∀ϵ>0,∃δ>0,(x1−x2)<δ⇒|f(x1)−f(x2)|<ϵ\forall \epsilon >0 \qquad, \exists \delta>0 \qquad, (x_1-x_2) < \delta \Rightarrow |f(x_1)-f(x_2)|<\epsilon$$

d) $$∀ϵ>0,∀δ>0,|x1−x2|<δ⇒|f(x1)−f(x2)|<ϵ\forall \epsilon >0 \qquad, \forall \delta>0 \qquad, |x_1-x_2| < \delta \Rightarrow |f(x_1)-f(x_2)|<\epsilon$$

e) $$∀ϵ>0,∃δ>0,|x1−x2|<δ⇒|f(x1)−f(x2)|>ϵ\forall \epsilon >0 \qquad, \exists \delta>0 \qquad, |x_1-x_2| < \delta \Rightarrow |f(x_1)-f(x_2)|>\epsilon$$

f) $$∀ϵ>0,∃δ>0,|x1−x2|<ϵ⇒|f(x1)−f(x2)|<δ\forall \epsilon >0 \qquad, \exists \delta>0 \qquad, |x_1-x_2| < \epsilon \Rightarrow |f(x_1)-f(x_2)|<\delta$$

g) $$∀ϵ>0,∃δ>0,|f(x1)−f(x2)|>ϵ⇒|x1−x2|>δ\forall \epsilon >0 \qquad, \exists \delta>0 \qquad, |f(x_1)-f(x_2)| > \epsilon \Rightarrow |x_1-x_2|> \delta$$

h) $$∃ϵ>0,∀δ>0,|x1−x2|<δ⇒|f(x1)−f(x2)|<ϵ\exists \epsilon >0 \qquad, \forall \delta>0 \qquad, |x_1-x_2| < \delta \Rightarrow |f(x_1)-f(x_2)|<\epsilon$$

i) $$∀ϵ>0,∃δ>0,x1−x2<δ⇒f(x1)−f(x2)<ϵ\forall \epsilon >0 \qquad, \exists \delta>0 \qquad, x_1-x_2 < \delta \Rightarrow f(x_1)-f(x_2)<\epsilon$$

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.

a) There are no functions for which $|f(x_1)-f(x_2)|<-1$ is true. So it is empty set.

b) Let us have $\delta=-1$ and the statement is true. Every function.

c) Every constant function is good. Suppose there are exist $x,y\; x. For every positive $\delta: \;x-y<\delta$ but the conclusion can't be true so only constants.

d) Suppose function is not a constant and the conclusion fails immediately. Only constants.

e) Just $x=y$ and no function can hold it. Empty set.

f) Let $f$ have a property:
We obtain for $x, y$ greater than $0$ and our $\delta$ is greater than $|a|\epsilon+c-b$. For negative $x, y$ it is the same. For $x,y$ of different signs the function is bounded on a closed interval, so the statement is true and we choose the maximum for our bound.
And we can now move $f$ along the $x$-axis - all the conclusions will be the same.

In general I suppose, f) is the case of functions with finite modulus of continuity — it is just a definition. But I do not know if this set has a special name or could it be simplified.

g) First of all, uniformly continuous functions have or (quite simple) or
So it is just the definition of uniformly continous functions.

h) Every bounded satisfies and for unbound one can create an example which will deny the existence of $\epsilon$. Bounded.

i) First of all it should be non-decreasing because $x_1-x_2\leq 0\implies f(x_1)-f(x_2)\leq 0$. And then suppose it is non-decreasing. So for $x>y$ it is $|x-y|>\delta \implies |f(x)-f(y)|<\epsilon$, which is the definition of uniform continuity. For $x\leq y$ it is even more simple. Non-decreasing and uniformly continuous.