**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.