UPDATE: Bounty awarded, but it is still shady about whatf)is.In Makarov’s

Selected Problems in Real Analysisthere’s this challenging problem:Describe the set of functions f:R→R having the following properties (ϵ,δ,x1,x2∈R) :

a) ∀ϵ,∃δ>0,|x1−x2|<δ⇒|f(x1)−f(x2)|<ϵ

b) ∀ϵ>0,∃δ,|x1−x2|<δ⇒|f(x1)−f(x2)|<ϵ

c) ∀ϵ>0,∃δ>0,(x1−x2)<δ⇒|f(x1)−f(x2)|<ϵ

d) ∀ϵ>0,∀δ>0,|x1−x2|<δ⇒|f(x1)−f(x2)|<ϵ

e) ∀ϵ>0,∃δ>0,|x1−x2|<δ⇒|f(x1)−f(x2)|>ϵ

f) ∀ϵ>0,∃δ>0,|x1−x2|<ϵ⇒|f(x1)−f(x2)|<δ

g) ∀ϵ>0,∃δ>0,|f(x1)−f(x2)|>ϵ⇒|x1−x2|>δ

h) ∃ϵ>0,∀δ>0,|x1−x2|<δ⇒|f(x1)−f(x2)|<ϵ

i) ∀ϵ>0,∃δ>0,x1−x2<δ⇒f(x1)−f(x2)<ϵ

Here's what

everybodygot 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**

a) There are no functions for which |f(x1)−f(x2)|<−1 is true. So it is **empty set**.

b) Let us have δ=−1 and the statement is true. **Every function**.

c) Every constant function is good. Suppose there are exist x,yx<y,f(x)≠f(y). For every positive δ:x−y<δ 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: ∀x>0ax+b≤f(x)≤ax+c,b≤c, ∀x<0Ax+B≤f(x)≤Ax+C,B≤C.

We obtain for x,y greater than 0 |f(x)−f(y)|≤|ax+c−ay−b|≤|a||x−y|+|c−b| and our δ is greater than |a|ϵ+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 ∀ε>0∃δ|x1−x2|<δ⇒|f(x1)−f(x2)|<ε or (quite simple) ∀ε>0∃δ|x1−x2|≤δ⇒|f(x1)−f(x2)|≤ε or ∀ε>0∃δ|f(x1)−f(x2)|>ε⇒|x1−x2|>δ

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 ϵ. **Bounded**.

i) First of all it should be non-decreasing because x1−x2≤0⟹f(x1)−f(x2)≤0. And then suppose it is non-decreasing. So for x>y it is |x−y|>δ⟹|f(x)−f(y)|<ϵ, which is the definition of uniform continuity. For x≤y it is even more simple. **Non-decreasing and uniformly continuous.**

**Attribution***Source : Link , Question Author : Gabriel Romon , Answer Author : sas*