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.
Best Answer
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 $x,y\; x<y,\; f(x)\neq f(y)$. 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: $$\forall x>0\; ax+b\leq f(x)\leq ax+c,\; b\leq c,$$ $$\forall x<0\; Ax+B\leq f(x)\leq Ax+C,\; B\leq C.$$ We obtain for $x, y$ greater than $0$ $$|f(x)-f(y)|\leq|ax+c-ay-b|\leq |a||x-y|+|c-b|$$ 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 $$\forall \varepsilon > 0 \; \exists \delta \; |x_1-x_2| < \delta \Rightarrow |f(x_1)-f(x_2)| < \varepsilon$$ or (quite simple) $$\forall \varepsilon > 0 \; \exists \delta \; |x_1-x_2| \leq \delta \Rightarrow |f(x_1)-f(x_2)| \leq \varepsilon$$ or $$\forall \varepsilon > 0 \; \exists \delta \; |f(x_1)-f(x_2)| > \varepsilon \Rightarrow |x_1-x_2| > \delta$$ 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.