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

Solution 1:

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.

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.

Solution 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.

Solution 3:

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 $|f(x_1) - f(x_2)| \gt \epsilon$

Solution 4:

I agree with you for (a), (c), (d), (g), (h)

For (b) I would say "all functions" because $\delta$ is allowed to be negative.

(e) is the empty set because you can take $x_1=x_2$.

For (i), I would say "nondecreasing and uniformly continuous". If $f$ satisfies (i), then it is nondecreasing because $x_1-x_2\leq 0\implies f(x_1)-f(x_2)<\varepsilon$ for any $\varepsilon >0$. Moreover, $f$ is also uniformly continuous because if $\vert x_1-x_2\vert\delta$, then you have both $f(x_1)-f(x_2)<\varepsilon$ and $f(x_2)-f(x_1)<\varepsilon$, i.e. $\vert f(x_1)-f(x_2)\vert <\varepsilon$. Conversely,assume that $f$ is nondecreasing and uniformly continuous. Given $\varepsilon >0$, choose $\delta$ according to the definition of uniform continuity. If $x_1-x_2<\delta$, then either $x_1\leq x_2$, in which case $f(x_1)-f(x_2)\leq 0<\varepsilon$, or $x_1>x_2$, in which case $\vert x_1-x_2\vert<\delta$ and hence $f(x_1)-f(x_2)\leq \vert f(x_1)-f(x_2)\vert<\varepsilon$.

For (f), I still don't know. As you noticed, (f) implies that the function must be bounded on any bounded interval. But in fact (f) says that for any $C$, the bound must be uniform on all intervals of length $\leq C$. This implies that $f$ has at most linear growth ($\vert f(x)\vert \leq a\, \vert x\vert+b$) but not conversely. Right now, I have no good description of the functions satisfying (f).

Edit. Thanks to @sas for pointing out a mistake in the first version of my answer; and have a look at his/her answer.

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

Solution 1:

Solution 2:

Solution 3:

Solution 4:

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