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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.

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    $\begingroup$ isn't a) the empty set? if you're allowed to choose $\epsilon<0$... $\endgroup$ – Ferra Feb 20 '14 at 14:54
  • $\begingroup$ g) looks like the set of "not too quickly rising functions". Not sure how to charactarize it, though. It contains at least all functions with limited derivatives, but I don't know how to describe te wholes set... $\endgroup$ – 5xum Feb 20 '14 at 14:54
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    $\begingroup$ Hint for g). Consider the contraposition. $\endgroup$ – Hanul Jeon Feb 20 '14 at 15:26
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    $\begingroup$ The missing item, (i), says that at every point $(x,f(x))$, discontinuous behavior of $f$ can happen only below the horizontal line through that point, $y=f(x)$. For example, as $x$ increases, $f$ can have discontinuous jumps down, but not up, and the "down side" of oscillations can cause discontinuities, but not the "up" side. $\endgroup$ – zyx Mar 6 '14 at 18:09
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    $\begingroup$ (b) is true for all functions. Regardless of the function, for every $\epsilon > 0$, there does exist $\delta$--namely $\delta=0$. A=>B is true if $A$ is always false. So $|x_{1}-x_{2}| < 0 \implies |f(x_{1})-f(x_{2})|< \epsilon$. $\endgroup$ – DisintegratingByParts Mar 10 '14 at 23:23
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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.

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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.

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

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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.

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  • $\begingroup$ > For (b) I would say "all functions" because δ is allowed to be negative. ... or zero. $\endgroup$ – leonbloy Mar 12 '14 at 11:48
  • $\begingroup$ For (f) - what if $|f(x)-f(y)|\leq a|x-y|^2 +b$. Or even $\exp(|x-y|)$? Since $|x-y|$ is bounded we will have no problem, won't we? $\endgroup$ – sas Mar 12 '14 at 12:26
  • $\begingroup$ @sas You're right. I've edited my answer accordingly. $\endgroup$ – Etienne Mar 12 '14 at 18:33
  • $\begingroup$ @leonbloy Of course. I always forget that "negative" means "strictly smaller than $0$". $\endgroup$ – Etienne Mar 12 '14 at 19:15
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I think (i) might be non-decreasing, uniformly continuous functions.

Suppose there exists $b \gt a$ such that $f(b) \lt f(a)$ (i.e., $f$ is, somewhere, decreasing).

Take $\epsilon = \frac{ f(a) - f(b)}{2}$. No matter what $\delta$ is chosen, $a - b \lt \delta$ because $a \lt b$ and so $a - b \lt 0 \lt \delta$. But $f(a) - f(b) = 2\epsilon > \epsilon$, so no suitable $\delta$ can exist. Therefore no function that decreases can satisfy the criteria.

If we have a non-decreasing function, any choice of $x_1 , x_2$ with $x_1 \lt x_2$ gives us a negative value for $f(x_1) - f(x_2)$ which will be less than any positive epsilon. It remains , then, only to be able to choose $\delta$ such that the $\epsilon$ condition holds when $x_1 \gt x_2$. But in that situation, both $x_1 - x_2 = |x_1 - x_2|$ and $f(x_1) - f(x_2) = |f(x_1) - f(x_2)|$, and we're just looking at the definition of uniform continuity.

I think.

It's been about 20 years since I've had my hands in this stuff :).

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  • $\begingroup$ It is better to say "non-decreasing". And, definitely, uniformly continuous - because $x^2$ when $x>0$ does not work $\endgroup$ – sas Mar 12 '14 at 12:18
  • $\begingroup$ Thanks--I don't remember learning about the non-uniform version of continuity (good summary here math.wisc.edu/~robbin/521dir/cont.pdf) I will edit to fix. Also, I thought "increasing" meant "increasing, or perhaps constant", and if you meant to exclude constant you would say "strictly increasing". But non-decreasing expresses that better, I will fix that too. $\endgroup$ – msouth Mar 13 '14 at 17:30
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First, I have to give a lecture about quantifiers. As you know I was nit-picking about quantifiers in the comments, but only because it is extremely important. Before I address the exercises, here's three examples which demonstrate the importance:

  • $\forall x_1\forall\epsilon>0\,\exists\delta>0\,\forall x_2\,\left(|x_1-x_2|<\delta\implies |f(x_1)-f(x_2)|<\epsilon\right)$
  • $\forall\epsilon>0\,\exists\delta>0\,\forall x_1\,\forall x_2\,\left(|x_1-x_2|<\delta\implies |f(x_1)-f(x_2)|<\epsilon\right)$
  • $\forall x_1\forall x_2\forall\epsilon>0\,\exists\delta>0\,\left(|x_1-x_2|<\delta\implies |f(x_1)-f(x_2)|<\epsilon\right)$

Note that the only difference in the statements is the location of the quantified variables $x_1$ and $x_2$; however, the first means 'continuous', the second means 'uniformly continuous', and the last is satisfied by every function.

In fact, if you interpret many of your statements to simply be lacking $\forall x_1\forall x_2$ out in front, then many of the exercises become trivial: the answer is every function. These exercises are b), c), f), g), and i). This is because of the existentially quantified conditional that then occurs in each of these problems. Existentially quantified conditionals do not represent anything meaningful we would usually say in speech, logic, or math as they are usually too easy to make true. The other exercises which have this problem: a) and e) do not automatically become true only because we can choose $x_1=x_2$.

With these concerns in mind, we have to actually try to determine what Makarov meant when he wrote this problem. I intepreted f) to mean

  • $\forall x_1\forall\delta>0\,\exists\epsilon>0\,\forall x_2\,\left(|x_1-x_2|<\delta\implies |f(x_1)-f(x_2)|<\epsilon\right)$

NOTE. I interchanged the role of $\delta$ and $\epsilon$ as $\delta$ is usually used to represent the neighborhood in the domain, and I actually found it hard to do the problem without this change. The answer here is 'sends bounded sets to bounded sets'. Once we choose $x_1$ and a $\delta$-neighborhood around $x_1$, the statement implies that there is an $\epsilon$ (no matter how big) such that $f$ sends everything in $(x_1-\delta, x_1+\delta)$ to $(f(x_1)-\epsilon, f(x_1)+\epsilon)$.

EDIT

We could interpret f) to mean

  • $\forall\delta>0\,\exists\epsilon>0\,\forall x_1\forall x_2\,\left(|x_1-x_2|<\delta\implies |f(x_1)-f(x_2)|<\epsilon\right)$

Logically, this is stronger (not immediately strictly stronger) than what I considered above. It means that for every $\delta$ there is an $\epsilon$ such that every $\delta$-neighborhood is sent to an $\epsilon$-neighborhood.

I don't immediately have a name for such functions. Linear functions satisfy this condition. Bounded functions do also. Lipschitz functions satisfy it. Quadratic functions do not (so it is strictly stronger). This suggests we might be able to characterize them as functions with at most linear growth and at most linear decay. Maybe this is what sas had in mind above.

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  • $\begingroup$ I think the term you want is nit-pick, but it won't let me make the change because it's less than six characters. I can see alot of things wrong with that policy... :) $\endgroup$ – msouth Mar 13 '14 at 17:40
  • $\begingroup$ That's a funny thing to point out $\endgroup$ – Robert Wolfe Mar 13 '14 at 18:00
  • $\begingroup$ Why's it shady about what f) is? Take any function $f$ which sends bounded sets to bounded sets (no restriction on its growth as opposed with above). Then it satisfies this property. Take a point $x_1$ and take a $\delta$-neighborhood around it. $f$ sends it to a bounded set. Thus there is an $\epsilon$-neighborhood that contains the image of the $\delta$-neighborhood. Conversely, suppose $f$ satisfies this property. Consider a bounded set. Take any element in it. Choose a $\delta$ big enough to cover it. Then the property says its image can be covered by an $\epsilon$-neighborhood. ie bounded $\endgroup$ – Robert Wolfe Mar 14 '14 at 16:32
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For (h), you are right.

If $f$ is in the set of all functions with property (h), then there exists $\varepsilon_f > 0$ s.t. $|x_1 - x_2| < \delta $ implies $|f(x_1)-f(x_2)| < \varepsilon_f $ for all $\delta >0$. Let $x \in \mathbb{R}$ be given. So, you can set $\delta =2|x|$. Then, we have $|x-0| < \delta$, and thus $|f(x)-f(0)| < \varepsilon_f $ and we have $$ |f(x)| < \varepsilon_f + |f(0)|.$$

On the other hand, if $f$ is bounded, then we can take $\varepsilon = 2M$ where $|f(x)| \leq M$ for all $x \in \mathbb{R}$.Then $f$ is in the set of all functions with property (h).

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Another crack at f)

I think it gives us a definition of something you might call "uniformly bounded on any finite interval"

The mere absence of any poles guarantees that a function is bounded on every finite interval. The class of functions that f) describes adds a further restriction--if I give you the width of the interval, you have to be able to tell me a bound on the range, independent of where the interval is located.

Thus, $f(x) = x^2$ is not a suitable function--as you move to higher $x$ values, the range (over an interval of the fixed $\epsilon$ width) increases. The $\delta$, in other words, can't just be a function of $\epsilon$ for $f(x) = x^2$, but would also depend on $x$.

$f(x) = ax+b$, on the other hand, is fine--give me the width of the interval and I can tell you the size of the range; $f(x) = ln(x),\,x>1$ works, too, I think. Both of these functions are unbounded, but their unboundedness is in some sense "contained".

It doesn't tell you anything about continuity, of course, since a function equal to 1 on the rationals and 0 on the irrationals clearly fits the definition with a $\delta$ of 2 for any $\epsilon$. But for functions with derivatives, it might tell you something about that derivative--perhaps the derivative must be bounded or something like that (this is speculation based on intuition from the $f(x)=x^2$ example--the unboundedness of the derivative seems to tell you that no $\epsilon$,$\delta$ relationship will be found that is independent of $x$. Seems like you could prove without too much work that if the function has a derivative and the derivative is bounded then you will be able to come up with a $\delta$ independent of $x$, since the function never changes "too much". Similarly with $ln(x) ,\, x \gt 1$, the $\frac 1 x$ has a nice, bounded absolute value everywhere, which seems like it would constrain the $\delta$ independent of where you locate the $\epsilon$ neighborhood.)

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  • $\begingroup$ "Bounded derivative" is not a good heuristic, because it fails on $\sqrt x$ and other uniformly continuous functions. A uniformly continuous function on an interval (or more generally a path-connected domain) satisfies $f$ because if $|f(x)-f(y)|<1$ when $|x-y|<\delta$, then $|f(x)-f(y)|<n$ when $|f(x)-f(y)|<n\delta$ for any $n$, so it satisfies (f). $\endgroup$ – Mario Carneiro Apr 25 '15 at 8:09
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Let us call a function $f:{\mathbb R}\to{\mathbb R}$ loosely Lipschitz if there exist positive constants $L$ and $K$ such that for all real $x$ and $y$, $$ |f(x)-f(y)| \le L|x-y|+K. $$ I claim $f$ is loosely Lipschitz if and only if condition f) holds.

Proof: Clearly, if $f$ is loosely Lipschitz then f) holds with $\delta=L\epsilon+K$ for every $\epsilon>0$. In the other direction, suppose f) holds. Let $\epsilon=2$ and choose $\delta>0$ accordingly. Then given any distinct reals $x$ and $y$, there is a positive integer $m$ such that $|x-y|\le m <|x-y|+1$. With $r=(y-x)/m$ and $x_j=x+jr$ for $j=0,\ldots m$, we have $|r|\le1$, $x_0=x$ and $x_m=y$, therefore $$ |f(x)-f(y)| \le \sum_{j=1}^m |f(x_j-r)-f(x_j)| \le m\delta <\delta|x-y|+\delta. $$ Hence $f$ is loosely Lipschitz.

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