Proof with a monotone function

In summary, we can use the definition of the limit to infinity to show that for any monotone function f and any c>0, the limit of f(cx)/f(x) is 1 as x approaches infinity. This holds true for both increasing and decreasing functions, and for any c>0.
  • #1
Felafel
171
0

Homework Statement


Let ##f:\mathbb{R}\to \mathbb{R}## a monotone function sucht that
## \displaystyle \lim_{x \to +\infty} \frac{f(2x)}{f(x)}=1##
show that for all c>0, we have
##\displaystyle \lim_{x \to +\infty} \frac{f(cx)}{f(x)}=1##

I think I'm almost there. Does it look okay to you? also, is it valid for 0<c<1 or just for c>1?
thank you very much

The Attempt at a Solution


For the definition of limit to infinity:
##\forall \epsilon >0## ##\exists S>0## ##:##

##|f(x)-l|<\epsilon## ##\forall x>S##

##\displaystyle \lim_{x \to +\infty} \frac{f(2x)}{f(x)}=1## ##\Rightarrow## ##|\frac{f(2x)}{f(x)}-1|<\epsilon$ $\forall x>S##

which means

##f(x)(-\epsilon+1)<f(2x)<(\epsilon+1)f(x)## (I see it's monotonically decreasing, and so 1 is the infimum)

But if ##\forall \epsilon>0## i get ##-\epsilon f(x)+f(x)<f(2x)<\epsilon f(x)+f(x)##

Being ##\epsilon \to 0## ##\Rightarrow## ##|f(2x)-f(x)|=0##

And so:

##f(2x) \leq (1+\epsilon)f(x)##

##f(3x) \leq (1+\epsilon)f(2x)##

##f(3x) \leq (1+\epsilon)^2f(x)##

##1 \leq f(cx) \leq (1+\epsilon)^{c-1} f(x)##

##\epsilon \to 0## ##\Rightarrow## ##1 \leq f(cx) \leq f(x)## and, for the squeeze rule:

##\displaystyle \lim_{x \to +\infty} f(cx)=1## so

##\displaystyle \lim_{x \to +\infty} \frac{f(cx)}{f(x)}=1##
 
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  • #2
Hi Felafel! :smile:
Felafel said:
##f(2x) \leq (1+\epsilon)f(x)##

##f(3x) \leq (1+\epsilon)f(2x)##

shouldn't that be f(4x) ? :confused:

anyway, how would you apply it to c = √2 ?
 
  • #3
tiny-tim said:
Hi Felafel! :smile:


shouldn't that be f(4x) ? :confused:

anyway, how would you apply it to c = √2 ?
hello! :)
no, I actually meant to write that 3x, doesn't it work to you?
also, i think that being ##cx=\sqrt{2x} > x## it should go, as the sequence is decreasing.
My doubt was more if it worked for ##c=\frac{1}{2}## being then cx<x
but maybe I am wrong..
 
  • #4
Felafel said:
no, I actually meant to write that 3x, doesn't it work to you?

no, i don't see where that line comes from :confused:
 
  • #5
oh, okay i'll just delete it then, it is also rather unnecessary if the other passages are right.
do you think the rest of the proof works, on the other hand?
 
  • #6
Is this monotone function decreasing, and is it positive?
 
  • #7
Felafel said:
oh, okay i'll just delete it then, it is also rather unnecessary if the other passages are right.
do you think the rest of the proof works, on the other hand?

For monotone f and ##c \in (1,2)## what are the relationships between f(x), f(cx) and f(2x)?
 
  • #8
I'll give it a thoroughly different try:

Definition of the limit to infinity:

##\forall \epsilon >0 \exists S>0:##
##|f(x)-l|<\epsilon \forall x>S##
##\displaystyle \lim_{x \to +\infty} \frac{f(2x)}{f(x)}=1\Rightarrow## ##|\frac{f(2x)}{f(x)}-1|<\epsilon## ##\forall x>S##

which means:

##\displaystyle \lim_{x \to \infty}f(x)= \displaystyle \lim_{x \to \infty}f(2x)= L##

Assuming the function is monotonically increasing, and c>1, L is the supremum, and we also have:

##f(cx)=\frac{f(cx)}{f((c-1)x)} \frac{f((c-1)x)}{f((c-2)x)}...\frac{f(2x)}{f(x)}## (c terms)
Each term is ## \leq \epsilon +1## thus:

##f(x)(1-\epsilon)^c \leq f(cx) \leq L## and doing the limit i get:
##\displaystyle \lim_{x \to \infty} f(cx)=L## ##\Rightarrow## ##\displaystyle \lim_{x} \frac{f(cx)}{f(x)}=1##

Assuming f is increasing and 0<c<1, i get:

##f(cx)=\frac{f(x)}{f(x-1)} \frac{f(x-1)}{f(x-2)}...\frac{f(s+1)}{f(s)}## (c-s terms)

So:
##f(s)(L-\epsilon)^{c-s} \leq f(cx) \leq f(x)(1+\epsilon)^c## doing the limit:
##f(cx)\to L##If the function is monotonically decreasing I'd follow the same procedure with inverted signs.
 
  • #9
Hi Felafel! :smile:
Felafel said:
##\displaystyle \lim_{x \to \infty}f(x)= \displaystyle \lim_{x \to \infty}f(2x)= L##

But what if L = ∞ ?

eg if f(x) = logx,

then limlogx = ∞, but limf(2x)/f(x) = lim(logx + log2)/logx = 1 :wink:
 
  • #10
yess, it should go even if L is not finite :)!
thanks for checking!
 
  • #11
i still don't see where you're getting eg f(cx)/f((c-1)x) ≤ ε + 1 from :confused:
 
  • #12
i did it because c>s and ##f(2x)/(f(x) \leq \epsilon +1## and thought it would apply to every element of the function. is it wrong?
 
  • #13
it only applies to f(a)/f(b) if a = 2b :redface:
 
  • #14
argh, thought a=b+1 was sufficient.
Is there any other way I can solve this problem then? :( or should I try a completely different reasoning?
 
  • #15
Felafel said:
I'll give it a thoroughly different try:

Definition of the limit to infinity:

##\forall \epsilon >0 \exists S>0:##
##|f(x)-l|<\epsilon \forall x>S##
##\displaystyle \lim_{x \to +\infty} \frac{f(2x)}{f(x)}=1\Rightarrow## ##|\frac{f(2x)}{f(x)}-1|<\epsilon## ##\forall x>S##

which means:

##\displaystyle \lim_{x \to \infty}f(x)= \displaystyle \lim_{x \to \infty}f(2x)= L##

Assuming the function is monotonically increasing, and c>1, L is the supremum, and we also have:

##f(cx)=\frac{f(cx)}{f((c-1)x)} \frac{f((c-1)x)}{f((c-2)x)}...\frac{f(2x)}{f(x)}## (c terms)
Each term is ## \leq \epsilon +1## thus:

##f(x)(1-\epsilon)^c \leq f(cx) \leq L## and doing the limit i get:
##\displaystyle \lim_{x \to \infty} f(cx)=L## ##\Rightarrow## ##\displaystyle \lim_{x} \frac{f(cx)}{f(x)}=1##

Assuming f is increasing and 0<c<1, i get:

##f(cx)=\frac{f(x)}{f(x-1)} \frac{f(x-1)}{f(x-2)}...\frac{f(s+1)}{f(s)}## (c-s terms)

So:
##f(s)(L-\epsilon)^{c-s} \leq f(cx) \leq f(x)(1+\epsilon)^c## doing the limit:
##f(cx)\to L##If the function is monotonically decreasing I'd follow the same procedure with inverted signs.

I think this is longer than necessary. First: [tex] \frac{f(4x)}{f(x)}= \frac{f(4x)}{f(2x)} \cdot \frac{f(2x)}{f(x)} \to 1 \text{ as } x \to \infty, [/tex] and similarly,
[tex] \lim_{x \to \infty} \frac{f(2^k x)}{f(x)} = 1, \: k = \pm 1, \pm 2, \ldots . [/tex]
Also, if x > 0 and ##c \in (1,2)## we have ## x < cx < 2x##, so for monotone f > 0 we have either ##f(x) \leq f(cx) \leq f(2x)## or ##f(x) \geq f(cx) \geq f(2x)##, and dividing by f(x) gives either ##1 \leq f(cx)/f(x) \leq f(2x)/f(x)## or ##1 \geq f(cx)/f(x) \geq f(2x)/f(x)##. Thus, ##f(cx)/f(x) \to 1.## Applying the same argument to ##f(2^k x)/f(x)## gives the result for any c > 0.

The same type of argument applies if f < 0.
 
  • #16
Ray Vickson said:
I think this is longer than necessary. First: [tex] \frac{f(4x)}{f(x)}= \frac{f(4x)}{f(2x)} \cdot \frac{f(2x)}{f(x)} \to 1 \text{ as } x \to \infty, [/tex] and similarly,
[tex] \lim_{x \to \infty} \frac{f(2^k x)}{f(x)} = 1, \: k = \pm 1, \pm 2, \ldots . [/tex]
Also, if x > 0 and ##c \in (1,2)## we have ## x < cx < 2x##, so for monotone f > 0 we have either ##f(x) \leq f(cx) \leq f(2x)## or ##f(x) \geq f(cx) \geq f(2x)##, and dividing by f(x) gives either ##1 \leq f(cx)/f(x) \leq f(2x)/f(x)## or ##1 \geq f(cx)/f(x) \geq f(2x)/f(x)##. Thus, ##f(cx)/f(x) \to 1.## Applying the same argument to ##f(2^k x)/f(x)## gives the result for any c > 0.

The same type of argument applies if f < 0.

Great! Thank you very much :)
 

Related to Proof with a monotone function

1. What is a monotone function?

A monotone function is a mathematical function that either always increases or always decreases as its input variable increases. This means that the function is either always moving up or always moving down, without any abrupt changes in direction.

2. How is a monotone function different from a non-monotone function?

A non-monotone function does not have a consistent direction of increase or decrease. It may have both increasing and decreasing sections, or it may have abrupt changes in direction. In contrast, a monotone function has a consistent direction of increase or decrease throughout its domain.

3. What is the significance of a monotone function in mathematics?

Monotone functions have several important properties that make them useful in mathematics. They are particularly helpful in proving the existence of limits and finding the maximum or minimum values of a function. They are also used in optimization problems and in the study of sequences and series.

4. How can you prove that a function is monotone?

To prove that a function is monotone, you can use the definition of monotonicity, which states that the function must either always increase or always decrease. You can also use the derivative of the function to show that it is always positive (increasing) or always negative (decreasing).

5. Can a monotone function have more than one point of inflection?

No, a monotone function can only have one point of inflection, which is a point where the function changes from increasing to decreasing (or vice versa). This is because a monotone function cannot have any abrupt changes in direction, and having multiple points of inflection would result in multiple changes in direction.

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