Proof with a monotone function

[Felafel]

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

$\displaystyle \lim_{x \to +\infty} \frac{f(cx)}{f(x)}=1$

 

[tiny-tim]

Hi Felafel!

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

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

shouldn't that be f(4x) ?

anyway, how would you apply it to c = √2 ?

 

[Felafel]

Hi Felafel!


shouldn't that be f(4x) ?

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..

 

[tiny-tim]

no, I actually meant to write that 3x, doesn't it work to you?

no, i don't see where that line comes from

 

[Felafel]

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?

 

[SammyS]

Is this monotone function decreasing, and is it positive?

 

[Ray Vickson]

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)?

 

[Felafel]

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.

 

[tiny-tim]

Hi Felafel!

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

But what if L = ∞ ?

eg if f(x) = logx,

then limlogx = ∞, but limf(2x)/f(x) = lim(logx + log2)/logx = 1

 

[Felafel]

yess, it should go even if L is not finite :)!
thanks for checking!

 

[tiny-tim]

i still don't see where you're getting eg f(cx)/f((c-1)x) ≤ ε + 1 from

 

[Felafel]

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?

 

[tiny-tim]

it only applies to f(a)/f(b) if a = 2b

 

[Felafel]

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?

 

[Ray Vickson]

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: \frac{f(4x)}{f(x)}= \frac{f(4x)}{f(2x)} \cdot \frac{f(2x)}{f(x)} \to 1 \text{ as } x \to \infty, and similarly,
\lim_{x \to \infty} \frac{f(2^k x)}{f(x)} = 1, \: k = \pm 1, \pm 2, \ldots .
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.

 

[Felafel]

I think this is longer than necessary. First: \frac{f(4x)}{f(x)}= \frac{f(4x)}{f(2x)} \cdot \frac{f(2x)}{f(x)} \to 1 \text{ as } x \to \infty, and similarly,
\lim_{x \to \infty} \frac{f(2^k x)}{f(x)} = 1, \: k = \pm 1, \pm 2, \ldots .
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 :)