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Old 2010-04-09, 17:17   #1
kakos22
 
Apr 2010

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Default extremely difficult problem with a function.

Let f:R->R is a function which its second derivative f'' is continuous function.
Prove that if lim(x->infinity) (f(x)+ f '(x) + f ''(x))=L then lim(x->infinity) f(x)=L. L is in R*.

Note: this exercise is not like a usually exercise for home, is like a challenge exercise and is very hard. If anyone can think anything that would help me just write it down.I will appreciate any kind of help. if you want you can ask me and i will post my thoughts about this exercise(how we can try to solve it).

please write back if anything is not clear or if you want to ask about something. thank you.
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Old 2010-04-09, 18:45   #2
R.D. Silverman
 
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"Bob Silverman"
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Quote:
Originally Posted by kakos22 View Post
Let f:R->R is a function which its second derivative f'' is continuous function.
Prove that if lim(x->infinity) (f(x)+ f '(x) + f ''(x))=L then lim(x->infinity) f(x)=L. L is in R*.

Note: this exercise is not like a usually exercise for home, is like a challenge exercise and is very hard. If anyone can think anything that would help me just write it down.I will appreciate any kind of help. if you want you can ask me and i will post my thoughts about this exercise(how we can try to solve it).

please write back if anything is not clear or if you want to ask about something. thank you.

Start with the following.

Suppose lim x-->oo f(x) = L. What does
this say about lim x-->oo f'(x)??? can lim x-->oo f'(x) = M, M !=0 and
still have lim x-->oo f(x) = L?? i.e. if the slope isn't 0 as x-->oo what
happens to f(x)? Similarly if f' -->0 as x-->oo, what can we say about
f''??

i.e. start by looking at the converse of the problem. Then consider the
following. if lim x-->oo f + f' --> (say) K, what can we say about f
and f' individually?

BTW, I recall seeing this exercize in my undergrad days. It may be in Rudin,
IIRC.
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Old 2010-04-09, 19:06   #3
R.D. Silverman
 
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Quote:
Originally Posted by R.D. Silverman View Post
Start with the following.

Suppose lim x-->oo f(x) = L. What does
this say about lim x-->oo f'(x)??? can lim x-->oo f'(x) = M, M !=0 and
still have lim x-->oo f(x) = L?? i.e. if the slope isn't 0 as x-->oo what
happens to f(x)? Similarly if f' -->0 as x-->oo, what can we say about
f''??

i.e. start by looking at the converse of the problem. Then consider the
following. if lim x-->oo f + f' --> (say) K, what can we say about f
and f' individually?

BTW, I recall seeing this exercize in my undergrad days. It may be in Rudin,
IIRC.

Further hint (I omit the lim x-->oo in my notation)

if (f + f') --> L, and f --> M, then f' --> L-M. but if f' --> L-M, what
happens to f? If f is not bounded, then f must go to oo and f' must go
to -oo. Can this happen???? [clearly if both go to oo, then their sum does
as well, and if one is bounded and the other is not then their sum is again
unbounded].
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Old 2010-04-09, 20:47   #4
rajula
 
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Quote:
Originally Posted by kakos22 View Post
Let f:R->R is a function which its second derivative f'' is continuous function.
Prove that if lim(x->infinity) (f(x)+ f '(x) + f ''(x))=L then lim(x->infinity) f(x)=L. L is in R*.
An interesting problem. I could not solve it immediately. I do not know if I should be ashamed.
The reason I replied is because I do not see how Silverman's hints might help in solving the problem. I might just be to sleepy to notice it. Here a small observation those hints alarmed me to point out.

One can not a priori assume the limits \lim_{x\to\infty}f(x), \lim_{x\to\infty}f'(x) and \lim_{x\to\infty}f''(x) to exist. Otherwise one could prove the claim
\lim_{x\to\infty}(f(x)+ f '(x) + f ''(x) + f '''(x)) = L \Longrightarrow \lim_{x\to\infty}f(x) = L
in a same way - a claim which is trivially false.
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Old 2010-04-09, 21:12   #5
kakos22
 
Apr 2010

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Quote:
Originally Posted by R.D. Silverman View Post
Further hint (I omit the lim x-->oo in my notation)

if (f + f') --> L, and f --> M, then f' --> L-M. but if f' --> L-M, what
happens to f? If f is not bounded, then f must go to oo and f' must go
to -oo. Can this happen???? [clearly if both go to oo, then their sum does
as well, and if one is bounded and the other is not then their sum is again
unbounded].

Thanks for your hints..you are just very near my thoughts..actually I am trying to prove the following: if a function g has continuous derivative g' and lim(x-->oo)g(x) is in R then lim(x-->oo)g'(x)=0.

I am quite sure that this statement is true( i hope so)..i couldn't find a counter example so at this moment i am trying to prove it..if i manage or anyone else from this forum manage to prove that lemma(it shouldn't be very difficult i believe) then our problem would take easier form, because if we assume that lim f(x) exists and is in R then from the lemma lim f'(x)=0 and hence lim f''(x)=0 so is true. So it would remain to prove the problem by contradiction assuming that lim f(x) doesn't exist.

I wait for your replies or any other thoughts..thanks
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Old 2010-04-09, 21:23   #6
rajula
 
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"Tapio Rajala"
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Quote:
Originally Posted by kakos22 View Post
I am trying to prove the following: if a function g has continuous derivative g' and lim(x-->oo)g(x) is in R then lim(x-->oo)g'(x)=0.
Unfortunately this is false. Take for example  g(x) = \frac{\sin(x^2)}{x}. One has to be careful when thinking about the existence of the limits.
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Old 2010-04-09, 21:44   #7
kakos22
 
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Quote:
Originally Posted by rajula View Post
Unfortunately this is false. Take for example  g(x) = \frac{\sin(x^2)}{x}. One has to be careful when thinking about the existence of the limits.
oh..thanks a lot rajula, you have just saved my life..i was just trying to prove something that is not true and you just saved my time..thank you..

I will come back with something new..
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Old 2010-04-10, 13:24   #8
kakos22
 
Apr 2010

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I am just trying to figure out. if the condition from the problem holds with L in R, then it has to be limf '(x)=limf ''(x)=0??..can anybody find an example such that limf '(x)= - limf ''(x)<>0 or
such that lim f '(x) and lim f ''(x) don't exist but does exist (lim f '(x) + lim f''(x)) and that is 0?? if nobody will find such an example i will begin trying to prove that limf '(x)=limf ''(x)=0..i wait for reply

Last fiddled with by kakos22 on 2010-04-10 at 13:28
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Old 2010-04-10, 13:58   #9
R.D. Silverman
 
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Quote:
Originally Posted by rajula View Post
Unfortunately this is false. Take for example  g(x) = \frac{\sin(x^2)}{x}. One has to be careful when thinking about the existence of the limits.
Please note that lim x-->oo g'(x) does not exist. It doesn't have a value.
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Old 2010-04-10, 14:02   #10
kakos22
 
Apr 2010

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Quote:
Originally Posted by R.D. Silverman View Post
Please note that lim x-->oo g'(x) does not exist. It doesn't have a value.
yes Silverman this is already clear..that is why the lemma i was trying to prove is not true..unfortunately rajula has found a counter example.
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Old 2010-04-10, 17:18   #11
kakos22
 
Apr 2010

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I am just trying to prove the following lemma:
Let function g has continuous derivative g' and lim x->oo g(x) exists and is in R.
Then lim x->oo g '(x) doesn't exist or exists and is 0.

I think that this time the lemma is true.If that so then the beginning problem will be easier. if anybody could found a counter example or something to help me with that i will appreciate it.
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