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If $fin C^2(mathbb R)$ is a probability density function, then $f'(x)to0$ implies $f(x)to0$ as $|x|toinfty$

1

2

$begingroup$

Let $f:mathbb Rto[0,infty)$ be twice continuously differentiable and assume $$int f(x):rm dx=1.$$

How can we conclude that as $|x|toinfty$ either

1. 
   $f(x)to0$ and $f'(x)to0$,


2. 
   $f(x)to0$ and $f'(x)notto0$ or


3. 
   $f(x)notto0$ and $f'(x)notto0$?

If I'm not missing anything, the only other case would be $f(x)notto0$ and $f'(x)to0$. So, I guess we need to show that this is impossible. Since $f'(x)to0$ as $|x|toinfty$ does not necessarily imply that $lim_toinftyf(x)$ even exists, this must have something to do with the integrability condition, right? If this is not sufficient, do we need to impose further conditions?

analysis probability-theory density-function

share|cite|improve this question

edited Apr 2 at 8:57

0xbadf00d

asked Mar 31 at 12:00

0xbadf00d0xbadf00d

1,62141534

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RRL The question is, whether we can conclude that either 1., 2. or 3. must hold (i.e. there is no other case).
  $endgroup$
  – 0xbadf00d
  Apr 3 at 9:57
  

  
  
  
  

add a comment | 

1

2

$begingroup$

Let $f:mathbb Rto[0,infty)$ be twice continuously differentiable and assume $$int f(x):rm dx=1.$$

How can we conclude that as $|x|toinfty$ either

1. 
   $f(x)to0$ and $f'(x)to0$,


2. 
   $f(x)to0$ and $f'(x)notto0$ or


3. 
   $f(x)notto0$ and $f'(x)notto0$?

If I'm not missing anything, the only other case would be $f(x)notto0$ and $f'(x)to0$. So, I guess we need to show that this is impossible. Since $f'(x)to0$ as $|x|toinfty$ does not necessarily imply that $lim_toinftyf(x)$ even exists, this must have something to do with the integrability condition, right? If this is not sufficient, do we need to impose further conditions?

analysis probability-theory density-function

share|cite|improve this question

edited Apr 2 at 8:57

0xbadf00d

asked Mar 31 at 12:00

0xbadf00d0xbadf00d

1,62141534

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RRL The question is, whether we can conclude that either 1., 2. or 3. must hold (i.e. there is no other case).
  $endgroup$
  – 0xbadf00d
  Apr 3 at 9:57
  

  
  
  
  

add a comment | 

$begingroup$

Let $f:mathbb Rto[0,infty)$ be twice continuously differentiable and assume $$int f(x):rm dx=1.$$

How can we conclude that as $|x|toinfty$ either

1. 
   $f(x)to0$ and $f'(x)to0$,


2. 
   $f(x)to0$ and $f'(x)notto0$ or


3. 
   $f(x)notto0$ and $f'(x)notto0$?

If I'm not missing anything, the only other case would be $f(x)notto0$ and $f'(x)to0$. So, I guess we need to show that this is impossible. Since $f'(x)to0$ as $|x|toinfty$ does not necessarily imply that $lim_toinftyf(x)$ even exists, this must have something to do with the integrability condition, right? If this is not sufficient, do we need to impose further conditions?

analysis probability-theory density-function

share|cite|improve this question

edited Apr 2 at 8:57

0xbadf00d

asked Mar 31 at 12:00

0xbadf00d0xbadf00d

1,62141534

$endgroup$

share|cite|improve this question

edited Apr 2 at 8:57

0xbadf00d

asked Mar 31 at 12:00

0xbadf00d0xbadf00d

1,62141534

share|cite|improve this question

edited Apr 2 at 8:57

0xbadf00d

asked Mar 31 at 12:00

0xbadf00d0xbadf00d

1,62141534

asked Mar 31 at 12:00

0xbadf00d0xbadf00d

1,62141534

asked Mar 31 at 12:00

0xbadf00d0xbadf00d

1,62141534

1 Answer
1

active

oldest

votes

2

$begingroup$

If $f:mathbbR to [0,infty)$ is integrable (over $mathbbR$) and
$f'(x) to 0$ as $|x| to +infty$, then it follows that $f(x) to 0$.

For proof by contradiction, assume WLOG that $f(x) notto 0$ as $x to +infty$. There exists $beta > 0$ and a sequence $x_n to +infty$ such that $f(x_n) geqslant beta$. Selecting a subsequence if necessary, we can assume that $x_n+1 > x_n + beta.$

Since $f'(x) to 0$, there exists $a > 0$ such that $-frac12 < f'(x) < frac12$ for all $x geqslant a$, and for all sufficiently large $n geqslant N$ we also have $x_n > a$.

Thus,

$$tag*int_a^infty f(x) , dx geqslant sum_n=N^inftyint_x_n^x_n+1f(x) , dx > sum_n=N^inftyint_x_n^x_n + betaf(x) , dx.$$

By the mean value theorem for $x in (x_n,x_n + beta$) there exists $c_n in (x_n,x)$ such that

$$f(x) = f(x_n) + f'(c_n)(x - x_n)> beta - frac12beta = fracbeta2$$

Therefore, we have

$$int_x_n^x_n + betaf(x) , dx > fracbeta^22,$$

and using (*) we obtain a contradiction,

$$int_a^infty f(x) , dx > sum_n=N^infty fracbeta^22 = + infty$$

share|cite|improve this answer

edited Apr 4 at 4:57

answered Apr 3 at 0:33

RRLRRL

53.9k52675

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do we know that the sequence $(x_n)_ninmathbb N$ exist? A continuous function doesn't to have a limit at infinity. Is the reason simply that as long as $(x_n)_ninmathbb N$ is any sequence with $x_ntoinfty$, we know by assumption $limsup_ntoinftyf(x_n)>0$ and hence we may select a subsequence which converges to the limit superior?
  $endgroup$
  – 0xbadf00d
  Apr 3 at 5:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes — since $f$ is positive, if it fails to converge to $0$, then the limit is not zero or fails to exist. In either case limsup is positive and is the limit of the images of a subsequence.
  $endgroup$
  – RRL
  Apr 3 at 5:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I guess that (in order for $(ast)$ to hold) we need to assume $x_1ge a$ (which is clearly no problem). Moreover, I guess the first inequality in $(ast)$ is due to the possible gap between $r$ and $x_1$, right? (Since if $x_1=r$, then $(ast)$ should be an equality.)
  $endgroup$
  – 0xbadf00d
  Apr 3 at 15:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes it could be an equality, but what is important is the second relation which is an inequality
  $endgroup$
  – RRL
  Apr 3 at 16:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  At which point did you use the integrability? I guess at the end, but even when $int f(x):rm dx=infty$ this would be a contradiction (since your last equation would read $infty>infty$).
  $endgroup$
  – 0xbadf00d
  Apr 3 at 16:13
  

  
  
  
  

 | 
show 12 more comments

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2

$begingroup$

If $f:mathbbR to [0,infty)$ is integrable (over $mathbbR$) and
$f'(x) to 0$ as $|x| to +infty$, then it follows that $f(x) to 0$.

For proof by contradiction, assume WLOG that $f(x) notto 0$ as $x to +infty$. There exists $beta > 0$ and a sequence $x_n to +infty$ such that $f(x_n) geqslant beta$. Selecting a subsequence if necessary, we can assume that $x_n+1 > x_n + beta.$

Since $f'(x) to 0$, there exists $a > 0$ such that $-frac12 < f'(x) < frac12$ for all $x geqslant a$, and for all sufficiently large $n geqslant N$ we also have $x_n > a$.

Thus,

$$tag*int_a^infty f(x) , dx geqslant sum_n=N^inftyint_x_n^x_n+1f(x) , dx > sum_n=N^inftyint_x_n^x_n + betaf(x) , dx.$$

By the mean value theorem for $x in (x_n,x_n + beta$) there exists $c_n in (x_n,x)$ such that

$$f(x) = f(x_n) + f'(c_n)(x - x_n)> beta - frac12beta = fracbeta2$$

Therefore, we have

$$int_x_n^x_n + betaf(x) , dx > fracbeta^22,$$

and using (*) we obtain a contradiction,

$$int_a^infty f(x) , dx > sum_n=N^infty fracbeta^22 = + infty$$

share|cite|improve this answer

edited Apr 4 at 4:57

answered Apr 3 at 0:33

RRLRRL

53.9k52675

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do we know that the sequence $(x_n)_ninmathbb N$ exist? A continuous function doesn't to have a limit at infinity. Is the reason simply that as long as $(x_n)_ninmathbb N$ is any sequence with $x_ntoinfty$, we know by assumption $limsup_ntoinftyf(x_n)>0$ and hence we may select a subsequence which converges to the limit superior?
  $endgroup$
  – 0xbadf00d
  Apr 3 at 5:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes — since $f$ is positive, if it fails to converge to $0$, then the limit is not zero or fails to exist. In either case limsup is positive and is the limit of the images of a subsequence.
  $endgroup$
  – RRL
  Apr 3 at 5:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I guess that (in order for $(ast)$ to hold) we need to assume $x_1ge a$ (which is clearly no problem). Moreover, I guess the first inequality in $(ast)$ is due to the possible gap between $r$ and $x_1$, right? (Since if $x_1=r$, then $(ast)$ should be an equality.)
  $endgroup$
  – 0xbadf00d
  Apr 3 at 15:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes it could be an equality, but what is important is the second relation which is an inequality
  $endgroup$
  – RRL
  Apr 3 at 16:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  At which point did you use the integrability? I guess at the end, but even when $int f(x):rm dx=infty$ this would be a contradiction (since your last equation would read $infty>infty$).
  $endgroup$
  – 0xbadf00d
  Apr 3 at 16:13
  

  
  
  
  

 | 
show 12 more comments

2

$begingroup$

If $f:mathbbR to [0,infty)$ is integrable (over $mathbbR$) and
$f'(x) to 0$ as $|x| to +infty$, then it follows that $f(x) to 0$.

For proof by contradiction, assume WLOG that $f(x) notto 0$ as $x to +infty$. There exists $beta > 0$ and a sequence $x_n to +infty$ such that $f(x_n) geqslant beta$. Selecting a subsequence if necessary, we can assume that $x_n+1 > x_n + beta.$

Since $f'(x) to 0$, there exists $a > 0$ such that $-frac12 < f'(x) < frac12$ for all $x geqslant a$, and for all sufficiently large $n geqslant N$ we also have $x_n > a$.

Thus,

$$tag*int_a^infty f(x) , dx geqslant sum_n=N^inftyint_x_n^x_n+1f(x) , dx > sum_n=N^inftyint_x_n^x_n + betaf(x) , dx.$$

By the mean value theorem for $x in (x_n,x_n + beta$) there exists $c_n in (x_n,x)$ such that

$$f(x) = f(x_n) + f'(c_n)(x - x_n)> beta - frac12beta = fracbeta2$$

Therefore, we have

$$int_x_n^x_n + betaf(x) , dx > fracbeta^22,$$

and using (*) we obtain a contradiction,

$$int_a^infty f(x) , dx > sum_n=N^infty fracbeta^22 = + infty$$

share|cite|improve this answer

edited Apr 4 at 4:57

answered Apr 3 at 0:33

RRLRRL

53.9k52675

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do we know that the sequence $(x_n)_ninmathbb N$ exist? A continuous function doesn't to have a limit at infinity. Is the reason simply that as long as $(x_n)_ninmathbb N$ is any sequence with $x_ntoinfty$, we know by assumption $limsup_ntoinftyf(x_n)>0$ and hence we may select a subsequence which converges to the limit superior?
  $endgroup$
  – 0xbadf00d
  Apr 3 at 5:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes — since $f$ is positive, if it fails to converge to $0$, then the limit is not zero or fails to exist. In either case limsup is positive and is the limit of the images of a subsequence.
  $endgroup$
  – RRL
  Apr 3 at 5:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I guess that (in order for $(ast)$ to hold) we need to assume $x_1ge a$ (which is clearly no problem). Moreover, I guess the first inequality in $(ast)$ is due to the possible gap between $r$ and $x_1$, right? (Since if $x_1=r$, then $(ast)$ should be an equality.)
  $endgroup$
  – 0xbadf00d
  Apr 3 at 15:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes it could be an equality, but what is important is the second relation which is an inequality
  $endgroup$
  – RRL
  Apr 3 at 16:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  At which point did you use the integrability? I guess at the end, but even when $int f(x):rm dx=infty$ this would be a contradiction (since your last equation would read $infty>infty$).
  $endgroup$
  – 0xbadf00d
  Apr 3 at 16:13
  

  
  
  
  

 | 
show 12 more comments

$begingroup$

If $f:mathbbR to [0,infty)$ is integrable (over $mathbbR$) and
$f'(x) to 0$ as $|x| to +infty$, then it follows that $f(x) to 0$.

For proof by contradiction, assume WLOG that $f(x) notto 0$ as $x to +infty$. There exists $beta > 0$ and a sequence $x_n to +infty$ such that $f(x_n) geqslant beta$. Selecting a subsequence if necessary, we can assume that $x_n+1 > x_n + beta.$

Since $f'(x) to 0$, there exists $a > 0$ such that $-frac12 < f'(x) < frac12$ for all $x geqslant a$, and for all sufficiently large $n geqslant N$ we also have $x_n > a$.

Thus,

$$tag*int_a^infty f(x) , dx geqslant sum_n=N^inftyint_x_n^x_n+1f(x) , dx > sum_n=N^inftyint_x_n^x_n + betaf(x) , dx.$$

By the mean value theorem for $x in (x_n,x_n + beta$) there exists $c_n in (x_n,x)$ such that

$$f(x) = f(x_n) + f'(c_n)(x - x_n)> beta - frac12beta = fracbeta2$$

Therefore, we have

$$int_x_n^x_n + betaf(x) , dx > fracbeta^22,$$

and using (*) we obtain a contradiction,

$$int_a^infty f(x) , dx > sum_n=N^infty fracbeta^22 = + infty$$

share|cite|improve this answer

edited Apr 4 at 4:57

answered Apr 3 at 0:33

RRLRRL

53.9k52675

$endgroup$

share|cite|improve this answer

edited Apr 4 at 4:57

answered Apr 3 at 0:33

RRLRRL

53.9k52675

answered Apr 3 at 0:33

RRLRRL

53.9k52675

answered Apr 3 at 0:33

RRLRRL

53.9k52675