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If $|f|

4

$begingroup$

Let $f: X to mathbb C$ be integrable in a measure space $(X, mathfrak M, mu)$, i.e. $int |f| , dmu < infty$. Suppose that $|f(x)| leq 1$ for all $x in X$. How can one compute the limit
$$ lim_{nto infty} int left ( frac{f ^n}{1 + n |f|} right ), dmu quad ? $$

My attempt:

I want to find a Lebesgue integrable $g$ that dominates the sequence $f_n = frac{f ^n}{1 + n |f|}$ and, then, I would conclude that

$$ lim_{nto infty} int left ( frac{f ^n}{1 + n |f|} right ) dmu = int left ( lim frac{f ^n}{1 + n |f|} right ) dmu = 0, $$
since $|f| < 1$ implies $lim frac{f ^n}{1 + n |f|} = 0$.

My problem is in find such function $g$, I can see that $|f_n| < 1$ for each $n$, however the function $g = 1$ does not need to be Lebesgue integrable since $mu(X)$ maybe $infty$.

Help?

real-analysis integration lebesgue-integral

share|cite|improve this question

asked Mar 22 at 21:23

user 242964user 242964

746413

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Isn't your sequence monotonically decreasing? In that case, you could use $frac{f^1}{1+|f|}$ as your $g$ and then apply dominated convergence theorem.
  $endgroup$
  – kkc
  Mar 22 at 21:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kkc why $frac{f}{1+ |f|}$ is Lebesgue integrable?
  $endgroup$
  – user 242964
  Mar 22 at 21:39
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Because $$left|frac {f^n} {1+n|f|}right|leq |f|^nleq|f|$$ In fact, you could take $f$ as your $g$.
  $endgroup$
  – Stefan Lafon
  Mar 22 at 21:49
  
  
  

  
  
  
  

add a comment | 

4

$begingroup$

Let $f: X to mathbb C$ be integrable in a measure space $(X, mathfrak M, mu)$, i.e. $int |f| , dmu < infty$. Suppose that $|f(x)| leq 1$ for all $x in X$. How can one compute the limit
$$ lim_{nto infty} int left ( frac{f ^n}{1 + n |f|} right ), dmu quad ? $$

My attempt:

I want to find a Lebesgue integrable $g$ that dominates the sequence $f_n = frac{f ^n}{1 + n |f|}$ and, then, I would conclude that

$$ lim_{nto infty} int left ( frac{f ^n}{1 + n |f|} right ) dmu = int left ( lim frac{f ^n}{1 + n |f|} right ) dmu = 0, $$
since $|f| < 1$ implies $lim frac{f ^n}{1 + n |f|} = 0$.

My problem is in find such function $g$, I can see that $|f_n| < 1$ for each $n$, however the function $g = 1$ does not need to be Lebesgue integrable since $mu(X)$ maybe $infty$.

Help?

real-analysis integration lebesgue-integral

share|cite|improve this question

asked Mar 22 at 21:23

user 242964user 242964

746413

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Isn't your sequence monotonically decreasing? In that case, you could use $frac{f^1}{1+|f|}$ as your $g$ and then apply dominated convergence theorem.
  $endgroup$
  – kkc
  Mar 22 at 21:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kkc why $frac{f}{1+ |f|}$ is Lebesgue integrable?
  $endgroup$
  – user 242964
  Mar 22 at 21:39
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Because $$left|frac {f^n} {1+n|f|}right|leq |f|^nleq|f|$$ In fact, you could take $f$ as your $g$.
  $endgroup$
  – Stefan Lafon
  Mar 22 at 21:49
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $f: X to mathbb C$ be integrable in a measure space $(X, mathfrak M, mu)$, i.e. $int |f| , dmu < infty$. Suppose that $|f(x)| leq 1$ for all $x in X$. How can one compute the limit
$$ lim_{nto infty} int left ( frac{f ^n}{1 + n |f|} right ), dmu quad ? $$

My attempt:

I want to find a Lebesgue integrable $g$ that dominates the sequence $f_n = frac{f ^n}{1 + n |f|}$ and, then, I would conclude that

$$ lim_{nto infty} int left ( frac{f ^n}{1 + n |f|} right ) dmu = int left ( lim frac{f ^n}{1 + n |f|} right ) dmu = 0, $$
since $|f| < 1$ implies $lim frac{f ^n}{1 + n |f|} = 0$.

My problem is in find such function $g$, I can see that $|f_n| < 1$ for each $n$, however the function $g = 1$ does not need to be Lebesgue integrable since $mu(X)$ maybe $infty$.

Help?

real-analysis integration lebesgue-integral

share|cite|improve this question

asked Mar 22 at 21:23

user 242964user 242964

746413

$endgroup$

share|cite|improve this question

asked Mar 22 at 21:23

user 242964user 242964

746413

share|cite|improve this question

asked Mar 22 at 21:23

user 242964user 242964

746413

asked Mar 22 at 21:23

user 242964user 242964

746413

asked Mar 22 at 21:23

user 242964user 242964

746413