Question about lim inf or lim sup Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Lim Sup and Lim InfQuestions about Levi Convergence Theorem in Real AnalysisShow that $lim inf a_nleliminf s_n.$Integrating the two-views of lim sup and lim infProperties of sup and lim inf.Fatou's lemma. Case of convergence in measureProve that $lim inf dfracx_n+1x_nleq lim inf x_n^1over n leq lim sup x_n^1over n leq lim sup dfracx_n+1x_n. $Theorem 3.37 in Baby Rudin: $liminffracc_n+1c_nleqliminfsqrt[n]c_nleqlimsupsqrt[n]c_nleq limsupfracc_n+1c_n$Monotone Convergence theorem for decreasing sequenceHow to deal with lim sup and lim inf?
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Question about lim inf or lim sup
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Lim Sup and Lim InfQuestions about Levi Convergence Theorem in Real AnalysisShow that $lim inf a_nleliminf s_n.$Integrating the two-views of lim sup and lim infProperties of sup and lim inf.Fatou's lemma. Case of convergence in measureProve that $lim inf dfracx_n+1x_nleq lim inf x_n^1over n leq lim sup x_n^1over n leq lim sup dfracx_n+1x_n. $Theorem 3.37 in Baby Rudin: $liminffracc_n+1c_nleqliminfsqrt[n]c_nleqlimsupsqrt[n]c_nleq limsupfracc_n+1c_n$Monotone Convergence theorem for decreasing sequenceHow to deal with lim sup and lim inf?
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When taking the course of real analysis, the professor wrote the following:
Let $f_n geq 0$ for all $n$.
Let $$a = lim inf int_X f_n dmu + epsilon$$
I think here $epsilon$ is any positive number.
Now, by the definition of $lim inf$, we have
$$forall N, exists, n >N, text such that int_X f_ndmu <a$$
My question is, if $f_n$ is decreasing, i.e., $$f_1geq f_2 geq f_3 cdots geq f_n geq cdots,$$ then how could the conclusion hold? (if choosing $epsilon rightarrow 0$)
real-analysis limsup-and-liminf
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add a comment |
$begingroup$
When taking the course of real analysis, the professor wrote the following:
Let $f_n geq 0$ for all $n$.
Let $$a = lim inf int_X f_n dmu + epsilon$$
I think here $epsilon$ is any positive number.
Now, by the definition of $lim inf$, we have
$$forall N, exists, n >N, text such that int_X f_ndmu <a$$
My question is, if $f_n$ is decreasing, i.e., $$f_1geq f_2 geq f_3 cdots geq f_n geq cdots,$$ then how could the conclusion hold? (if choosing $epsilon rightarrow 0$)
real-analysis limsup-and-liminf
$endgroup$
$begingroup$
Did you mean $exists N forall n>N$?
$endgroup$
– Reveillark
Mar 27 at 3:38
$begingroup$
@Reveillark No. But I am not sure if this is a typo made from that professor.
$endgroup$
– sleeve chen
Mar 27 at 4:38
add a comment |
$begingroup$
When taking the course of real analysis, the professor wrote the following:
Let $f_n geq 0$ for all $n$.
Let $$a = lim inf int_X f_n dmu + epsilon$$
I think here $epsilon$ is any positive number.
Now, by the definition of $lim inf$, we have
$$forall N, exists, n >N, text such that int_X f_ndmu <a$$
My question is, if $f_n$ is decreasing, i.e., $$f_1geq f_2 geq f_3 cdots geq f_n geq cdots,$$ then how could the conclusion hold? (if choosing $epsilon rightarrow 0$)
real-analysis limsup-and-liminf
$endgroup$
When taking the course of real analysis, the professor wrote the following:
Let $f_n geq 0$ for all $n$.
Let $$a = lim inf int_X f_n dmu + epsilon$$
I think here $epsilon$ is any positive number.
Now, by the definition of $lim inf$, we have
$$forall N, exists, n >N, text such that int_X f_ndmu <a$$
My question is, if $f_n$ is decreasing, i.e., $$f_1geq f_2 geq f_3 cdots geq f_n geq cdots,$$ then how could the conclusion hold? (if choosing $epsilon rightarrow 0$)
real-analysis limsup-and-liminf
real-analysis limsup-and-liminf
asked Mar 27 at 3:18
sleeve chensleeve chen
3,19342256
3,19342256
$begingroup$
Did you mean $exists N forall n>N$?
$endgroup$
– Reveillark
Mar 27 at 3:38
$begingroup$
@Reveillark No. But I am not sure if this is a typo made from that professor.
$endgroup$
– sleeve chen
Mar 27 at 4:38
add a comment |
$begingroup$
Did you mean $exists N forall n>N$?
$endgroup$
– Reveillark
Mar 27 at 3:38
$begingroup$
@Reveillark No. But I am not sure if this is a typo made from that professor.
$endgroup$
– sleeve chen
Mar 27 at 4:38
$begingroup$
Did you mean $exists N forall n>N$?
$endgroup$
– Reveillark
Mar 27 at 3:38
$begingroup$
Did you mean $exists N forall n>N$?
$endgroup$
– Reveillark
Mar 27 at 3:38
$begingroup$
@Reveillark No. But I am not sure if this is a typo made from that professor.
$endgroup$
– sleeve chen
Mar 27 at 4:38
$begingroup$
@Reveillark No. But I am not sure if this is a typo made from that professor.
$endgroup$
– sleeve chen
Mar 27 at 4:38
add a comment |
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$begingroup$
Did you mean $exists N forall n>N$?
$endgroup$
– Reveillark
Mar 27 at 3:38
$begingroup$
@Reveillark No. But I am not sure if this is a typo made from that professor.
$endgroup$
– sleeve chen
Mar 27 at 4:38