some confusion about supremum and infimum
$begingroup$
Let $a_{mn}$ be a double array or real numbers. Define $$A=liminf_{nrightarrow infty} limsup_{mrightarrowinfty}a_{mn}\B=limsup_{nrightarrow infty} liminf_{mrightarrowinfty}a_{mn},$$
Then which one is true?
- $Ale B$
- $Age B $
My attempt :
I think option $1$ that $Ale B$ is true because the supremum is always greater than then infimum.
Is this correct ? Any hints/solution would be appreciated.
real-analysis
edited Dec 15 '18 at 7:50
Jonathan
16312
asked Dec 15 '18 at 6:02
jasminejasmine
1,781417
$endgroup$
add a comment |
$begingroup$
Let $a_{mn}$ be a double array or real numbers. Define $$A=liminf_{nrightarrow infty} limsup_{mrightarrowinfty}a_{mn}\B=limsup_{nrightarrow infty} liminf_{mrightarrowinfty}a_{mn},$$
Then which one is true?
- $Ale B$
- $Age B $
My attempt :
I think option $1$ that $Ale B$ is true because the supremum is always greater than then infimum.
Is this correct ? Any hints/solution would be appreciated.
real-analysis
edited Dec 15 '18 at 7:50
Jonathan
16312
asked Dec 15 '18 at 6:02
jasminejasmine
1,781417
$endgroup$
add a comment |
$begingroup$
Let $a_{mn}$ be a double array or real numbers. Define $$A=liminf_{nrightarrow infty} limsup_{mrightarrowinfty}a_{mn}\B=limsup_{nrightarrow infty} liminf_{mrightarrowinfty}a_{mn},$$
Then which one is true?
- $Ale B$
- $Age B $
My attempt :
I think option $1$ that $Ale B$ is true because the supremum is always greater than then infimum.
Is this correct ? Any hints/solution would be appreciated.
real-analysis
edited Dec 15 '18 at 7:50
Jonathan
16312
asked Dec 15 '18 at 6:02
jasminejasmine
1,781417
$endgroup$
Let $a_{mn}$ be a double array or real numbers. Define $$A=liminf_{nrightarrow infty} limsup_{mrightarrowinfty}a_{mn}\B=limsup_{nrightarrow infty} liminf_{mrightarrowinfty}a_{mn},$$
Then which one is true?
- $Ale B$
- $Age B $
My attempt :
I think option $1$ that $Ale B$ is true because the supremum is always greater than then infimum.
Is this correct ? Any hints/solution would be appreciated.
real-analysis
real-analysis
edited Dec 15 '18 at 7:50
Jonathan
16312
asked Dec 15 '18 at 6:02
jasminejasmine
1,781417
edited Dec 15 '18 at 7:50
Jonathan
16312
asked Dec 15 '18 at 6:02
jasminejasmine
1,781417
edited Dec 15 '18 at 7:50
Jonathan
16312
edited Dec 15 '18 at 7:50
Jonathan
16312
edited Dec 15 '18 at 7:50
Jonathan
16312
16312
asked Dec 15 '18 at 6:02
jasminejasmine
1,781417
asked Dec 15 '18 at 6:02
jasminejasmine
1,781417
asked Dec 15 '18 at 6:02
jasminejasmine
1,781417
1,781417
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Of course the supremum is always greater than or equal to the infimum. However, if one has mixtures of $limsup$ and $liminf$, it is indecisive which one is greater. For example, let $$
a_{mn} = (-1)^n (1-frac{1}{2^m}).
$$ Then we have
$$
1=liminf_{mtoinfty}limsup_{ntoinfty}a_{mn} >limsup_{mtoinfty}liminf_{ntoinfty}a_{mn} =-1
$$ while
$$
-1=liminf_{ntoinfty}limsup_{mtoinfty}a_{mn}<limsup_{ntoinfty}liminf_{mtoinfty}a_{mn}=1.
$$
answered Dec 15 '18 at 7:11
SongSong
14.1k1633
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Of course the supremum is always greater than or equal to the infimum. However, if one has mixtures of $limsup$ and $liminf$, it is indecisive which one is greater. For example, let $$
a_{mn} = (-1)^n (1-frac{1}{2^m}).
$$ Then we have
$$
1=liminf_{mtoinfty}limsup_{ntoinfty}a_{mn} >limsup_{mtoinfty}liminf_{ntoinfty}a_{mn} =-1
$$ while
$$
-1=liminf_{ntoinfty}limsup_{mtoinfty}a_{mn}<limsup_{ntoinfty}liminf_{mtoinfty}a_{mn}=1.
$$
answered Dec 15 '18 at 7:11
SongSong
14.1k1633
$endgroup$
add a comment |
$begingroup$
Of course the supremum is always greater than or equal to the infimum. However, if one has mixtures of $limsup$ and $liminf$, it is indecisive which one is greater. For example, let $$
a_{mn} = (-1)^n (1-frac{1}{2^m}).
$$ Then we have
$$
1=liminf_{mtoinfty}limsup_{ntoinfty}a_{mn} >limsup_{mtoinfty}liminf_{ntoinfty}a_{mn} =-1
$$ while
$$
-1=liminf_{ntoinfty}limsup_{mtoinfty}a_{mn}<limsup_{ntoinfty}liminf_{mtoinfty}a_{mn}=1.
$$
answered Dec 15 '18 at 7:11
SongSong
14.1k1633
$endgroup$
add a comment |
$begingroup$
Of course the supremum is always greater than or equal to the infimum. However, if one has mixtures of $limsup$ and $liminf$, it is indecisive which one is greater. For example, let $$
a_{mn} = (-1)^n (1-frac{1}{2^m}).
$$ Then we have
$$
1=liminf_{mtoinfty}limsup_{ntoinfty}a_{mn} >limsup_{mtoinfty}liminf_{ntoinfty}a_{mn} =-1
$$ while
$$
-1=liminf_{ntoinfty}limsup_{mtoinfty}a_{mn}<limsup_{ntoinfty}liminf_{mtoinfty}a_{mn}=1.
$$
answered Dec 15 '18 at 7:11
SongSong
14.1k1633
$endgroup$
Of course the supremum is always greater than or equal to the infimum. However, if one has mixtures of $limsup$ and $liminf$, it is indecisive which one is greater. For example, let $$
a_{mn} = (-1)^n (1-frac{1}{2^m}).
$$ Then we have
$$
1=liminf_{mtoinfty}limsup_{ntoinfty}a_{mn} >limsup_{mtoinfty}liminf_{ntoinfty}a_{mn} =-1
$$ while
$$
-1=liminf_{ntoinfty}limsup_{mtoinfty}a_{mn}<limsup_{ntoinfty}liminf_{mtoinfty}a_{mn}=1.
$$
answered Dec 15 '18 at 7:11
SongSong
14.1k1633
answered Dec 15 '18 at 7:11
SongSong
14.1k1633
answered Dec 15 '18 at 7:11
SongSong
14.1k1633
answered Dec 15 '18 at 7:11
SongSong
14.1k1633
14.1k1633
add a comment |
add a comment |
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