Mixing
Existence of dominating, monotone decreasing square summable series
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Suppose ${a_i}_{iinBbb N}$ is an $ell^2$ sequence. Does there necessarily exist a monotone decreasing sequence ${b_i}_{iinBbb N}$ in $ell^2$ with $b_k geq |a_k|$ for all $k$?
sequences-and-series analysis
asked Jan 23 at 6:43
user541020user541020
735
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Suppose ${a_i}_{iinBbb N}$ is an $ell^2$ sequence. Does there necessarily exist a monotone decreasing sequence ${b_i}_{iinBbb N}$ in $ell^2$ with $b_k geq |a_k|$ for all $k$?
sequences-and-series analysis
asked Jan 23 at 6:43
user541020user541020
735
$endgroup$
add a comment |
$begingroup$
Suppose ${a_i}_{iinBbb N}$ is an $ell^2$ sequence. Does there necessarily exist a monotone decreasing sequence ${b_i}_{iinBbb N}$ in $ell^2$ with $b_k geq |a_k|$ for all $k$?
sequences-and-series analysis
asked Jan 23 at 6:43
user541020user541020
735
$endgroup$
Suppose ${a_i}_{iinBbb N}$ is an $ell^2$ sequence. Does there necessarily exist a monotone decreasing sequence ${b_i}_{iinBbb N}$ in $ell^2$ with $b_k geq |a_k|$ for all $k$?
sequences-and-series analysis
sequences-and-series analysis
asked Jan 23 at 6:43
user541020user541020
735
asked Jan 23 at 6:43
user541020user541020
735
asked Jan 23 at 6:43
user541020user541020
735
asked Jan 23 at 6:43
user541020user541020
735
asked Jan 23 at 6:43
user541020user541020
735
735
add a comment |
add a comment |
1 Answer
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No. To construct a counterexample think of a sequence which is non-zero only very rarely (with increasing distances).
Say $a_{2^n}=1/n$ and $a_i=0$ if $i$ is not a power of $2.$ Then ${a_i}in ell^2.$ Now assume there is a monotone decreasing sequence ${b_i}$ with $b_i geq |a_i|$ for all $i.$ Then
$$
sum_{i=1}^infty b_i^2 = sum_{n=1}^infty sum_{i=2^{n-1}}^{2^n} b_i^2 ge sum_{n=1}^infty sum_{i=2^{n-1}}^{2^n} |a_{2^n}|^2 = sum_{n=1}^infty frac{2^{n-1}}{n^2},
$$
which clearly diverges, so ${b_i}notin ell^2.$
answered Jan 23 at 6:47
Reiner MartinReiner Martin
3,509414
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No. To construct a counterexample think of a sequence which is non-zero only very rarely (with increasing distances).
Say $a_{2^n}=1/n$ and $a_i=0$ if $i$ is not a power of $2.$ Then ${a_i}in ell^2.$ Now assume there is a monotone decreasing sequence ${b_i}$ with $b_i geq |a_i|$ for all $i.$ Then
$$
sum_{i=1}^infty b_i^2 = sum_{n=1}^infty sum_{i=2^{n-1}}^{2^n} b_i^2 ge sum_{n=1}^infty sum_{i=2^{n-1}}^{2^n} |a_{2^n}|^2 = sum_{n=1}^infty frac{2^{n-1}}{n^2},
$$
which clearly diverges, so ${b_i}notin ell^2.$
answered Jan 23 at 6:47
Reiner MartinReiner Martin
3,509414
$endgroup$
add a comment |
$begingroup$
No. To construct a counterexample think of a sequence which is non-zero only very rarely (with increasing distances).
Say $a_{2^n}=1/n$ and $a_i=0$ if $i$ is not a power of $2.$ Then ${a_i}in ell^2.$ Now assume there is a monotone decreasing sequence ${b_i}$ with $b_i geq |a_i|$ for all $i.$ Then
$$
sum_{i=1}^infty b_i^2 = sum_{n=1}^infty sum_{i=2^{n-1}}^{2^n} b_i^2 ge sum_{n=1}^infty sum_{i=2^{n-1}}^{2^n} |a_{2^n}|^2 = sum_{n=1}^infty frac{2^{n-1}}{n^2},
$$
which clearly diverges, so ${b_i}notin ell^2.$
answered Jan 23 at 6:47
Reiner MartinReiner Martin
3,509414
$endgroup$
add a comment |
$begingroup$
No. To construct a counterexample think of a sequence which is non-zero only very rarely (with increasing distances).
Say $a_{2^n}=1/n$ and $a_i=0$ if $i$ is not a power of $2.$ Then ${a_i}in ell^2.$ Now assume there is a monotone decreasing sequence ${b_i}$ with $b_i geq |a_i|$ for all $i.$ Then
$$
sum_{i=1}^infty b_i^2 = sum_{n=1}^infty sum_{i=2^{n-1}}^{2^n} b_i^2 ge sum_{n=1}^infty sum_{i=2^{n-1}}^{2^n} |a_{2^n}|^2 = sum_{n=1}^infty frac{2^{n-1}}{n^2},
$$
which clearly diverges, so ${b_i}notin ell^2.$
answered Jan 23 at 6:47
Reiner MartinReiner Martin
3,509414
$endgroup$
No. To construct a counterexample think of a sequence which is non-zero only very rarely (with increasing distances).
Say $a_{2^n}=1/n$ and $a_i=0$ if $i$ is not a power of $2.$ Then ${a_i}in ell^2.$ Now assume there is a monotone decreasing sequence ${b_i}$ with $b_i geq |a_i|$ for all $i.$ Then
$$
sum_{i=1}^infty b_i^2 = sum_{n=1}^infty sum_{i=2^{n-1}}^{2^n} b_i^2 ge sum_{n=1}^infty sum_{i=2^{n-1}}^{2^n} |a_{2^n}|^2 = sum_{n=1}^infty frac{2^{n-1}}{n^2},
$$
which clearly diverges, so ${b_i}notin ell^2.$
answered Jan 23 at 6:47
Reiner MartinReiner Martin
3,509414
edited Jan 23 at 6:55
answered Jan 23 at 6:47
Reiner MartinReiner Martin
3,509414
answered Jan 23 at 6:47
Reiner MartinReiner Martin
3,509414
answered Jan 23 at 6:47
Reiner MartinReiner Martin
3,509414
3,509414
add a comment |
add a comment |
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