Mixing
How this inequality is derived?
$begingroup$
Let $T$ ∶ $ℓ_2$ → $ℓ_2$ be defined by
$T((x_1,x_2,...,x_n...))$=$(X_2-X_1, X_3-X_2,...,X_{n+1}-x_n,...)$
Then I have find the norm of $T$.
Here is the answer to this question: https://math.stackexchange.com/a/1647794/581242
I am not able to see how this first inequality is derived.
$|(Tx)| = sqrt{sum_{i=1}^infty |x_{i+1}-x_i|^2} leq
sqrt{sum_{i=1}^infty |x_{i+1}|^2 + sum_{i=1}^infty|x_i|^2} leq 2|x|$
functional-analysis normed-spaces
asked Jan 26 at 0:43
StammeringMathematicianStammeringMathematician
2,5881324
$endgroup$
add a comment |
$begingroup$
Let $T$ ∶ $ℓ_2$ → $ℓ_2$ be defined by
$T((x_1,x_2,...,x_n...))$=$(X_2-X_1, X_3-X_2,...,X_{n+1}-x_n,...)$
Then I have find the norm of $T$.
Here is the answer to this question: https://math.stackexchange.com/a/1647794/581242
I am not able to see how this first inequality is derived.
$|(Tx)| = sqrt{sum_{i=1}^infty |x_{i+1}-x_i|^2} leq
sqrt{sum_{i=1}^infty |x_{i+1}|^2 + sum_{i=1}^infty|x_i|^2} leq 2|x|$
functional-analysis normed-spaces
asked Jan 26 at 0:43
StammeringMathematicianStammeringMathematician
2,5881324
$endgroup$
1
$begingroup$
Hint: apply Minkowski's inequality.
$endgroup$
– Victoria M
Jan 26 at 0:47
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The first inequality is wrong. @StammeringMathematician
$endgroup$
– Kavi Rama Murthy
Jan 26 at 0:49
$begingroup$
@KaviRamaMurthy Thanks for pointing that out. Now I can see it after your answer below.
$endgroup$
– StammeringMathematician
Jan 26 at 0:53
$begingroup$
@KaviRamaMurthy Is this an application of Minkowski's inequality.
$endgroup$
– StammeringMathematician
Jan 26 at 0:54
add a comment |
$begingroup$
Let $T$ ∶ $ℓ_2$ → $ℓ_2$ be defined by
$T((x_1,x_2,...,x_n...))$=$(X_2-X_1, X_3-X_2,...,X_{n+1}-x_n,...)$
Then I have find the norm of $T$.
Here is the answer to this question: https://math.stackexchange.com/a/1647794/581242
I am not able to see how this first inequality is derived.
$|(Tx)| = sqrt{sum_{i=1}^infty |x_{i+1}-x_i|^2} leq
sqrt{sum_{i=1}^infty |x_{i+1}|^2 + sum_{i=1}^infty|x_i|^2} leq 2|x|$
functional-analysis normed-spaces
asked Jan 26 at 0:43
StammeringMathematicianStammeringMathematician
2,5881324
$endgroup$
Let $T$ ∶ $ℓ_2$ → $ℓ_2$ be defined by
$T((x_1,x_2,...,x_n...))$=$(X_2-X_1, X_3-X_2,...,X_{n+1}-x_n,...)$
Then I have find the norm of $T$.
Here is the answer to this question: https://math.stackexchange.com/a/1647794/581242
I am not able to see how this first inequality is derived.
$|(Tx)| = sqrt{sum_{i=1}^infty |x_{i+1}-x_i|^2} leq
sqrt{sum_{i=1}^infty |x_{i+1}|^2 + sum_{i=1}^infty|x_i|^2} leq 2|x|$
functional-analysis normed-spaces
functional-analysis normed-spaces
asked Jan 26 at 0:43
StammeringMathematicianStammeringMathematician
2,5881324
asked Jan 26 at 0:43
StammeringMathematicianStammeringMathematician
2,5881324
asked Jan 26 at 0:43
StammeringMathematicianStammeringMathematician
2,5881324
asked Jan 26 at 0:43
StammeringMathematicianStammeringMathematician
2,5881324
asked Jan 26 at 0:43
StammeringMathematicianStammeringMathematician
2,5881324
2,5881324
1
$begingroup$
Hint: apply Minkowski's inequality.
$endgroup$
– Victoria M
Jan 26 at 0:47
$begingroup$
The first inequality is wrong. @StammeringMathematician
$endgroup$
– Kavi Rama Murthy
Jan 26 at 0:49
$begingroup$
@KaviRamaMurthy Thanks for pointing that out. Now I can see it after your answer below.
$endgroup$
– StammeringMathematician
Jan 26 at 0:53
$begingroup$
@KaviRamaMurthy Is this an application of Minkowski's inequality.
$endgroup$
– StammeringMathematician
Jan 26 at 0:54
add a comment |
1
$begingroup$
Hint: apply Minkowski's inequality.
$endgroup$
– Victoria M
Jan 26 at 0:47
$begingroup$
The first inequality is wrong. @StammeringMathematician
$endgroup$
– Kavi Rama Murthy
Jan 26 at 0:49
$begingroup$
@KaviRamaMurthy Thanks for pointing that out. Now I can see it after your answer below.
$endgroup$
– StammeringMathematician
Jan 26 at 0:53
$begingroup$
@KaviRamaMurthy Is this an application of Minkowski's inequality.
$endgroup$
– StammeringMathematician
Jan 26 at 0:54
1
1
$begingroup$
Hint: apply Minkowski's inequality.
$endgroup$
– Victoria M
Jan 26 at 0:47
$begingroup$
Hint: apply Minkowski's inequality.
$endgroup$
– Victoria M
Jan 26 at 0:47
$begingroup$
The first inequality is wrong. @StammeringMathematician
$endgroup$
– Kavi Rama Murthy
Jan 26 at 0:49
$begingroup$
The first inequality is wrong. @StammeringMathematician
$endgroup$
– Kavi Rama Murthy
Jan 26 at 0:49
$begingroup$
@KaviRamaMurthy Thanks for pointing that out. Now I can see it after your answer below.
$endgroup$
– StammeringMathematician
Jan 26 at 0:53
$begingroup$
@KaviRamaMurthy Thanks for pointing that out. Now I can see it after your answer below.
$endgroup$
– StammeringMathematician
Jan 26 at 0:53
$begingroup$
@KaviRamaMurthy Is this an application of Minkowski's inequality.
$endgroup$
– StammeringMathematician
Jan 26 at 0:54
$begingroup$
@KaviRamaMurthy Is this an application of Minkowski's inequality.
$endgroup$
– StammeringMathematician
Jan 26 at 0:54
add a comment |
1 Answer
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$begingroup$
It appears that there was a small typo. Use the inequality $|a-b|^{2} leq 2(a^{2}+b^{2})$ . The final result is correct but the first inequality is wrongly stated.
answered Jan 26 at 0:47
Kavi Rama MurthyKavi Rama Murthy
68.8k53169
$endgroup$
add a comment |
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$begingroup$
It appears that there was a small typo. Use the inequality $|a-b|^{2} leq 2(a^{2}+b^{2})$ . The final result is correct but the first inequality is wrongly stated.
answered Jan 26 at 0:47
Kavi Rama MurthyKavi Rama Murthy
68.8k53169
$endgroup$
add a comment |
$begingroup$
It appears that there was a small typo. Use the inequality $|a-b|^{2} leq 2(a^{2}+b^{2})$ . The final result is correct but the first inequality is wrongly stated.
answered Jan 26 at 0:47
Kavi Rama MurthyKavi Rama Murthy
68.8k53169
$endgroup$
add a comment |
$begingroup$
It appears that there was a small typo. Use the inequality $|a-b|^{2} leq 2(a^{2}+b^{2})$ . The final result is correct but the first inequality is wrongly stated.
answered Jan 26 at 0:47
Kavi Rama MurthyKavi Rama Murthy
68.8k53169
$endgroup$
It appears that there was a small typo. Use the inequality $|a-b|^{2} leq 2(a^{2}+b^{2})$ . The final result is correct but the first inequality is wrongly stated.
answered Jan 26 at 0:47
Kavi Rama MurthyKavi Rama Murthy
68.8k53169
answered Jan 26 at 0:47
Kavi Rama MurthyKavi Rama Murthy
68.8k53169
answered Jan 26 at 0:47
Kavi Rama MurthyKavi Rama Murthy
68.8k53169
answered Jan 26 at 0:47
Kavi Rama MurthyKavi Rama Murthy
68.8k53169
68.8k53169
add a comment |
add a comment |
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1
$begingroup$
Hint: apply Minkowski's inequality.
$endgroup$
– Victoria M
Jan 26 at 0:47
$begingroup$
The first inequality is wrong. @StammeringMathematician
$endgroup$
– Kavi Rama Murthy
Jan 26 at 0:49
$begingroup$
@KaviRamaMurthy Thanks for pointing that out. Now I can see it after your answer below.
$endgroup$
– StammeringMathematician
Jan 26 at 0:53
$begingroup$
@KaviRamaMurthy Is this an application of Minkowski's inequality.
$endgroup$
– StammeringMathematician
Jan 26 at 0:54