Suppression-unconditional constant can replace the unconditional constant
I am trying to solve the following question.
Let $(u_n)_{n=1}^{infty}$ be an unconditional basis for a Banach space $X$ with suppression-unconditional constant $K_{su}$. Prove that for all $N$, whenever $a_1,...,a_N,b_1,...,b_N$ are scalars such that $|a_n| leq |b_n|$ for all $1 leq n leq N$ and $a_nb_n>0$, we have
$|sum_{n=1}^N a_n u_n| leq K_{su} |sum_{n=1}^{N}b_n u_n|$
I am not sure how do you use the fact that $(u_n)$ is an unconditional basis here. Any hint will be appreciated.
functional-analysis banach-spaces
asked Nov 28 '18 at 2:26
Topology
770314
add a comment |
I am trying to solve the following question.
Let $(u_n)_{n=1}^{infty}$ be an unconditional basis for a Banach space $X$ with suppression-unconditional constant $K_{su}$. Prove that for all $N$, whenever $a_1,...,a_N,b_1,...,b_N$ are scalars such that $|a_n| leq |b_n|$ for all $1 leq n leq N$ and $a_nb_n>0$, we have
$|sum_{n=1}^N a_n u_n| leq K_{su} |sum_{n=1}^{N}b_n u_n|$
I am not sure how do you use the fact that $(u_n)$ is an unconditional basis here. Any hint will be appreciated.
functional-analysis banach-spaces
asked Nov 28 '18 at 2:26
Topology
770314
add a comment |
I am trying to solve the following question.
Let $(u_n)_{n=1}^{infty}$ be an unconditional basis for a Banach space $X$ with suppression-unconditional constant $K_{su}$. Prove that for all $N$, whenever $a_1,...,a_N,b_1,...,b_N$ are scalars such that $|a_n| leq |b_n|$ for all $1 leq n leq N$ and $a_nb_n>0$, we have
$|sum_{n=1}^N a_n u_n| leq K_{su} |sum_{n=1}^{N}b_n u_n|$
I am not sure how do you use the fact that $(u_n)$ is an unconditional basis here. Any hint will be appreciated.
functional-analysis banach-spaces
asked Nov 28 '18 at 2:26
Topology
770314
I am trying to solve the following question.
Let $(u_n)_{n=1}^{infty}$ be an unconditional basis for a Banach space $X$ with suppression-unconditional constant $K_{su}$. Prove that for all $N$, whenever $a_1,...,a_N,b_1,...,b_N$ are scalars such that $|a_n| leq |b_n|$ for all $1 leq n leq N$ and $a_nb_n>0$, we have
$|sum_{n=1}^N a_n u_n| leq K_{su} |sum_{n=1}^{N}b_n u_n|$
I am not sure how do you use the fact that $(u_n)$ is an unconditional basis here. Any hint will be appreciated.
functional-analysis banach-spaces
functional-analysis banach-spaces
asked Nov 28 '18 at 2:26
Topology
770314
asked Nov 28 '18 at 2:26
Topology
770314
asked Nov 28 '18 at 2:26
Topology
770314
asked Nov 28 '18 at 2:26
Topology
770314
asked Nov 28 '18 at 2:26
Topology
770314
770314
add a comment |
add a comment |
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