Prove that $S_{n} : X to X$ is continuous mapping.
$begingroup$
Consider $X$ - Banach space with Schauder basis : $forall x in X$ $exists!$ $x = sum_{i=1}^{infty} x_{i}e_{i}$.
Let's consider $S_{n} = sum_{i = 1}^{n} x_{i}e_{i}$. We want to show that it is a continous mapping (i.e. $exists C > 0: $ $|S_n| le C |x|$, for all $x in X$).
How should I start with? Should I use Cauchy-sequences?
real-analysis functional-analysis banach-spaces
$endgroup$
add a comment |
$begingroup$
Consider $X$ - Banach space with Schauder basis : $forall x in X$ $exists!$ $x = sum_{i=1}^{infty} x_{i}e_{i}$.
Let's consider $S_{n} = sum_{i = 1}^{n} x_{i}e_{i}$. We want to show that it is a continous mapping (i.e. $exists C > 0: $ $|S_n| le C |x|$, for all $x in X$).
How should I start with? Should I use Cauchy-sequences?
real-analysis functional-analysis banach-spaces
$endgroup$
add a comment |
$begingroup$
Consider $X$ - Banach space with Schauder basis : $forall x in X$ $exists!$ $x = sum_{i=1}^{infty} x_{i}e_{i}$.
Let's consider $S_{n} = sum_{i = 1}^{n} x_{i}e_{i}$. We want to show that it is a continous mapping (i.e. $exists C > 0: $ $|S_n| le C |x|$, for all $x in X$).
How should I start with? Should I use Cauchy-sequences?
real-analysis functional-analysis banach-spaces
$endgroup$
Consider $X$ - Banach space with Schauder basis : $forall x in X$ $exists!$ $x = sum_{i=1}^{infty} x_{i}e_{i}$.
Let's consider $S_{n} = sum_{i = 1}^{n} x_{i}e_{i}$. We want to show that it is a continous mapping (i.e. $exists C > 0: $ $|S_n| le C |x|$, for all $x in X$).
How should I start with? Should I use Cauchy-sequences?
real-analysis functional-analysis banach-spaces
real-analysis functional-analysis banach-spaces
asked Dec 20 '18 at 17:45
openspaceopenspace
3,4512822
3,4512822
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$begingroup$
$x = sum_{i=1}^infty x_i e_i$ means that the partial sums converge to $x$ in norm. So, if you add and subtract $x$ inside $|S_n|$, then apply the triangle inequality, you should be able to finish the proof.
$endgroup$
add a comment |
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$begingroup$
$x = sum_{i=1}^infty x_i e_i$ means that the partial sums converge to $x$ in norm. So, if you add and subtract $x$ inside $|S_n|$, then apply the triangle inequality, you should be able to finish the proof.
$endgroup$
add a comment |
$begingroup$
$x = sum_{i=1}^infty x_i e_i$ means that the partial sums converge to $x$ in norm. So, if you add and subtract $x$ inside $|S_n|$, then apply the triangle inequality, you should be able to finish the proof.
$endgroup$
add a comment |
$begingroup$
$x = sum_{i=1}^infty x_i e_i$ means that the partial sums converge to $x$ in norm. So, if you add and subtract $x$ inside $|S_n|$, then apply the triangle inequality, you should be able to finish the proof.
$endgroup$
$x = sum_{i=1}^infty x_i e_i$ means that the partial sums converge to $x$ in norm. So, if you add and subtract $x$ inside $|S_n|$, then apply the triangle inequality, you should be able to finish the proof.
answered Dec 20 '18 at 19:05
DunhamDunham
2,074614
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