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Hot answers tagged banach-spaces

4

$\left\Vert p_{n}-p_{m}\right\Vert =\sup_{\left[ 0,1\right] }\left\vert %TCIMACRO{\dsum \limits_{i=0}^{n}}% %BeginExpansion {\displaystyle\sum\limits_{i=0}^{n}} %EndExpansion \frac{x^{i}}{i!}-% %TCIMACRO{\dsum \limits_{i=0}^{m}}% %BeginExpansion {\displaystyle\sum\limits_{i=0}^{m}} %EndExpansion \frac{x^{i}}{i!}\right\vert =\sup_{\left[ 0,1\right] ... 2 Nothing would go wrong per se, but in an incomplete normed space, there are simply much fewer weakly compact sets. Let$X$be a normed space, and$\tilde{X}$its completion. Then we have a canonical isometric isomorphism$\rho \colon \tilde{X}' \to X'$, namely the restriction of a functional to$X\$. Thus we may identify the two spaces (that is habitually ...

1

Approximate Unit Regard ball cone: $$\mathcal{B}_+:=\{A\in\mathcal{A}:\|A\|<1:A\geq0\}$$ Order elements: $$E,E'\in\mathcal{I}\cap\mathcal{B}_+:\quad E\leq E'$$ Then one has: $$I\in\mathcal{I}:\quad\|I-IE\|,\|I-EI\|\stackrel{E\to1}{\longrightarrow}0$$ (That is the hard part!) Quotient Norm Note that it holds: $$1-\sigma(E)\geq0\implies\|1-E\|\leq1$$ ...

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