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For each $ x= (x_n)_{n=1}^\infty \in \ell_1$ set

$$\| x \| = \sup\{ \big| \sum_{k=1}^n x_k \big|\colon n\in \mathbb{N}\} $$

Does this define a norm in $\ell_1$?

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1 Answer 1

$\left\| x \right\| = \sup \left| \sum\limits_{k=1}^n x_k \right|$

$\forall \lambda \in \Bbb C, \left\| \lambda x \right\|=\sup \left| \sum\limits_{k=1}^n \lambda x_k \right|=\sup \left| \lambda\sum\limits_{k=1}^n x_k \right|=\sup \left|\lambda\right|\left| \sum\limits_{k=1}^n x_k \right|=\left|\lambda\right|\sup \left| \sum\limits_{k=1}^n x_k \right|=\left|\lambda\right|\left\| x \right\|$

$\forall x, y \in \ell^1,\left\| x+y \right\|=\sup \left| \sum\limits_{k=1}^n x_k+y_k \right|=\sup \left| \sum\limits_{k=1}^n x_k+\sum\limits_{k=1}^ny_k \right|\le \sup\left (\left| \sum\limits_{k=1}^n x_k\right| + \left|\sum\limits_{k=1}^ny_k \right| \right)\le \sup \left| \sum\limits_{k=1}^n x_k\right| + \sup \left|\sum\limits_{k=1}^ny_k \right|=\left\| x \right\|+\left\| y \right\|$

$\forall x \in \ell ^1, \left [\left\| x \right\|=0\right ] \Leftrightarrow \left[\forall n,\sum\limits_{k=1}^n x_k =0 \right] \Leftrightarrow \left[\forall n, \sum\limits_{k=1}^{n+1} x_k- \sum\limits_{k=1}^n x_k = 0\right]\Leftrightarrow \left[\forall n, x_n = 0\right]\Leftrightarrow x = 0$

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