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By duality and Hahn Banach theorem, we know that for $x\in \ell^1$, its norm can be computed as

$$\|x\|_1=\sup_{\|\beta\|_\infty=1} \left|\sum_k x_k \beta_k\right|.$$

To obtain the norm, in that supremum, is it enough to consider elements $\beta$ such that $\beta_k=\pm 1$?

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2 Answers 2

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Yes it is, for negative $x_k$ you'd want $\beta_k = -1$, for positive $x_k$, you'd want $\beta_k = 1$. This gives you the maximum contribution to the sum for sequence $\beta$ whos maximum is $1$.

It will boil down to $||x||_1 = \sum_k |x_k|$ as you'd expect.

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  • $\begingroup$ Thanks! There is a way to adapt that idea when the sequences take values in $\mathbb{C} $? Would be forced to take $\frac{\overline{x_k} } {|x_k|}$ i guess? $\endgroup$
    – Mark_Hoffman
    Apr 17, 2019 at 20:56
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    $\begingroup$ Yeah sure take $\beta_k = \bar{x_k}/|x_k|$ $\endgroup$
    – fGDu94
    Apr 17, 2019 at 20:57
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Note that $$\left| \sum_k x_k \beta_k \right| \le \sum_k |x_k \beta_k|\le \sum_k |x_k|$$

Assuming you're working with real numbers (rather than complex), $|x_k| = x_k \beta_k$ where $\beta_k = 1$ if $x_k \ge 0$, $-1$ if $x_k < 0$.

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  • $\begingroup$ Thanks to you too! $\endgroup$
    – Mark_Hoffman
    Apr 17, 2019 at 21:05

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