# About $l^1$ norm

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 $$\beta_k=\pm 1$$?

## 2 Answers

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.

• 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? – Mark_Hoffman Apr 17 at 20:56
• Yeah sure take $\beta_k = \bar{x_k}/|x_k|$ – George Dewhirst Apr 17 at 20:57

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 $$x_k < 0$$.

• Thanks to you too! – Mark_Hoffman Apr 17 at 21:05