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Let $f : l^{1} \to {\mathbb R}$ and

$f(x) = \sum (1-1/n) x_n$

Where $x = (x_1, x_2 , \ldots)$

Show that $f$ is a bounded linear functional and find its norm.

My work : $| f | = | \sum (1-1/n) x_n |$

$\leq \sum | (1-1/n) | \cdot | x_n |$

$<\leq \| x \| \sum (1-1/n)$

But what I can do after that?

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  • $\begingroup$ Since the series $\sum_{n=1}^\infty\left(1-\frac1n\right)$ diverges, you did nothing. $\endgroup$ – José Carlos Santos Dec 19 '18 at 13:42
  • $\begingroup$ use the triangle inequality $\endgroup$ – tonydo Dec 19 '18 at 13:45
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We have $|f(x)| \le \sum_{n=1}^{\infty}(1- \frac{1}{n})|x_n| \le \sum_{n=1}^{\infty}|x_n| = ||x||_1$.

Hence $f$ is bounded and $||f|| \le 1$. It is your turn to determine $||f||$.

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$x\in l^1\Rightarrow \|x\|=\sum |x_n|<\infty.$ $|f(x)|\leq\sum (1-1/n)|x_n|\leq \sum |x_n|=\|x\|.$ So, $\|f\|\leq 1.$ For $\|f\|=\sup_{\|x\|\leq 1}|f(x)|.$ Taking $x_i=0$ for $i=1(1)n$ with $\|x\|=1$ and $x_i\geq 0$ we get $\|f\|\geq |f(x)|=\sum (1-1/i)x_i\geq (1-1/n)\|x\|=1-1/n.$ So, $\|f\|=1.$

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    $\begingroup$ Space is free :) Your answer would be a lot more readable if you moved distinct steps onto separate lines. $\endgroup$ – postmortes Dec 19 '18 at 14:16

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