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

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?

• Since the series $\sum_{n=1}^\infty\left(1-\frac1n\right)$ diverges, you did nothing. – José Carlos Santos Dec 19 '18 at 13:42
• use the triangle inequality – tonydo Dec 19 '18 at 13:45

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 $$||f||$$.

$$x\in l^1\Rightarrow \|x\|=\sum |x_n|<\infty.$$ $$|f(x)|\leq\sum (1-1/n)|x_n|\leq \sum |x_n|=\|x\|.$$ 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, $$\|f\|=1.$$

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