# Prove that F is a bounded linear functional and $||F|| _{X^*}=||w||_{\infty}$. [closed]

Let $$(X,||\cdot||)=(l^1,||\cdot||_1)$$, $$w=(w_n)_{n \geq 1} \in l^{\infty}$$. Define for all $$x=(a_n)_{n \geq 1}\in l^{1}$$, $$F(x)=\sum_{n \geq 1}w_na_n.$$ Prove that $$F$$ is a bounded linear functional on $$(X,||\cdot||)$$ and that $$||F|| _{X^*}=||w||_{\infty}$$.

## closed as off-topic by user10354138, Nosrati, Saad, mrtaurho, Lord_FarinDec 23 '18 at 11:05

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• What exactly do you mean by $l^1$,$\Vert\cdot\Vert_1$, and $(w_n)_{n\geq1}$? – R. Burton Dec 9 '18 at 17:46
• @R.Burton These are classical functional analysis notations. – Rebellos Dec 9 '18 at 17:49
• By $X^*$ you denote the dual space of $X$, right ? – Rebellos Dec 9 '18 at 17:53
• Yes, that's right. – vladr10 Dec 9 '18 at 17:55

For the case of linearity, take $$x=(a_n) \in \ell^1$$ and $$y = (b_n) \in \ell^2$$ and calculate $$F(\lambda x+y)$$ with $$\lambda \in \mathbb R$$. That is left as an exercise for you.
$$|F(x)| = \bigg|\sum w_n a_n \bigg| \leq \sum|w_na_n| \leq \|w_n\|_\infty\sum |a_n| = \|w_n\|_\infty\|x\|_1$$
Thus, $$F$$ is a bounded linear operator with $$\|F\| \leq \|w_n\|_\infty \equiv \|w\|_\infty$$.
Now, the space $$X^*$$ aka the dual space of $$X$$ is the space of all the bounded linear functionals on $$X$$. But recall that a dual normed space is a Banach space When equipped with the norm : $$||f||=\sup\{|f(x)|:||x||=1\}.$$ But $$\ell^1$$ is a Banach space which means that $$\|F\|_{X^*} = \sup\{ |F(x)| : \|x\|_{X^*}=1\} = \|w_n\|_\infty \equiv \|w\|_\infty$$