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_Farin Dec 23 '18 at 11:05

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

For boundedness :

$$|F(x)| = \bigg|\sum w_n a_n \bigg| \leq \sum|w_na_n| \leq \|w_n\|_\infty\sum |a_n| = \|w_n\|_\infty\|x\|_1 $$

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 $$\|F\|_{X^*} = \sup\{ |F(x)| : \|x\|_{X^*}=1\} = \|w_n\|_\infty \equiv \|w\|_\infty$$


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