I have this question in Functional Analysis class I could only solve partially:
a. for a sequence $ g \in l^{\infty} $ and define $ F_g : l^1 \to C $ ; $ F_g = \sum_{n=1}^{\infty} g_nx_n $. We are to show that $ F_g $ is a properly defined bounded linear functional on $ l^1 $ and its norm is $ ||F_g||=||g||_{\infty} $. THIS I COULD DO EXCEPT CALCULATE THE NORM I.E. SHOW IT IS HE SUPREMUM SO I NEED HELP FOR SHOWING THE NORM
b. We are to show that the map $ g \to F_g $ is a one to one surjective isometry from $ l^{\infty} $ to the dual of the space $l^{1} $ NO IDEA HERE
c. Now we assume $ g = (g_n) $ is a given sequence (no additional assumptions) and we assume the mapping $ (x_n) = x \to \lim_{N \to \infty} \sum_{n=1}^{N} x_n g_n $ defines a bounded properly defined linear functional on $ l^2 $. We are to show that g must be a sequence in $ l^2 $. NO IDEA HERE
d. We are to prove the last part's conclusion holds while assuming only that the functional is only properly defined without assuming it is bounded. NO IDEA HERE.
******* PROGRESS: CAN DO PARTS A AND B FULLY NOW I AM STUCK ONLY ON C AND D
It is a rather difficult question for me and the last of my question list to do and I could only manage most of part a. so I would desperately need the help on it please. Thank you kindly all
For part (c) consider the following sequence $x_n = (g_1,g_2,.....,g_n, 0,0,...,0,...)$. This is certainly in $l^2$. Now under the given map $x_n \rightarrow \Sigma_{i=1}^n |g_i|^2$. Hence we get
$\Sigma_{i=1}^n |g_i|^2 \leq M \|x_n\|$
where $M$ is the norm of the operator. Computing the norm we have
$\Sigma_{i=1}^n |g_i|^2 \leq M (\Sigma_{i=1}^n |g_i|^2)^{1/2}$
$(\Sigma_{i=1}^n |g_i|^2)^{1/2} \leq M$ for all n. This implies that $g \in l^2$.
Also if you want to use banach stienhaus theorem then here is a solution to part d.
Let $T_i(x) := \Sigma_{j=0}^i x_jg_j$. Then clearly $T_i$ is a sequence of pointwise bounded operator. This follows from the fact that limit in part c exists. Hence the theorem says that norms $\| T_i \| \big(\geq (\Sigma_{j=1}^i |g_j|^2)^{1/2}\big)$ is bounded. Thus $(\Sigma_{j=1}^i |g_j|^2)^{1/2}$ is bounded for all $j$. Thus $g \in l^2$.