Linear Limited Functional

Question

how can I find the norm of functional $f(x)=\sum_{j=1}^{\infty}\xi_j \frac{1}{\sqrt{j(j+1)}}$ defined by $f:l^2\rightarrow \mathbb{R}$ for $\forall x=(\xi)\in l^2$ ?

Answer 1

$|f(\xi)| \leq \sqrt {\sum |\xi_j|^{2}} \sqrt {\sum \frac 1 {j(j+1)}}$ by Cauchy Schwarz inequlaity. Hence $\|f\| \leq \sqrt {\sum \frac 1 {j(j+1)}}$. Equality golds q=when $\xi_j= \frac 1 {\sqrt {\sum \frac 1 {j(j+1)}}}$. Hence $\|f\| = \sqrt {\sum \frac 1 {j(j+1)}}$.

Incidentally $\sum \frac 1 {j(j+1)}=\sum (\frac 1 j -\frac 1 {j+1}) =1$.