I'm working through Conways Functional analysis book right now, and I got a bit stumped, I just want to make sure I'm approaching the problem correctly. The problem statement goes as follows: Let $H = l^2(\mathbb{N} \cup \{0\}) $. Suppose $\{a_n\} \in H$. If $|\lambda| <1$ and $L: H \to \mathbb{F}$ is defined by $L( \{a_n\}) = \sum_{n=1}^\infty n a_n\lambda^{n-1} $ find the vector $h_0$ in $H$ such that $L(h)= \langle{h},h_0\rangle$ for all $h$.
Just making sure all I really have to do here is show that $L$ is bounded and linear, and then use the Riesz theorem here? Its clear to me that this is linear, and showing that it is bounded shouldn't take too much work, my concern comes from finding the correct $h_0$. I was just looking for some guidance on this problem.
This is not a full answer, but from what I gathered, its clearly linear so I won't bother showing that to be true. To show that it is bounded we have that $$| L(a_n)|^2 = \left\|\sum_{n=1}^\infty (na_n\lambda^{n-1})^2 \right\| \leq \sum_{n=1}^\infty \|(na_n)^2\lambda^{2(n-1)}\|= \sum_{n=1}^\infty |na_n|^2|\lambda|^{2(n-1)} = \|a \|_2\sum_{n=1}^\infty \|n^2\lambda^{2(n-1)}\|$$
I'd like to argue that this is bounded and will go to $0$ since $\lambda$ will to go $0$ faster than $n^2$ to infinity but I'm not sure if what I have now is enough to conclude that. If so, then $L$ is a bounded linear functional and by Riesz theorem we have that $L = \langle h,h_0 \rangle$. I would like to think that we can have $h_0 = \frac{\overline{\lambda^{n-1}}}{n}$ since this would be in $l^2$. Upon picking this $h_0$ we would obtain, with the geometric series that $$\|h_0 \|^2 = \sum_{n=1}^\infty |\lambda^{2(n-1)}| = \frac{1}{|1-\lambda^2|} \implies \|h_0\| = \frac{1}{\sqrt{1-|\lambda^2|}}$$