Linear Operator on $\mathcal{H} = l^2(\mathbb{N}\cup \{0\})$ This is taken from Conway's a course in functional analysis section 1.3 problem 3. It's a 3 part question I wanted to see if what I had below was correct and to ask about how I would approach part (c).
Let $\mathcal{H} = l^2(\mathbb{N}\cup \{0\})$.
(a) Show that if $\{\alpha_n\}\in \mathcal{H}$, then the power series $\sum_{n=0}^\infty\alpha_nz^n$ has radius of convergence $\geq 1$.
Since $\{\alpha_n\} \in \mathcal{H}$, we have $\sum_{n=0}^\infty |\alpha_n|^2 < \infty$. We hope to show that $\left\lvert\sum_{n=0}^\infty \alpha_nz^n\right\rvert< \infty$ for any $z<1$. For $z = 0$ the series is $0$ so let us show the series converges where $z \neq 0$. By Cauchy Schwarz inequality:
\begin{align}
    \left(\sum_{n=0}^\infty \alpha_nz^n\right)^2 &\leq \left(\sum_{n=0}^\infty |\alpha_n|^2\right)\left(\sum_{n=0}^\infty z^{2n}\right)\\
    &=\left(\sum_{n=0}^\infty |\alpha_n|^2\right)\left(\frac{1}{1 - z^2}\right)\\
    &<\infty
\end{align}
where the last inequality comes from $\{\alpha_n\} \in \mathcal{H}$, and the result follows.
(b) If $|\lambda|<1$ and $L:\mathcal{H}\rightarrow \mathbb{F}$ is defined by $L(\{\alpha_n\}) = \sum_{n=0}^\infty \alpha_n\lambda^n$ find th vector $h_0$ in $\mathcal{H}$ such that $L(h) = \langle h, h_0\rangle$ for every $h$ in $\mathcal{H}$.
For this we can let $h_0(n) = \overline{\lambda^n}$.
(c) What is the norm of the linear functional $L$ defined in (b)
We have $||L|| = \sup\{|L(h)|: ||h||= 1\}$, and here I"m thinking that I should use the Cauchy Schwarz inequality from part (a), but it doesn't seem to suggest a good candidate for $||L||$.

Parts (a) and (b) are good. For part (c), the norm of a linear functional $h \mapsto \langle h, h_0\rangle$ is precisely $\|h_0\|$. And yes, this is proven by Cauchy-Schwarz: $|\langle h, h_0\rangle| \le \|h_0\|\|h\|$, with equality occurring when taking $h = h_0$.
So, we just need to compute $\left\|\left(\overline{\lambda^n}\right)_n\right\|$, which is a straightforward geometric series:
$$\left\|\left(\overline{\lambda^n}\right)_n\right\|^2 = \sum_{n=0}^\infty|\lambda|^{2n} = \frac{1}{1 - |\lambda|^2} \implies\|L\| = \frac{1}{\sqrt{1 - |\lambda|^2}}.$$
