Find the vector $h_0$ is $\mathcal{H}$ such that $L(h)=\langle h,h_0\rangle$

Question

Let $\mathcal{H}=l^2(\mathbb{N}\cup \{0\})$, and $L:\mathcal{H}\to\mathbb{C}$ is defined by $L(\{\alpha_n\})=\sum_{n=0}^{\infty}{n\alpha_n}\lambda^{n-1}$, where $|\lambda|<1$. Find the vector $h_0$ in $\mathcal{H}$ such that $L(h)=\langle h,h_0\rangle$ for every $h$ in $\mathcal{H}$.(Exercise 3, Riesz Representation, Conway)

My attemp: Clearly $L:\mathcal{H}\to\mathbb{C}$ is a linear functional. And we can prove that is bounded. So, $$|L(\{\alpha_n\})|=|\sum_{n=0}^{\infty}{n\alpha_n}\lambda^{n-1}|\leq\sum_{n=0}^{\infty}{n|\alpha_{n}\lambda^{n-1}|}$$

But it bothers me that $n$ in the summatory. Any hints, thanks!

Answer 1

You can prove that $L$ is bounded on its own. But if you write $L(h)=\langle h,h_0\rangle$, boundedness of $L$ follows automatically.

The form of $h_0$ is obvious from the definition of $L$: if you take $$ h_0=\{n\lambda^{n-1}\}, $$ you get $\langle h,h_0\rangle=L(h)$. The only question is whether $h_0\in\ell^2$. For this, $$ \sum_n|n\lambda^{n-1}|^2=\sum_nn^2|\lambda|^{2n-2}=\frac{|\lambda|^2+1}{(1-|\lambda|^2)^3}<\infty. $$ If you've never calculated such a sum, you consider it a power series, and as such it is almost the second derivative of the geometric series.

Answer 2

It should bother you because it might not be clear why the sum converges, but by the Cauchy-Schwartz inequality, you have

$$ \sum_{n=0}^{\infty} n|\lambda|^{n-1} |\alpha_n| \leq \left( \sum_{n=0}^{\infty} n^2 |\lambda|^{2n-2}\right)^{\frac{1}{2}} \left( \sum_{n=0}^{\infty} |\alpha_n|^2 \right)^{\frac{1}{2}} $$

where the first term converges since $|\lambda| < 1$ and the second term converges since $\{ \alpha_n \} \in \ell^2(\mathbb{N} \cup \{ 0 \})$ and this shows that $L$ is indeed continuous.

The question asks you to find a vector $h_0 = \{ \beta_n \}$ such that for all $h = \{ \alpha_n \}$ we have

$$ L(h) = \sum_{n=0}^{\infty} n \alpha_n \lambda^{n-1} = \left< \{ \alpha_n \}, \{ \beta_n \} \right> = \sum_{n=0}^{\infty} \alpha_n \overline{\beta_n} $$

so a good guess would be to take $\beta_n = n \overline{\lambda}^{n-1}$. What is left to do is to check that indeed $\{ \beta_n \} \in \ell^2(\mathbb{N} \cup \{ 0 \})$ and again, the Cauchy-Schwatz inequality will be helpful.