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Let $c_{00}$ be the vector space of all complex sequences having finitely many non-zero terms. Define the inner product $\langle x,y \rangle= \sum_{n=1}^\infty x_n \bar{y_n}$. Define a linear functional $f:c_{00} \rightarrow \mathbb{C}$ by $f(x)=\sum_{n=1}^\infty \frac{x_n}{n}$. If $N$ is the kernel of $f$. Prove the following:

$(A)$ $||f|| \leq \frac{\pi}{\sqrt{6}}$

$(B)$ $N^{\perp}= \{0\}$

How do I do this? I am a little rusty in functional analysis. I have looked up the required definitions but am confused as to how to proceed. Help!

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    @Aweygan Yes that is correct – 2017-02-04

1 Answers 1

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Part (A) follows directly from Cauchy-Schwartz inequality.

For part (B): Choose any $(y_n)\in N^\perp$. Then we know that there exists a natrual number $m$ such that $y_n=0$ for all $n>m$. Now let $k\leq m$ and $k'>m$, define the sequence $(x_n)$ by $$x_n=\begin{cases}k, &n=k\\ -k', &n=k'\\ 0, &\text{elsewhere}\end{cases}$$ Then by straight forward computation we can see that $(x_n)\in N$, and $$0=\sum_{n=1}^\infty x_n\overline{y_n}=x_k\overline{y_k}=k\overline{y_k},$$ thus we have $y_k=0$. As $k\leq m$ is arbitrary we conclude that $y_k=0$ for all $k\leq m$. Therefore $y_n=0$ for all $n\in\mathbb{N}$, which means $N^\perp=\{0\}$.