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Consider the subspace $c_{00}$ of $\mathbb{C}^{\mathbb{N}}$ consisting of complex sequences with at most finitely many non-zero entries, and define $\langle\cdot,\cdot\rangle$ by

\begin{equation} \langle x,y \rangle = \sum_{n=1}^{\infty} x_n \bar{y_n}, \ \ \ x = (x_n)_{n \geq 1}, y = (y_n)_{n \geq 1} \in c_{00} \end{equation}

We may define a bounded linear functional $\phi$ on this space by

\begin{equation} \phi(x) = \sum_{n=1}^{\infty} \frac{x_n}{n}, \ \ \ x = (x_n)_{n \geq 1}\in c_{00} \end{equation} Show that $\lVert \phi \rVert = \frac{\pi}{\sqrt{6}}$


So I've been able to show that $\lVert \phi \rVert = \sup\{ \vert \phi(x) \vert : x \in c_{00}, \lVert x \rVert_{2} \leq 1\} \leq \frac{\pi}{\sqrt{6}}$ i.e., $\lVert \phi \rVert$ is bounded above by $\frac{\pi}{\sqrt{6}}$.

Now to show that it is bounded below by the same:

Let $y^{(n)} = (1, 1/2, ... , 1/n, 0, 0, ...)$ and $x^{(n)} = \frac{y^{(n)}}{\lVert y^{(n)}\rVert_{2}}$, $n \geq 1$.

Then, $x^{(n)} \in c_{00}$ with $\lVert x^{(n)} \rVert_{2} = 1$ and

$$\begin{align} \vert \phi(x^{(n)}) \vert &= \vert \sum_{n=1}^{\infty} \frac{x_n}{n} \vert \end{align}$$

I've been told that I should be able to show that this sum is equal to $\left( \sum_{k=1}^{n} \frac{1}{k^2}\right)^{\frac{1}{2}}$ which clearly gives me the result I need but I cannot figure out how to show this equality?

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  • $\begingroup$ If $y\in \ell^2(\mathbb{N})$ and the functional is given by $$\varphi(x)=\sum_{n=1}^\infty x_n\overline{y_n},\quad x\in c_{00}$$ then $\|\varphi\|=\|y\|_2.$ The upper estimate follows from the Cauchy-Schwarz inequality. The lower bound can be obtained by taking $x^{(n)}_k=y_k$ for $1\le k\le n$ and letting $n\to \infty.$ $\endgroup$ Commented Dec 1, 2023 at 12:18

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$\phi (y^{(n)})=\sum\limits_{k=1}^{n}\frac 1{k^{2}}$ by the definition of $\phi$. Hence, $\phi (x^{n})=\sum\limits_{k=1}^{n}\frac 1{k^{2}}/\|y^{(n)}\|=(\sum\limits_{k=1}^{n}\frac 1{k^{2}}) /\sqrt {\sum\limits_{k=1}^{n}\frac 1{k^{2}}}=\sqrt {\sum\limits_{k=1}^{n}\frac 1{k^{2}}}$

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