# Finding a lower bound of an operator norm

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

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

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

$$$$\phi(x) = \sum_{n=1}^{\infty} \frac{x_n}{n}, \ \ \ x = (x_n)_{n \geq 1}\in c_{00}$$$$ Show that $$\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?

• 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 $n\to \infty.$ Commented Dec 1, 2023 at 12:18

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