Q:

Norm of a bounded functional

I want to calculate the norm of the functional $f:\ell^2(\mathbb{N})\to\mathbb{R}$ given by
$$f(x)=\sum^{\infty}_{n=1}\frac{x_n}{n}$$
where $x=(x_1,x_2,...)$.
My work so far:
Using Holder's inequality we know that $|f(x)|\leq||(\frac{1}{n})||_2||x||_2$.
But $||(\frac{1}{n})||_2=\frac{\pi^2}{6}$, so $f$ is a bounded functional and $||f||\leq\frac{\pi^2}{6}$.
But how to show that $||f||=\frac{\pi^2}{6}$?

A:

Minor typo: it should be $\sqrt{\pi^2/6}$.
The Cauchy-Schwarz inequality is tight when the two vectors are parallel.
In this case, choosing $x_n = 1/n$ gives you the desired equality.

