Norm of a bounded functional

Question

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 $||f||=\frac{\pi^2}{6}$?

Answer 1

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.