How to find the norm of $ \Lambda x = \sum_{n=1}^{\infty} \frac{x_n}{n\sqrt{6}}$ in $\ell^2$

Question

Suppose that $(x_n)$ is a sequence in $\ell^2$, i.e. $\displaystyle \sum_{i=1}^{\infty} x_n^2 < \infty$. Define:

$$\Lambda x = \sum_{n=1}^{\infty} \frac{x_n}{n\sqrt{6}}$$

Find $\| \Lambda \|_2$

Using the Cauchy-Schwarz inequality, it's easy to see that for any sequence $(x_n)$ in $\ell^2$ such that $\|(x_n)\|=1$, we have $\displaystyle |\Lambda x| \leq \frac{\pi}{6}$:

$$|\Lambda x|^2 = |\sum_{n=1}^{\infty} \frac{x_n}{n\sqrt{6}}|^2 \leq |\sum_{n=1}^{\infty} x_n^2|\cdot|\sum_{n=1}^{\infty}\frac{1}{6n^2}|=\frac{\pi^2}{36}$$

How can I show that $\| \Lambda \|$ is indeed equal to $\displaystyle \frac{\pi}{6}$?

Answer 1

It is useful to recall a general corollary of the Cauchy-Schwarz inequality: $|\langle x,y \rangle | \leq \| x \| \| y \|$ and the inequality is strict except when $x$ is a scalar multiple of $y$. So choose $x_n$ to be a (nonzero) scalar multiple of the sequence $y_n = \frac{1}{n \sqrt{6}}$ and you will get your result.