Trouble proving the equality when asked to compute the operator norm $\phi : \ell^{2} \to \mathbb R$ where $\phi(x)=\sum \frac{x_{n}}{n}$

Question

Compute the operator norm $\phi : \ell^{2} \to \mathbb R$ where $\phi(x)=\sum\limits_{n \in \mathbb N} \frac{x_{n}}{n}$

My proof so far:

$\lvert \phi(x)\rvert=\lvert\sum\limits_{n \in \mathbb N} \frac{x_{n}}{n}\rvert\leq\sum\limits_{n \in \mathbb N} \rvert\frac{x_{n}}{n}\rvert \leq (\sum\limits_{n \in \mathbb N} x_{n}^2)^{\frac{1}{2}}\cdot (\sum\limits_{n \in \mathbb N} \frac{1}{n^2})^{\frac{1}{2}}\implies \lvert \lvert\phi\rvert\rvert_{*}\leq (\sum\limits_{n \in \mathbb N} \frac{1}{n^2})^{\frac{1}{2}}=:M$

Now, for me it always generally difficult to prove the reverse inequality, as I always need to normalize the the sequence, i.e. $(x^{n})_{n}\subseteq B_{1}^{\lvert \lvert \cdot \rvert \rvert_{2}}(0)$. I cannot find a way to satisfy the restriction in the unit ball while still approximating $M$. In other spaces like $\ell^{1}, \ell^{\infty}$ this is a lot easier.

Any ideas/hints?

Answer 1

Consider the sequence $\left(\frac1{Mn}\right)_{n\in\Bbb N}$. Then its norm is $1$ and $\phi\left(\left(\frac1{Mn}\right)_{n\in\Bbb N}\right)=M$. So $\|\phi\|\geqslant M$.

Answer 2

Let me show how one would find Jose's solution. Consider $v=(1,1/2,1/3,\cdots)$. Then for $x \in \ell^2$ we simply have $$ \phi(x)=x\cdot v \, . $$ Now by properties of dot product (true in infinite space as well) one has $$ |\phi(x)| =|x\cdot v| \leq \|x\|_{\ell^2} \|v\|_{\ell^2} \, . $$ Furthermore, the maximum value for $x\cdot v$ is attained when $x$ is parallel to $v$. That is why $x=v/\|v\|$ (normalized b/c for operator norm we restrict to the unit sphere) maximizes $\phi(x)$ on the unit ball. Jose's vector is exactly $v/\|v\|$.

Note: properties of dot product still hold because we can truncate everything at first $N$ coordinates, apply usual finite-dimenional inequalities, and then take limit as $N\to \infty$.