Let $$\ell^2 = \left \{x = \left (x_1,x_2, \cdots \right )\ :\ x_i \in \Bbb C,\sum\limits_{i=1}^{\infty} \left |x_i \right |^2 < \infty \right \}$$ be a normed linear space with the norm $$\|x\|_2 = \left ( \sum\limits_{i=1}^{\infty} \left |x_i \right |^2 \right )^{\frac 1 2}.$$ Let $g : \ell^2 \longrightarrow \Bbb C$ be the bounded linear functional defined by $$g(x) = \sum\limits_{n=1}^{\infty} \frac {x_n} {3^n}\ \text{for all}\ x = \left (x_1,x_2, \cdots \right ) \in \ell^2.$$
Then find $\|g\|_{\text {op}} = \sup \left \{\left |g(x) \right |\ :\ \|x\|_2 \leq 1 \right \}.$
I am trying by taking $x \in \ell^2$ with $\|x \|_2 = 1$ like for every $n \in \Bbb N$ we can consider the sequence $x^{(n)} = \left (\frac {1} {\sqrt n}, \frac {1} {\sqrt n}, \cdots , \frac {1} {\sqrt n}, 0,0, \cdots \right ) \in \ell^2$ having $\frac {1} {\sqrt n}$ as it's first $n$ components and all other zero. Then they have the property that $\left \|x^{(n)} \right \|_2 = 1$ and for each $n \in \Bbb N$ we have $$g \left ( x^{(n)} \right ) = \frac {1} {2 \sqrt n} \left ( 1 - \frac {1} {3^n} \right ).$$ As $n$ increases the sequence $\left \{g \left ( x^{(n)} \right ) \right \}_{n=1}^{\infty}$ decreases to $0.$ So $\sup\limits_{n \geq 1} g \left (x^{(n)} \right ) = g \left (x^{(1)} \right ) = \frac 1 3.$ So $\|g\|_{\text {op}} \geq \frac 1 3.$ Can we make this bound more precise? Any help in this regard will be highly appreciated.
Thank you very much.
If $(a_n) \in \ell^{2}$ and $g(x_n)=\sum a_n x_n$ the $\|g\|_{op}=\|(a_n)\|_2$. To see that use Cauchy-Schwarz inequality to see that $\|g\|_{op}\leq \|(a_n)\|$. Then note that $g(\overline {a_n})=\sum |a_n|^{2}$ so $\sum |a_n|^{2} \leq \|g\|_{op} \|(a_n)\|$. This implies that $\|g\|_{op}\geq \|(a_n)\|$.
In our case we get $\|g\|_{op}=\sqrt {\sum \frac 1 9^{n}}=\frac 1{2 \sqrt 2}$.