$f : \ell^2 \to\mathbb R$ is a function and for every $x=(x_1,x_2,\ldots) \in \ell^2$ is defined by $\sum_{n=1}^\infty\frac{x_n}{\sqrt{6}n}$.
- Show that $f$ is a bounded linear functional on $\ell^2$
- find $\lVert{f}\rVert$
By using triangle inequality for $\lVert\cdot\rVert_2$ we can have an inequality in which $1/ \sqrt{6}n$ appears in every term, but how this can be written as $\lVert{x}\rVert_2$ and a real number; in order to show that $f$ is bounded. Any help would be appreciated.
The problem is to show that the image of the closed unit ball centered at $0\in\ell^2$ is bounded.
Suppose $x$ is in the closed unit ball. Then $\sum_{n=1}^\infty x_n^2 =\|x\|^2 \le 1.$ Then
$$ \|f(x)\|^2 = \sum_{n=1}^\infty \frac{x_n^2}{6n^2} \le \sum_{n=1}^\infty x_n^2 = \|x\|^2 \le 1. $$ So $1$ is an upper bound.