Proving that $f( x ):=\sum_{i=1}^{\infty}x( i ) / i $ is continuous functional in sequence space l2

Question

$f( x ):=\sum_{i=1}^{\infty}x( i ) / i $

I would like to prove that f is a continuous functional in sequence space l2.

I tried using the usual epsilon-delta approach, but I cannot seem to find a relation between epsilon and delta so that

$\left \| f(x_{n})-f(x) \right \|_2<g(\delta)$

(where g() is some function of $\delta$, which allows relating $\delta$ to $\epsilon$.

The only relation I get is the unbounded

$\left \| f(x_{n})-f(x) \right \|_2<\sum_{i=1}^{\infty}\delta$

Any advice on how to achieve this or what I am doing wrong?

Answer 1

following the useful suggestions in the comments) Using the comments from @Gae. S and @amsmath and I arrived at the following:

$\left \| f(x_{n})-f(x) \right \|_1=\left \|\sum_{i=1}^{\infty}(x_n( i )-x( i )) / i \right \|_1 \leq \sum_{i=1}^{\infty}\left|x_n( i )-x( i ))\right| / i \leq (\sum_{i=1}^{\infty}\left|x_n( i )-x( i ))\right|^{2})^{1/2} (\sum_{i=1}^{\infty}\left| 1/i\right|^{2})^{1/2}\leq \delta (\pi/6)^{1/2} $

Hence we can set $\delta$ to equal $\delta=\epsilon(6/\pi)^{1/2} $ and showed that f is a continuous functional in l2.

Answer 2

It is trivial. Let $\mathcal{H}=l^{2}=\{x\in\mathbb{R}^{N}\mid\sum_{k=1}^{\infty}x_{k}^{2}<\infty\}$ be the usual Hilbert space. Let $T:\mathcal{H}\rightarrow\mathbb{R}$ be defined by $Tx=\sum_{k=1}^{\infty}\frac{x_{k}}{k}$. Note that $\sum_{k=1}^{\infty}|\frac{x_{k}}{k}|\leq\left\{ \sum_{k=1}^{\infty}x_{k}^{2}\right\} ^{\frac{1}{2}}\left\{ \sum_{k=1}^{\infty}\frac{1}{k^{2}}\right\} ^{\frac{1}{2}}\leq||x||_{2}\cdot\frac{\pi^{2}}{6}<\infty$. Therefore, the series $\sum_{k=1}^\infty \frac{x_k}{k}$ converges and hence $T$ is well-defined. It it routine to verify that $T$ is linear. Let $x\in\mathcal{H}$. From the previous discussion, $|Tx|\leq\sum_{k=1}^{\infty}|\frac{x_{k}}{k}|\leq||x||_{2}\cdot\frac{\pi^{2}}{6}$. Hence, $T$ is bounded with $||T||\leq\frac{\pi^{2}}{6}$. It follows that $T$ is continuous (because a bounded linear functional is continuous).