I am working on a problem where I am at crossroads with showing that the operator $T: \ell_2 \rightarrow \mathbb{C}$ defined as $$T(x)=\sum_{n=1}^{\infty} \frac{x_n}{\sqrt{n}}$$ is unbounded. Since the sequence $x=\frac{1}{\sqrt{n}}$ is not in $\ell_2$, I want to find a sequence $\{x^{(j)}\}\subset\ell_2$ such that $$T(x^{(j)})\ge j||x^{(j)}||_{\ell_2}$$ for every $j$.
All my constructions have failed so far.
On
Consider $\mathbf{x}_n:m\mapsto\frac{1}{\sqrt{n}}\mathbb{1}_{\{n+1,\ldots,2n\}}(m)$, $n\in\mathbb{Z}_+$.
Clearly $\|\mathbf{x}_n\|_2=1$ for all $n$ and $T\mathbf{x}_n=\sum^{2n}_{k=n+1}\frac{1}{\sqrt{nk}}\geq \sum^{2n}_{k=n+1}\frac{1}{k}\xrightarrow{n\rightarrow\infty}\infty$
since $\sum_n\frac1n$ diverges to $\infty$.
Let $x_n^{(j)} = \begin{cases} 1/\sqrt{n} & n \le j \\ 0 & \text{otherwise} \end{cases}$.
Then $|T x^{(j)}| = \sum_{n=1}^j \frac{1}{n} = \|x^{(j)}\|_{\ell_2}^2$ so $$|T x^{(j)}| / \|x^{(j)}\|_{\ell^2} = \sqrt{\sum_{n=1}^j \frac{1}{n}} \overset{j \to \infty}{\longrightarrow} \infty.$$