# Showing an operator is unbounded by definition

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.

• Does the sequence $\{x^{(j)}\}$ stand for a sequence that is zero beyond the j-th term? Apr 20, 2020 at 17:04
• No, it does not. $\{x^{(j)}\}$ is a sequence of sequences. That is, each $x^{(j)}$ is a complex--valued sequence. Apr 20, 2020 at 19:24

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 $$|T x^{(j)}| / \|x^{(j)}\|_{\ell^2} = \sqrt{\sum_{n=1}^j \frac{1}{n}} \overset{j \to \infty}{\longrightarrow} \infty.$$
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 $$\infty$$.
• Your $Tx_n$ is $\sum_{k=1}^n \frac{1}{\sqrt{nk}}$ Apr 20, 2020 at 20:18