Given a linear functional $T: C_{00}\to C_{00}$. Where $C_{00}$ is space sequences with finitely many non-zero terms with $\ell_2$ norm.
$T$ is defined as $$T(x)=\sum_{n=1}^{\infty} \frac{x_n}{\sqrt{n}},$$
Show that this functional is unbounded.
Initially Idea was to choose $x_n=\{1\}_{1\leq k\leq n}\in C_{00}$
This gives $$T(x_n)=\sum_{k=1}^{n} \frac{1}{\sqrt{k}},$$
This gives $T(x_n)$ diverges hence functional is bounded.
Is this a correct approach.
Best Answer
More generally let $\{a_k\}\notin \ell^2.$ Then the linear functional defined on $C_{0,0}$ by $$Tx=\sum_{k=1}^\infty a_kx_k$$ is unbounded. Indeed let $$x_k=\begin{cases}\overline{a_k} & 1\le k\le n\\ 0&k>n\end{cases}$$ Then $$\|x\|_2=\left (\sum_{k=1}^n |a_k|^2\right)^{1/2},\quad Tx=\sum_{k=1}^n|a_k|^2$$ and $${Tx\over \|x\|_2}=\left (\sum_{k=1}^n |a_k|^2\right)^{1/2}\underset{n\to \infty}{\longrightarrow} \infty$$ In the OP case $a_k=k^{-1/2}.$