In case the functional is bounded on the space $C_0,L_1,L_2$, find its norm
$$f(x)=\sum_{n=1}^{\infty} \frac{x}{n \sqrt[n]{n}}$$
In $C_0$
$$ |f(x)| \leqslant \sum_{n=1}^{\infty}\left|\frac{x_n}{n \sqrt[n]{n}}\right| \leqslant \sum\left|x_n\right| $$
Therefore, the functional $f(x)$ is not bounded on $C_0$
- Is it this evalution and conclusion valid in $L_1$ ?
In $L_2$ $$ |f(x)| \leqslant \sqrt{\sum^{\infty}\left|x_n\right|^2} \cdot \sqrt{\sum^{\infty} \frac{1}{\left.|n \sqrt[n]{n}\right|^2}} \leqslant ||x_n||_{L_2}\cdot \frac{\pi}{\sqrt6} $$