= Solution
Let $e_m$ denote the sequence with a one in coordinate $m$ and zeros elsewhere.
For $T_1$, one has $\lVert e_m\rVert_p=1$ but $\lVert T_1e_m\rVert_p=m$ for every $p$, including $p=\infty$. Thus $T_1$ is unbounded for every $p$.
For $T_2$, the sequence $T_2e_m$ has entries $-(m-1)$ and $m$ in coordinates $m-1$ and $m$, respectively. Hence $\lVert T_2e_m\rVert_p\geq m$, so $T_2$ is also unbounded for every $p$.
For $T_3$, coordinatewise division gives
$$
\lVert T_3x\rVert_p\leq\lVert x\rVert_p
$$
for every $1\leq p\leq\infty$. Thus $T_3$ is a bounded <diagonal operator on sequence space> for every $p$.
For $T_4$, write $a=(n^{-1/2})_{n\geq1}$. Then $T_4x=x_1a$, and coordinate evaluation satisfies $|x_1|\leq\lVert x\rVert_p$. By the <p-series> criterion, $a\in\ell^p$ exactly when $p>2$, while $a\in\ell^\infty$. Therefore $T_4$ is bounded precisely for
$$
2<p\leq\infty.
$$
For $p\leq2$, even $T_4e_1$ does not belong to $\ell^p$.
For $T_5$, let $q$ be the <Holder conjugate exponent> of $p$. <Holder inequality> gives, for finite $p$,
$$
\left|\sum_{j=1}^nx_j\right|
\leq n^{1/q}\lVert x\rVert_p,
$$
and consequently
$$
\lVert T_5x\rVert_p^p
\leq\lVert x\rVert_p^p
\sum_{n=1}^{\infty}\frac{n^{p/q}}{2^{np}}<\infty.
$$
For $p=\infty$,
$$
|(T_5x)_n|\leq\frac n{2^n}\lVert x\rVert_\infty
\leq\lVert x\rVert_\infty.
$$
Thus $T_5$ is bounded for every $1\leq p\leq\infty$. Altogether,
$$
\boxed{
\begin{array}{c|ccccc}
&T_1&T_2&T_3&T_4&T_5\\ \hline
\text{bounded on }\ell^p
&\text{never}&\text{never}&\text{all }p&2<p\leq\infty&\text{all }p
\end{array}}
$$
Back to article page
