 Solution

Let $e_m$ denote the sequence with a one in coordinate $m$ and zeros elsewhere.

For $T_1$, one has $\Vert e_m{\Vert }_p=1$ but $\Vert T_1{e}_m{\Vert }_p=m$ for every $p$, including $p=\infty$. Thus $T_1$ is unbounded for every $p$.

For $T_2$, the sequence $T_2{e}_m$ has entries $-(m-1)$ and $m$ in coordinates $m-1$ and $m$, respectively. Hence $\Vert T_2{e}_m{\Vert }_p\ge m$, so $T_2$ is also unbounded for every $p$.

For $T_3$, coordinatewise division givesfor every $1\le p\le \infty$. Thus $T_3$ is a bounded diagonal operator on sequence space for every $p$.

For $T_4$, write $a=(n^{-1/2}{)}_{n\ge 1}$. Then $T_{4}x=x_{1}a$, and coordinate evaluation satisfies $|x_1|\le \Vert x{\Vert }_p$. By the p-series criterion, $a\in {\ell }^p$ exactly when $p>2$, while $a\in {\ell }^{\infty }$. Therefore $T_4$ is bounded precisely forFor $p\le 2$, even $T_4{e}_1$ does not belong to ${\ell }^p$.

For $T_5$, let $q$ be the Holder conjugate exponent of $p$. Holder inequality gives, for finite $p$,and consequentlyFor $p=\infty$,Thus $T_5$ is bounded for every $1\le p\le \infty$. Altogether,