For the operator $U\colon \ell_{p}\to\ell_{p},\;\left( 1\leqslant p<\infty \right) :$ \begin{equation*} Ux=U\left( x_{1},x_{2},\dots \right) =\left( 0,x_{1},\frac{x_{2}}{2},\frac{% x_{3}}{3},\dots \right) ,\text{ for all }x=\left( x_{1},x_{2},\ldots \right) \in \ell_{p} \end{equation*} Calculate the operator norm $U$. Is the operator $U$ compact? (prove your answer).
I know that the norm operator of $U$ is $1$, and that the operator is compact, but do not know how to show it. More than this I do not know. Solution with an explanation would be good occurred. Thanks.
1) Regarding the norm: $$ \|Ux\|_p^p=\sum_{j=1}^{+\infty}\Big| \frac{x_j}{j}\Big|^p\leq \sum_{j=1}^{+\infty}|x_j|^p=\|x\|_p^p\qquad \forall x\in\ell^p\quad\Rightarrow\quad\|U\|\leq 1. $$ And for $x=(1,0,0,\ldots)$, we have $1=\|Ux\|_p=\|x\|_p$, whence $ \|U\|\geq 1$. So finally $\|U\|=1$.
2) For compactness, we will use the following sufficient condition (which is also necessary when $X=H$ is a Hilbert space):
Fact: let $X$ be a Banach space and $B(X)$ the algebra of bounded linear operators on $X$. If $T_n$ is a sequence of finite rank operators which converges to $T$ in $B(X)$ for the operator norm, then $T$ is compact.
In this case, we can consider the truncations $$ U_n:(x_j)\longmapsto \left(0,x_1,\frac{x_2}{2},\ldots,\frac{x_n}{n},0,\ldots\right). $$ These have finite rank, and $$ \|Ux-U_nx\|_p^p=\sum_{j=n+1}^{+\infty}\Big| \frac{x_j}{j}\Big|^p\leq\frac{1}{n^p}\sum_{j=n+1}^{+\infty}| x_j|^p=\frac{\|x\|_p^p}{n^p}\quad \forall x\in \ell^p. $$ Hence $\|U-U_n\|\leq \frac{1}{n}$ so $U_n$ converges to $U$ for the operator norm of $B(\ell^p)$. So $U$ is compact.