Let $f:l_{\infty} \rightarrow l_{\infty}$ defined by $f(x_1,x_2,...)=(x_1,\frac{1}{2}x_2,...,\frac{1}{2}x_n,...)$. Then we need to (1) prove that $f$ is continuous operator and (2) find $p\geq 1$ such that $Im(f) \subset l_p$.
My attempt: (1) I proved that $f$ is bounded, namely, $||f(x)||_{\infty} \leq ||x||_{\infty}$, and so it continuous
(2) We need to find all $p\geq 1$ such that: every $x \in l_{\infty}$ we have $f(x) \in l_p$, i.e. $||f(x)||_p < \infty$, i.e. $|x_1|^p +\sum_{n \geq 2} |\frac{1}{2}x_n|^p < \infty$. Taking $x=(1,1,...) \in l_{\infty}$ we have $|1|^p +\sum_{n \geq 2} |\frac{1}{2}|^p < \infty$ which is a contradiction, so can we say there is no such $p$ exists.