Finite measure operator norm

Question

Let $T^n: L^1(\mu) \rightarrow L^{\infty}(\mu)$ be a bounded operator for any $n$ and $\mu$ a probability measure.

Is it then true that $||T^2||_{1 \rightarrow \infty} \le ||T||_{1 \rightarrow 2} ||T||_{2 \rightarrow \infty}$? where

$||T||_{a \rightarrow b}$ is defined as $sup_{||g||_a=1} ||Tg||_b$?

Answer 1

More generally, for any bounded operators on Banach spaces $T: A \to B$ and $S: B \to C$, $ST: A \to C$ is bounded with $\|ST\| \le \|S\| \|T\|$, because $\|STx\|_C \le \|S\| \|Tx\|_B \le \|S\| \|T\| \|x\|_A$.