Suppose we have a measure preserving probability space $(X,\mathcal{B},\mu,T)$. Consider the operator $K:L^1(X)\to L^1(X)$ by $Kv=v\circ T$, also known as the Koopman operator. Note that $\|Kv\|_1=\|v\|_1$ since $\mu$ is $T$-invariant.
Define $\mathcal{L}:L^\infty(X)\to L^\infty(X)$ by the following: $\mathcal{L}v$ is the unique element in $L^\infty(X)$ such that $\int \mathcal{L}v\, u\,d\mu=\int v\,Ku\,d\mu$ for all $u\in L^1(X)$.
Apparently, for every $1\leq p\leq \infty$ and $v\in L^\infty(X)$, $\|\mathcal{L}v\|_p\leq \|v\|_p$. However, I am struggling to show this.
I can show this for $p=1$: Taking $u=\operatorname{sgn}(\mathcal{L}v)$, we have $\|\mathcal{L}v\|_1=\int |\mathcal{L}v|\,d\mu=\int \mathcal{L}v\, u\,d\mu=\int v\,Ku\,d\mu\leq \|v\|_1\|Ku\|_\infty=\|v\|_1\|u\|_\infty=\|v\|_1$.
Moreover, if I know the result for all $p<\infty$, I can get the result for $p=\infty$ by noting that $\|\mathcal{L}v\|_\infty=\lim\limits_{p\to\infty}\|\mathcal{L}v\|_p\leq\lim\limits_{p\to\infty}\|v\|_p=\|v\|_\infty$.
My question is, how can I show this for $1<p<\infty$? Moreover, is there a nicer way to show this for $p=\infty$?
$L$ is defined to be the adjoint of $K$. It's a general fact that the norm of adjoint operator is the same as the norm of the original operator. You only need inequality in one direction, the easy one: given $K:X\to X$, let $L:X^*\to X^*$ be its adjoint (i.e., $Lx^* = x^*\circ K$ for $x^*\in X^*$); then $$ \|Lx^*\| = \|x^*\circ K\| \le \|x\|\|K\| $$ since the operator norm is submultiplicative under composition.
So, it suffices to show that $\|K\|_{L^p\to L^p} = 1$ for $1\le p < \infty$. This follows in the same way as for $L^1$: namely, $$ \int |Kv|^p = \int |v|^p \circ T = \int |v|^p $$